Best way to see if a number is fractional or not - python

I'm looking to differentiate between a number like
2.0 or 2 and an actual fractional number such as 2.4. What would be the best way to do this? Currently I'm doing:
def is_fractional(num):
if not str(num).replace('.','').isdigit(): return
return float(num) != int(num)
>>> is_fractional(2)
False
>>> is_fractional(2.1)
True
>>> is_fractional(2.0)
False
>>> is_fractional('a')
>>>

That operation is built-in:
>>> 5.0.is_integer()
True
>>> 5.00000001.is_integer()
False
>>> 4.9999999.is_integer()
False
Documentation is here.
ADDENDUM
The initial solution only works for float. Here's a more complete answer, with tests:
from decimal import Decimal
def is_integer(x):
if isinstance(x, int):
return True
elif isinstance(x, float):
return x.is_integer()
elif isinstance(x, Decimal):
return x.as_integer_ratio()[1] == 1
return False
good = [
0,
0.0,
3,
-9999999999999999999999,
-2.0000000000000,
Decimal("3.000000"),
Decimal("-9")
]
bad = [
-9.99999999999999,
"dogs",
Decimal("-4.00000000000000000000000000000000001"),
Decimal("0.99999999999999999999999999999999999")
]
for x in good:
assert is_integer(x)
for x in bad:
assert not is_integer(x)
print("All tests passed")

If some of your numbers are decimal.Decimals, they might have range issues where conversion to float fails, or drops the fractional part that actually exists, depending on their precision:
>>> import decimal
>>> x = decimal.Decimal('1.00000000000000000000000000000000000001')
>>> str(x)
'1.00000000000000000000000000000000000001'
>>> float(x).is_integer()
True
>>> y = decimal.Decimal('1e5000')
>>> str(y)
'1E+5000'
>>> float(y)
inf
The str method will generally work (modulo problem cases like the one illustrated above), so you could stick with that, but it might be better to attempt to use is_integer and use a fallback if that fails:
try:
return x.is_integer()
except AttributeError:
pass
(as others note, you'll need to check for int and long here as well, if those are allowed types, since they are integers by definition but lack an is_integer attribute).
At this point, it's worth considering all of the other answers, but here's a specific decimal.Decimal handler:
# optional: special case decimal.Decimal here
try:
as_tuple = x.as_tuple()
trailing0s = len(list(itertools.takewhile(lambda i: i == 0, reversed(as_tuple[1]))))
return as_tuple[2] + trailing0s < 0
except (AttributeError, IndexError): # no as_tuple, or not 3 elements long, etc
pass

Why do not check if the difference between the truncation to integer and the exact value is not zero?
is_frac = lambda x: int(x)-x != 0

Python includes a fractions module that generates fractions (rational numbers) from strings, floats, integers, and much more. Just create a Fraction and check whether its denominator is other than 1 (the Fraction constructor will automatically reduce the number to lowest terms):
from fractions import Fraction
def is_fractional(num):
return Fraction(num).denominator != 1
Note that the method above may raise an exception if the conversion to a Fraction fails. In this case, it's not known whether the object is fractional.

If you are dealing with decimal module or with a float object, you can do this easily:
def is_factional(num):
return isinstance(num, (float, Decimal))

Here is one way to do it (assuming e.g. 2/2 is not "fractional" in the sense you have in mind):
# could also extend to other numeric types numpy.float32
from decimal import Decimal
def is_frac(n):
numeric_types = (int, float, Decimal)
assert isinstance(n, numeric_types), 'n must be numeric :/'
# (ints are never fractions)
if type(n) is int: return False
return n != float(int(n))
# various sorts of numbers
ns = [-1, -1.0, 0, 0.1, 1, 1.0, 1., 2.3, 1e0, 1e3, 1.1e3,
Decimal(3), Decimal(3.0), Decimal(3.1)]
# confirm that values are as expected
dict(zip(ns, [is_frac(n) for n in ns]))
This will only work if n is an int or a float or decimal.Decimal. But you could extend it to handle other numeric types such as numpy.float64 or numpy.int32 by just including them in numeric_types.

Related

How does Python compare 'int' to 'float' objects?

The documentation about numeric types states that:
Python fully supports mixed arithmetic: when a binary arithmetic operator has operands of different numeric types, the operand with the “narrower” type is widened to that of the other, where integer is narrower than floating point, which is narrower than complex. Comparisons between numbers of mixed type use the same rule.
This is supported by the following behavior:
>>> int.__eq__(1, 1.0)
NotImplemented
>>> float.__eq__(1.0, 1)
True
However for large integer numbers something else seems to happen since they won't compare equal unless explicitly converted to float:
>>> n = 3**64
>>> float(n) == n
False
>>> float(n) == float(n)
True
On the other hand, for powers of 2, this doesn't seem to be a problem:
>>> n = 2**512
>>> float(n) == n
True
Since the documentation implies that int is "widened" (I assume converted / cast?) to float I'd expect float(n) == n and float(n) == float(n) to be similar but the above example with n = 3**64 suggests differently. So what rules does Python use to compare int to float (or mixed numeric types in general)?
Tested with CPython 3.7.3 from Anaconda and PyPy 7.3.0 (Python 3.6.9).
The language specification on value comparisons contains the following paragraph:
Numbers of built-in numeric types (Numeric Types — int, float, complex) and of the standard library types fractions.Fraction and decimal.Decimal can be compared within and across their types, with the restriction that complex numbers do not support order comparison. Within the limits of the types involved, they compare mathematically (algorithmically) correct without loss of precision.
This means when two numeric types are compared, the actual (mathematical) numbers that are represented by these objects are compared. For example the numeral 16677181699666569.0 (which is 3**34) represents the number 16677181699666569 and even though in "float-space" there is no difference between this number and 16677181699666568.0 (3**34 - 1) they do represent different numbers. Due to limited floating point precision, on a 64-bit architecture, the value float(3**34) will be stored as 16677181699666568 and hence it represents a different number than the integer numeral 16677181699666569. For that reason we have float(3**34) != 3**34 which performs a comparison without loss of precision.
This property is important in order to guarantee transitivity of the equivalence relation of numeric types. If int to float comparison would give similar results as if the int object would be converted to a float object then the transitive relation would be invalidated:
>>> class Float(float):
... def __eq__(self, other):
... return super().__eq__(float(other))
...
>>> a = 3**34 - 1
>>> b = Float(3**34)
>>> c = 3**34
>>> a == b
True
>>> b == c
True
>>> a == c # transitivity demands that this holds true
False
The float.__eq__ implementation on the other hand, which considers the represented mathematical numbers, doesn't infringe that requirement:
>>> a = 3**34 - 1
>>> b = float(3**34)
>>> c = 3**34
>>> a == b
True
>>> b == c
False
>>> a == c
False
As a result of missing transitivity the order of the following list won't be changed by sorting (since all consecutive numbers appear to be equal):
>>> class Float(float):
... def __lt__(self, other):
... return super().__lt__(float(other))
... def __eq__(self, other):
... return super().__eq__(float(other))
...
>>> numbers = [3**34, Float(3**34), 3**34 - 1]
>>> sorted(numbers) == numbers
True
Using float on the other hand, the order is reversed:
>>> numbers = [3**34, float(3**34), 3**34 - 1]
>>> sorted(numbers) == numbers[::-1]
True

call different functions for list elements [duplicate]

How can I convert a str to float?
"545.2222" → 545.2222
How can I convert a str to int?
"31" → 31
For the reverse, see Convert integer to string in Python and Converting a float to a string without rounding it.
Please instead use How can I read inputs as numbers? to close duplicate questions where OP received a string from user input and immediately wants to convert it, or was hoping for input (in 3.x) to convert the type automatically.
>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545
Python2 method to check if a string is a float:
def is_float(value):
if value is None:
return False
try:
float(value)
return True
except:
return False
For the Python3 version of is_float see: Checking if a string can be converted to float in Python
A longer and more accurate name for this function could be: is_convertible_to_float(value)
What is, and is not a float in Python may surprise you:
The below unit tests were done using python2. Check it that Python3 has different behavior for what strings are convertable to float. One confounding difference is that any number of interior underscores are now allowed: (float("1_3.4") == float(13.4)) is True
val is_float(val) Note
-------------------- ---------- --------------------------------
"" False Blank string
"127" True Passed string
True True Pure sweet Truth
"True" False Vile contemptible lie
False True So false it becomes true
"123.456" True Decimal
" -127 " True Spaces trimmed
"\t\n12\r\n" True whitespace ignored
"NaN" True Not a number
"NaNanananaBATMAN" False I am Batman
"-iNF" True Negative infinity
"123.E4" True Exponential notation
".1" True mantissa only
"1_2_3.4" False Underscores not allowed
"12 34" False Spaces not allowed on interior
"1,234" False Commas gtfo
u'\x30' True Unicode is fine.
"NULL" False Null is not special
0x3fade True Hexadecimal
"6e7777777777777" True Shrunk to infinity
"1.797693e+308" True This is max value
"infinity" True Same as inf
"infinityandBEYOND" False Extra characters wreck it
"12.34.56" False Only one dot allowed
u'四' False Japanese '4' is not a float.
"#56" False Pound sign
"56%" False Percent of what?
"0E0" True Exponential, move dot 0 places
0**0 True 0___0 Exponentiation
"-5e-5" True Raise to a negative number
"+1e1" True Plus is OK with exponent
"+1e1^5" False Fancy exponent not interpreted
"+1e1.3" False No decimals in exponent
"-+1" False Make up your mind
"(1)" False Parenthesis is bad
You think you know what numbers are? You are not so good as you think! Not big surprise.
Don't use this code on life-critical software!
Catching broad exceptions this way, killing canaries and gobbling the exception creates a tiny chance that a valid float as string will return false. The float(...) line of code can failed for any of a thousand reasons that have nothing to do with the contents of the string. But if you're writing life-critical software in a duck-typing prototype language like Python, then you've got much larger problems.
def num(s):
try:
return int(s)
except ValueError:
return float(s)
This is another method which deserves to be mentioned here, ast.literal_eval:
This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself.
That is, a safe 'eval'
>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31
Localization and commas
You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.
Example 1 -- United States number conventions
In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:
>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>
Example 2 -- European number conventions
In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:
>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222
The method locale.atoi is also available, but the argument should be an integer.
float(x) if '.' in x else int(x)
If you aren't averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation:
>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int
Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int("545") without first casting to float.
If your strings are in a list, you could use the map function as well.
>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>
It is only good if they're all the same type.
In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31?
I just want to know how to parse a float string to a float, and (separately) an int string to an int.
It's good that you ask to do these separately. If you're mixing them, you may be setting yourself up for problems later. The simple answer is:
"545.2222" to float:
>>> float("545.2222")
545.2222
"31" to an integer:
>>> int("31")
31
Other conversions, ints to and from strings and literals:
Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:
>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31
If you don't know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:
>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31
Non-Decimal (i.e. Integer) Literals from other Bases
If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases - you can let Python do it for you automatically with the correct syntax.
You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:
Binary, prefix 0b
>>> 0b11111
31
Octal, prefix 0o
>>> 0o37
31
Hexadecimal, prefix 0x
>>> 0x1f
31
This can be useful when describing binary flags, file permissions in code, or hex values for colors - for example, note no quotes:
>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215
Making ambiguous Python 2 octals compatible with Python 3
If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.
>>> 037
31
It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:
>>> 037
File "<stdin>", line 1
037
^
SyntaxError: invalid token
Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:
>>> 0o37
31
The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:
>>> import string
>>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
... int(x) or x.isalnum() and x or \
... len(set(string.punctuation).intersection(x)) == 1 and \
... x.count('.') == 1 and float(x) or x
>>> parseStr('123')
123
>>> parseStr('123.3')
123.3
>>> parseStr('3HC1')
'3HC1'
>>> parseStr('12.e5')
1200000.0
>>> parseStr('12$5')
'12$5'
>>> parseStr('12.2.2')
'12.2.2'
float("545.2222") and int(float("545.2222"))
The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type:
>>> import yaml
>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>
>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>
>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>
I use this function for that
import ast
def parse_str(s):
try:
return ast.literal_eval(str(s))
except:
return
It will convert the string to its type
value = parse_str('1') # Returns Integer
value = parse_str('1.5') # Returns Float
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else number_as_float
def num(s):
"""num(s)
num(3),num(3.7)-->3
num('3')-->3, num('3.7')-->3.7
num('3,700')-->ValueError
num('3a'),num('a3'),-->ValueError
num('3e4') --> 30000.0
"""
try:
return int(s)
except ValueError:
try:
return float(s)
except ValueError:
raise ValueError('argument is not a string of number')
You could use json.loads:
>>> import json
>>> json.loads('123.456')
123.456
>>> type(_)
<class 'float'>
>>>
As you can see it becomes a type of float.
You need to take into account rounding to do this properly.
i.e. - int(5.1) => 5
int(5.6) => 5 -- wrong, should be 6 so we do int(5.6 + 0.5) => 6
def convert(n):
try:
return int(n)
except ValueError:
return float(n + 0.5)
To typecast in Python use the constructor functions of the type, passing the string (or whatever value you are trying to cast) as a parameter.
For example:
>>>float("23.333")
23.333
Behind the scenes, Python is calling the objects __float__ method, which should return a float representation of the parameter. This is especially powerful, as you can define your own types (using classes) with a __float__ method so that it can be casted into a float using float(myobject).
Handles hex, octal, binary, decimal, and float
This solution will handle all of the string conventions for numbers (all that I know about).
def to_number(n):
''' Convert any number representation to a number
This covers: float, decimal, hex, and octal numbers.
'''
try:
return int(str(n), 0)
except:
try:
# Python 3 doesn't accept "010" as a valid octal. You must use the
# '0o' prefix
return int('0o' + n, 0)
except:
return float(n)
This test case output illustrates what I'm talking about.
======================== CAPTURED OUTPUT =========================
to_number(3735928559) = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0") = 0 == 0
to_number(100) = 100 == 100
to_number("42") = 42 == 42
to_number(8) = 8 == 8
to_number("0o20") = 16 == 16
to_number("020") = 16 == 16
to_number(3.14) = 3.14 == 3.14
to_number("2.72") = 2.72 == 2.72
to_number("1e3") = 1000.0 == 1000
to_number(0.001) = 0.001 == 0.001
to_number("0xA") = 10 == 10
to_number("012") = 10 == 10
to_number("0o12") = 10 == 10
to_number("0b01010") = 10 == 10
to_number("10") = 10 == 10
to_number("10.0") = 10.0 == 10
to_number("1e1") = 10.0 == 10
Here is the test:
class test_to_number(unittest.TestCase):
def test_hex(self):
# All of the following should be converted to an integer
#
values = [
# HEX
# ----------------------
# Input | Expected
# ----------------------
(0xDEADBEEF , 3735928559), # Hex
("0xFEEDFACE", 4277009102), # Hex
("0x0" , 0), # Hex
# Decimals
# ----------------------
# Input | Expected
# ----------------------
(100 , 100), # Decimal
("42" , 42), # Decimal
]
values += [
# Octals
# ----------------------
# Input | Expected
# ----------------------
(0o10 , 8), # Octal
("0o20" , 16), # Octal
("020" , 16), # Octal
]
values += [
# Floats
# ----------------------
# Input | Expected
# ----------------------
(3.14 , 3.14), # Float
("2.72" , 2.72), # Float
("1e3" , 1000), # Float
(1e-3 , 0.001), # Float
]
values += [
# All ints
# ----------------------
# Input | Expected
# ----------------------
("0xA" , 10),
("012" , 10),
("0o12" , 10),
("0b01010" , 10),
("10" , 10),
("10.0" , 10),
("1e1" , 10),
]
for _input, expected in values:
value = to_number(_input)
if isinstance(_input, str):
cmd = 'to_number("{}")'.format(_input)
else:
cmd = 'to_number({})'.format(_input)
print("{:23} = {:10} == {:10}".format(cmd, value, expected))
self.assertEqual(value, expected)
Pass your string to this function:
def string_to_number(str):
if("." in str):
try:
res = float(str)
except:
res = str
elif(str.isdigit()):
res = int(str)
else:
res = str
return(res)
It will return int, float or string depending on what was passed.
String that is an int
print(type(string_to_number("124")))
<class 'int'>
String that is a float
print(type(string_to_number("12.4")))
<class 'float'>
String that is a string
print(type(string_to_number("hello")))
<class 'str'>
String that looks like a float
print(type(string_to_number("hel.lo")))
<class 'str'>
There is also regex, because sometimes string must be prepared and normalized before casting to a number:
import re
def parseNumber(value, as_int=False):
try:
number = float(re.sub('[^.\-\d]', '', value))
if as_int:
return int(number + 0.5)
else:
return number
except ValueError:
return float('nan') # or None if you wish
Usage:
parseNumber('13,345')
> 13345.0
parseNumber('- 123 000')
> -123000.0
parseNumber('99999\n')
> 99999.0
And by the way, something to verify you have a number:
import numbers
def is_number(value):
return isinstance(value, numbers.Number)
# Will work with int, float, long, Decimal
a = int(float(a)) if int(float(a)) == float(a) else float(a)
This is a corrected version of Totoro's answer.
This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else
number_as_float
If you are dealing with mixed integers and floats and want a consistent way to deal with your mixed data, here is my solution with the proper docstring:
def parse_num(candidate):
"""Parse string to number if possible
It work equally well with negative and positive numbers, integers and floats.
Args:
candidate (str): string to convert
Returns:
float | int | None: float or int if possible otherwise None
"""
try:
float_value = float(candidate)
except ValueError:
return None
# Optional part if you prefer int to float when decimal part is 0
if float_value.is_integer():
return int(float_value)
# end of the optional part
return float_value
# Test
candidates = ['34.77', '-13', 'jh', '8990', '76_3234_54']
res_list = list(map(parse_num, candidates))
print('Before:')
print(candidates)
print('After:')
print(res_list)
Output:
Before:
['34.77', '-13', 'jh', '8990', '76_3234_54']
After:
[34.77, -13, None, 8990, 76323454]
Use:
def num(s):
try:
for each in s:
yield int(each)
except ValueError:
yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()
This is the most Pythonic way I could come up with.
If you don't want to use third party modules the following might be the most robust solution:
def string_to_int_or_float(s):
try:
f = float(s) # replace s with str(s) if you are not sure that s is a string
except ValueError:
print("Provided string '" + s + "' is not interpretable as a literal number.")
raise
try:
i = int(str(f).rstrip('0').rstrip('.'))
except:
return f
return i
It might not be the fastest, but it handles correctly literal numbers where many other solutions fail, such as:
>>> string_to_int_or_float('789.')
789
>>> string_to_int_or_float('789.0')
789
>>> string_to_int_or_float('12.3e2')
1230
>>> string_to_int_or_float('12.3e-2')
0.123
>>> string_to_int_or_float('4560e-1')
456
>>> string_to_int_or_float('4560e-2')
45.6
You can simply do this by
s = '542.22'
f = float(s) # This converts string data to float data with a decimal point
print(f)
i = int(f) # This converts string data to integer data by just taking the whole number part of it
print(i)
For more information on parsing of data types check on python documentation!
This is a function which will convert any object (not just str) to int or float, based on if the actual string supplied looks like int or float. Further if it's an object which has both __float and __int__ methods, it defaults to using __float__
def conv_to_num(x, num_type='asis'):
'''Converts an object to a number if possible.
num_type: int, float, 'asis'
Defaults to floating point in case of ambiguity.
'''
import numbers
is_num, is_str, is_other = [False]*3
if isinstance(x, numbers.Number):
is_num = True
elif isinstance(x, str):
is_str = True
is_other = not any([is_num, is_str])
if is_num:
res = x
elif is_str:
is_float, is_int, is_char = [False]*3
try:
res = float(x)
if '.' in x:
is_float = True
else:
is_int = True
except ValueError:
res = x
is_char = True
else:
if num_type == 'asis':
funcs = [int, float]
else:
funcs = [num_type]
for func in funcs:
try:
res = func(x)
break
except TypeError:
continue
else:
res = x
By using int and float methods we can convert a string to integer and floats.
s="45.8"
print(float(s))
y='67'
print(int(y))
For numbers and characters together:
string_for_int = "498 results should get"
string_for_float = "498.45645765 results should get"
First import re:
import re
# For getting the integer part:
print(int(re.search(r'\d+', string_for_int).group())) #498
# For getting the float part:
print(float(re.search(r'\d+\.\d+', string_for_float).group())) #498.45645765
For easy model:
value1 = "10"
value2 = "10.2"
print(int(value1)) # 10
print(float(value2)) # 10.2

How to display decimals python 3.6.0 [duplicate]

How can I convert a str to float?
"545.2222" → 545.2222
How can I convert a str to int?
"31" → 31
For the reverse, see Convert integer to string in Python and Converting a float to a string without rounding it.
Please instead use How can I read inputs as numbers? to close duplicate questions where OP received a string from user input and immediately wants to convert it, or was hoping for input (in 3.x) to convert the type automatically.
>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545
Python2 method to check if a string is a float:
def is_float(value):
if value is None:
return False
try:
float(value)
return True
except:
return False
For the Python3 version of is_float see: Checking if a string can be converted to float in Python
A longer and more accurate name for this function could be: is_convertible_to_float(value)
What is, and is not a float in Python may surprise you:
The below unit tests were done using python2. Check it that Python3 has different behavior for what strings are convertable to float. One confounding difference is that any number of interior underscores are now allowed: (float("1_3.4") == float(13.4)) is True
val is_float(val) Note
-------------------- ---------- --------------------------------
"" False Blank string
"127" True Passed string
True True Pure sweet Truth
"True" False Vile contemptible lie
False True So false it becomes true
"123.456" True Decimal
" -127 " True Spaces trimmed
"\t\n12\r\n" True whitespace ignored
"NaN" True Not a number
"NaNanananaBATMAN" False I am Batman
"-iNF" True Negative infinity
"123.E4" True Exponential notation
".1" True mantissa only
"1_2_3.4" False Underscores not allowed
"12 34" False Spaces not allowed on interior
"1,234" False Commas gtfo
u'\x30' True Unicode is fine.
"NULL" False Null is not special
0x3fade True Hexadecimal
"6e7777777777777" True Shrunk to infinity
"1.797693e+308" True This is max value
"infinity" True Same as inf
"infinityandBEYOND" False Extra characters wreck it
"12.34.56" False Only one dot allowed
u'四' False Japanese '4' is not a float.
"#56" False Pound sign
"56%" False Percent of what?
"0E0" True Exponential, move dot 0 places
0**0 True 0___0 Exponentiation
"-5e-5" True Raise to a negative number
"+1e1" True Plus is OK with exponent
"+1e1^5" False Fancy exponent not interpreted
"+1e1.3" False No decimals in exponent
"-+1" False Make up your mind
"(1)" False Parenthesis is bad
You think you know what numbers are? You are not so good as you think! Not big surprise.
Don't use this code on life-critical software!
Catching broad exceptions this way, killing canaries and gobbling the exception creates a tiny chance that a valid float as string will return false. The float(...) line of code can failed for any of a thousand reasons that have nothing to do with the contents of the string. But if you're writing life-critical software in a duck-typing prototype language like Python, then you've got much larger problems.
def num(s):
try:
return int(s)
except ValueError:
return float(s)
This is another method which deserves to be mentioned here, ast.literal_eval:
This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself.
That is, a safe 'eval'
>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31
Localization and commas
You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.
Example 1 -- United States number conventions
In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:
>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>
Example 2 -- European number conventions
In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:
>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222
The method locale.atoi is also available, but the argument should be an integer.
float(x) if '.' in x else int(x)
If you aren't averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation:
>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int
Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int("545") without first casting to float.
If your strings are in a list, you could use the map function as well.
>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>
It is only good if they're all the same type.
In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31?
I just want to know how to parse a float string to a float, and (separately) an int string to an int.
It's good that you ask to do these separately. If you're mixing them, you may be setting yourself up for problems later. The simple answer is:
"545.2222" to float:
>>> float("545.2222")
545.2222
"31" to an integer:
>>> int("31")
31
Other conversions, ints to and from strings and literals:
Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:
>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31
If you don't know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:
>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31
Non-Decimal (i.e. Integer) Literals from other Bases
If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases - you can let Python do it for you automatically with the correct syntax.
You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:
Binary, prefix 0b
>>> 0b11111
31
Octal, prefix 0o
>>> 0o37
31
Hexadecimal, prefix 0x
>>> 0x1f
31
This can be useful when describing binary flags, file permissions in code, or hex values for colors - for example, note no quotes:
>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215
Making ambiguous Python 2 octals compatible with Python 3
If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.
>>> 037
31
It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:
>>> 037
File "<stdin>", line 1
037
^
SyntaxError: invalid token
Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:
>>> 0o37
31
The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:
>>> import string
>>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
... int(x) or x.isalnum() and x or \
... len(set(string.punctuation).intersection(x)) == 1 and \
... x.count('.') == 1 and float(x) or x
>>> parseStr('123')
123
>>> parseStr('123.3')
123.3
>>> parseStr('3HC1')
'3HC1'
>>> parseStr('12.e5')
1200000.0
>>> parseStr('12$5')
'12$5'
>>> parseStr('12.2.2')
'12.2.2'
float("545.2222") and int(float("545.2222"))
The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type:
>>> import yaml
>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>
>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>
>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>
I use this function for that
import ast
def parse_str(s):
try:
return ast.literal_eval(str(s))
except:
return
It will convert the string to its type
value = parse_str('1') # Returns Integer
value = parse_str('1.5') # Returns Float
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else number_as_float
def num(s):
"""num(s)
num(3),num(3.7)-->3
num('3')-->3, num('3.7')-->3.7
num('3,700')-->ValueError
num('3a'),num('a3'),-->ValueError
num('3e4') --> 30000.0
"""
try:
return int(s)
except ValueError:
try:
return float(s)
except ValueError:
raise ValueError('argument is not a string of number')
You could use json.loads:
>>> import json
>>> json.loads('123.456')
123.456
>>> type(_)
<class 'float'>
>>>
As you can see it becomes a type of float.
You need to take into account rounding to do this properly.
i.e. - int(5.1) => 5
int(5.6) => 5 -- wrong, should be 6 so we do int(5.6 + 0.5) => 6
def convert(n):
try:
return int(n)
except ValueError:
return float(n + 0.5)
To typecast in Python use the constructor functions of the type, passing the string (or whatever value you are trying to cast) as a parameter.
For example:
>>>float("23.333")
23.333
Behind the scenes, Python is calling the objects __float__ method, which should return a float representation of the parameter. This is especially powerful, as you can define your own types (using classes) with a __float__ method so that it can be casted into a float using float(myobject).
Handles hex, octal, binary, decimal, and float
This solution will handle all of the string conventions for numbers (all that I know about).
def to_number(n):
''' Convert any number representation to a number
This covers: float, decimal, hex, and octal numbers.
'''
try:
return int(str(n), 0)
except:
try:
# Python 3 doesn't accept "010" as a valid octal. You must use the
# '0o' prefix
return int('0o' + n, 0)
except:
return float(n)
This test case output illustrates what I'm talking about.
======================== CAPTURED OUTPUT =========================
to_number(3735928559) = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0") = 0 == 0
to_number(100) = 100 == 100
to_number("42") = 42 == 42
to_number(8) = 8 == 8
to_number("0o20") = 16 == 16
to_number("020") = 16 == 16
to_number(3.14) = 3.14 == 3.14
to_number("2.72") = 2.72 == 2.72
to_number("1e3") = 1000.0 == 1000
to_number(0.001) = 0.001 == 0.001
to_number("0xA") = 10 == 10
to_number("012") = 10 == 10
to_number("0o12") = 10 == 10
to_number("0b01010") = 10 == 10
to_number("10") = 10 == 10
to_number("10.0") = 10.0 == 10
to_number("1e1") = 10.0 == 10
Here is the test:
class test_to_number(unittest.TestCase):
def test_hex(self):
# All of the following should be converted to an integer
#
values = [
# HEX
# ----------------------
# Input | Expected
# ----------------------
(0xDEADBEEF , 3735928559), # Hex
("0xFEEDFACE", 4277009102), # Hex
("0x0" , 0), # Hex
# Decimals
# ----------------------
# Input | Expected
# ----------------------
(100 , 100), # Decimal
("42" , 42), # Decimal
]
values += [
# Octals
# ----------------------
# Input | Expected
# ----------------------
(0o10 , 8), # Octal
("0o20" , 16), # Octal
("020" , 16), # Octal
]
values += [
# Floats
# ----------------------
# Input | Expected
# ----------------------
(3.14 , 3.14), # Float
("2.72" , 2.72), # Float
("1e3" , 1000), # Float
(1e-3 , 0.001), # Float
]
values += [
# All ints
# ----------------------
# Input | Expected
# ----------------------
("0xA" , 10),
("012" , 10),
("0o12" , 10),
("0b01010" , 10),
("10" , 10),
("10.0" , 10),
("1e1" , 10),
]
for _input, expected in values:
value = to_number(_input)
if isinstance(_input, str):
cmd = 'to_number("{}")'.format(_input)
else:
cmd = 'to_number({})'.format(_input)
print("{:23} = {:10} == {:10}".format(cmd, value, expected))
self.assertEqual(value, expected)
Pass your string to this function:
def string_to_number(str):
if("." in str):
try:
res = float(str)
except:
res = str
elif(str.isdigit()):
res = int(str)
else:
res = str
return(res)
It will return int, float or string depending on what was passed.
String that is an int
print(type(string_to_number("124")))
<class 'int'>
String that is a float
print(type(string_to_number("12.4")))
<class 'float'>
String that is a string
print(type(string_to_number("hello")))
<class 'str'>
String that looks like a float
print(type(string_to_number("hel.lo")))
<class 'str'>
There is also regex, because sometimes string must be prepared and normalized before casting to a number:
import re
def parseNumber(value, as_int=False):
try:
number = float(re.sub('[^.\-\d]', '', value))
if as_int:
return int(number + 0.5)
else:
return number
except ValueError:
return float('nan') # or None if you wish
Usage:
parseNumber('13,345')
> 13345.0
parseNumber('- 123 000')
> -123000.0
parseNumber('99999\n')
> 99999.0
And by the way, something to verify you have a number:
import numbers
def is_number(value):
return isinstance(value, numbers.Number)
# Will work with int, float, long, Decimal
a = int(float(a)) if int(float(a)) == float(a) else float(a)
This is a corrected version of Totoro's answer.
This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else
number_as_float
If you are dealing with mixed integers and floats and want a consistent way to deal with your mixed data, here is my solution with the proper docstring:
def parse_num(candidate):
"""Parse string to number if possible
It work equally well with negative and positive numbers, integers and floats.
Args:
candidate (str): string to convert
Returns:
float | int | None: float or int if possible otherwise None
"""
try:
float_value = float(candidate)
except ValueError:
return None
# Optional part if you prefer int to float when decimal part is 0
if float_value.is_integer():
return int(float_value)
# end of the optional part
return float_value
# Test
candidates = ['34.77', '-13', 'jh', '8990', '76_3234_54']
res_list = list(map(parse_num, candidates))
print('Before:')
print(candidates)
print('After:')
print(res_list)
Output:
Before:
['34.77', '-13', 'jh', '8990', '76_3234_54']
After:
[34.77, -13, None, 8990, 76323454]
Use:
def num(s):
try:
for each in s:
yield int(each)
except ValueError:
yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()
This is the most Pythonic way I could come up with.
If you don't want to use third party modules the following might be the most robust solution:
def string_to_int_or_float(s):
try:
f = float(s) # replace s with str(s) if you are not sure that s is a string
except ValueError:
print("Provided string '" + s + "' is not interpretable as a literal number.")
raise
try:
i = int(str(f).rstrip('0').rstrip('.'))
except:
return f
return i
It might not be the fastest, but it handles correctly literal numbers where many other solutions fail, such as:
>>> string_to_int_or_float('789.')
789
>>> string_to_int_or_float('789.0')
789
>>> string_to_int_or_float('12.3e2')
1230
>>> string_to_int_or_float('12.3e-2')
0.123
>>> string_to_int_or_float('4560e-1')
456
>>> string_to_int_or_float('4560e-2')
45.6
You can simply do this by
s = '542.22'
f = float(s) # This converts string data to float data with a decimal point
print(f)
i = int(f) # This converts string data to integer data by just taking the whole number part of it
print(i)
For more information on parsing of data types check on python documentation!
This is a function which will convert any object (not just str) to int or float, based on if the actual string supplied looks like int or float. Further if it's an object which has both __float and __int__ methods, it defaults to using __float__
def conv_to_num(x, num_type='asis'):
'''Converts an object to a number if possible.
num_type: int, float, 'asis'
Defaults to floating point in case of ambiguity.
'''
import numbers
is_num, is_str, is_other = [False]*3
if isinstance(x, numbers.Number):
is_num = True
elif isinstance(x, str):
is_str = True
is_other = not any([is_num, is_str])
if is_num:
res = x
elif is_str:
is_float, is_int, is_char = [False]*3
try:
res = float(x)
if '.' in x:
is_float = True
else:
is_int = True
except ValueError:
res = x
is_char = True
else:
if num_type == 'asis':
funcs = [int, float]
else:
funcs = [num_type]
for func in funcs:
try:
res = func(x)
break
except TypeError:
continue
else:
res = x
By using int and float methods we can convert a string to integer and floats.
s="45.8"
print(float(s))
y='67'
print(int(y))
For numbers and characters together:
string_for_int = "498 results should get"
string_for_float = "498.45645765 results should get"
First import re:
import re
# For getting the integer part:
print(int(re.search(r'\d+', string_for_int).group())) #498
# For getting the float part:
print(float(re.search(r'\d+\.\d+', string_for_float).group())) #498.45645765
For easy model:
value1 = "10"
value2 = "10.2"
print(int(value1)) # 10
print(float(value2)) # 10.2

Python how to check if a string represent a float? [duplicate]

How do I check if a string represents a numeric value in Python?
def is_number(s):
try:
float(s)
return True
except ValueError:
return False
The above works, but it seems clunky.
If what you are testing comes from user input, it is still a string even if it represents an int or a float. See How can I read inputs as numbers? for converting the input, and Asking the user for input until they give a valid response for ensuring that the input represents an int or float (or other requirements) before proceeding.
For non-negative (unsigned) integers only, use isdigit():
>>> a = "03523"
>>> a.isdigit()
True
>>> b = "963spam"
>>> b.isdigit()
False
Documentation for isdigit(): Python2, Python3
For Python 2 Unicode strings:
isnumeric().
Which, not only is ugly and slow
I'd dispute both.
A regex or other string parsing method would be uglier and slower.
I'm not sure that anything much could be faster than the above. It calls the function and returns. Try/Catch doesn't introduce much overhead because the most common exception is caught without an extensive search of stack frames.
The issue is that any numeric conversion function has two kinds of results
A number, if the number is valid
A status code (e.g., via errno) or exception to show that no valid number could be parsed.
C (as an example) hacks around this a number of ways. Python lays it out clearly and explicitly.
I think your code for doing this is perfect.
TL;DR The best solution is s.replace('.','',1).isdigit()
I did some benchmarks comparing the different approaches
def is_number_tryexcept(s):
""" Returns True if string is a number. """
try:
float(s)
return True
except ValueError:
return False
import re
def is_number_regex(s):
""" Returns True if string is a number. """
if re.match("^\d+?\.\d+?$", s) is None:
return s.isdigit()
return True
def is_number_repl_isdigit(s):
""" Returns True if string is a number. """
return s.replace('.','',1).isdigit()
If the string is not a number, the except-block is quite slow. But more importantly, the try-except method is the only approach that handles scientific notations correctly.
funcs = [
is_number_tryexcept,
is_number_regex,
is_number_repl_isdigit
]
a_float = '.1234'
print('Float notation ".1234" is not supported by:')
for f in funcs:
if not f(a_float):
print('\t -', f.__name__)
Float notation ".1234" is not supported by:
is_number_regex
scientific1 = '1.000000e+50'
scientific2 = '1e50'
print('Scientific notation "1.000000e+50" is not supported by:')
for f in funcs:
if not f(scientific1):
print('\t -', f.name)
print('Scientific notation "1e50" is not supported by:')
for f in funcs:
if not f(scientific2):
print('\t -', f.name)
Scientific notation "1.000000e+50" is not supported by:
is_number_regex
is_number_repl_isdigit
Scientific notation "1e50" is not supported by:
is_number_regex
is_number_repl_isdigit
EDIT: The benchmark results
import timeit
test_cases = ['1.12345', '1.12.345', 'abc12345', '12345']
times_n = {f.__name__:[] for f in funcs}
for t in test_cases:
for f in funcs:
f = f.__name__
times_n[f].append(min(timeit.Timer('%s(t)' %f,
'from __main__ import %s, t' %f)
.repeat(repeat=3, number=1000000)))
where the following functions were tested
from re import match as re_match
from re import compile as re_compile
def is_number_tryexcept(s):
""" Returns True if string is a number. """
try:
float(s)
return True
except ValueError:
return False
def is_number_regex(s):
""" Returns True if string is a number. """
if re_match("^\d+?\.\d+?$", s) is None:
return s.isdigit()
return True
comp = re_compile("^\d+?\.\d+?$")
def compiled_regex(s):
""" Returns True if string is a number. """
if comp.match(s) is None:
return s.isdigit()
return True
def is_number_repl_isdigit(s):
""" Returns True if string is a number. """
return s.replace('.','',1).isdigit()
There is one exception that you may want to take into account: the string 'NaN'
If you want is_number to return FALSE for 'NaN' this code will not work as Python converts it to its representation of a number that is not a number (talk about identity issues):
>>> float('NaN')
nan
Otherwise, I should actually thank you for the piece of code I now use extensively. :)
G.
how about this:
'3.14'.replace('.','',1).isdigit()
which will return true only if there is one or no '.' in the string of digits.
'3.14.5'.replace('.','',1).isdigit()
will return false
edit: just saw another comment ...
adding a .replace(badstuff,'',maxnum_badstuff) for other cases can be done. if you are passing salt and not arbitrary condiments (ref:xkcd#974) this will do fine :P
Updated after Alfe pointed out you don't need to check for float separately as complex handles both:
def is_number(s):
try:
complex(s) # for int, long, float and complex
except ValueError:
return False
return True
Previously said: Is some rare cases you might also need to check for complex numbers (e.g. 1+2i), which can not be represented by a float:
def is_number(s):
try:
float(s) # for int, long and float
except ValueError:
try:
complex(s) # for complex
except ValueError:
return False
return True
Which, not only is ugly and slow, seems clunky.
It may take some getting used to, but this is the pythonic way of doing it. As has been already pointed out, the alternatives are worse. But there is one other advantage of doing things this way: polymorphism.
The central idea behind duck typing is that "if it walks and talks like a duck, then it's a duck." What if you decide that you need to subclass string so that you can change how you determine if something can be converted into a float? Or what if you decide to test some other object entirely? You can do these things without having to change the above code.
Other languages solve these problems by using interfaces. I'll save the analysis of which solution is better for another thread. The point, though, is that python is decidedly on the duck typing side of the equation, and you're probably going to have to get used to syntax like this if you plan on doing much programming in Python (but that doesn't mean you have to like it of course).
One other thing you might want to take into consideration: Python is pretty fast in throwing and catching exceptions compared to a lot of other languages (30x faster than .Net for instance). Heck, the language itself even throws exceptions to communicate non-exceptional, normal program conditions (every time you use a for loop). Thus, I wouldn't worry too much about the performance aspects of this code until you notice a significant problem.
For int use this:
>>> "1221323".isdigit()
True
But for float we need some tricks ;-). Every float number has one point...
>>> "12.34".isdigit()
False
>>> "12.34".replace('.','',1).isdigit()
True
>>> "12.3.4".replace('.','',1).isdigit()
False
Also for negative numbers just add lstrip():
>>> '-12'.lstrip('-')
'12'
And now we get a universal way:
>>> '-12.34'.lstrip('-').replace('.','',1).isdigit()
True
>>> '.-234'.lstrip('-').replace('.','',1).isdigit()
False
This answer provides step by step guide having function with examples to find the string is:
Positive integer
Positive/negative - integer/float
How to discard "NaN" (not a number) strings while checking for number?
Check if string is positive integer
You may use str.isdigit() to check whether given string is positive integer.
Sample Results:
# For digit
>>> '1'.isdigit()
True
>>> '1'.isalpha()
False
Check for string as positive/negative - integer/float
str.isdigit() returns False if the string is a negative number or a float number. For example:
# returns `False` for float
>>> '123.3'.isdigit()
False
# returns `False` for negative number
>>> '-123'.isdigit()
False
If you want to also check for the negative integers and float, then you may write a custom function to check for it as:
def is_number(n):
try:
float(n) # Type-casting the string to `float`.
# If string is not a valid `float`,
# it'll raise `ValueError` exception
except ValueError:
return False
return True
Sample Run:
>>> is_number('123') # positive integer number
True
>>> is_number('123.4') # positive float number
True
>>> is_number('-123') # negative integer number
True
>>> is_number('-123.4') # negative `float` number
True
>>> is_number('abc') # `False` for "some random" string
False
Discard "NaN" (not a number) strings while checking for number
The above functions will return True for the "NAN" (Not a number) string because for Python it is valid float representing it is not a number. For example:
>>> is_number('NaN')
True
In order to check whether the number is "NaN", you may use math.isnan() as:
>>> import math
>>> nan_num = float('nan')
>>> math.isnan(nan_num)
True
Or if you don't want to import additional library to check this, then you may simply check it via comparing it with itself using ==. Python returns False when nan float is compared with itself. For example:
# `nan_num` variable is taken from above example
>>> nan_num == nan_num
False
Hence, above function is_number can be updated to return False for "NaN" as:
def is_number(n):
is_number = True
try:
num = float(n)
# check for "nan" floats
is_number = num == num # or use `math.isnan(num)`
except ValueError:
is_number = False
return is_number
Sample Run:
>>> is_number('Nan') # not a number "Nan" string
False
>>> is_number('nan') # not a number string "nan" with all lower cased
False
>>> is_number('123') # positive integer
True
>>> is_number('-123') # negative integer
True
>>> is_number('-1.12') # negative `float`
True
>>> is_number('abc') # "some random" string
False
PS: Each operation for each check depending on the type of number comes with additional overhead. Choose the version of is_number function which fits your requirement.
For strings of non-numbers, try: except: is actually slower than regular expressions. For strings of valid numbers, regex is slower. So, the appropriate method depends on your input.
If you find that you are in a performance bind, you can use a new third-party module called fastnumbers that provides a function called isfloat. Full disclosure, I am the author. I have included its results in the timings below.
from __future__ import print_function
import timeit
prep_base = '''\
x = 'invalid'
y = '5402'
z = '4.754e3'
'''
prep_try_method = '''\
def is_number_try(val):
try:
float(val)
return True
except ValueError:
return False
'''
prep_re_method = '''\
import re
float_match = re.compile(r'[-+]?\d*\.?\d+(?:[eE][-+]?\d+)?$').match
def is_number_re(val):
return bool(float_match(val))
'''
fn_method = '''\
from fastnumbers import isfloat
'''
print('Try with non-number strings', timeit.timeit('is_number_try(x)',
prep_base + prep_try_method), 'seconds')
print('Try with integer strings', timeit.timeit('is_number_try(y)',
prep_base + prep_try_method), 'seconds')
print('Try with float strings', timeit.timeit('is_number_try(z)',
prep_base + prep_try_method), 'seconds')
print()
print('Regex with non-number strings', timeit.timeit('is_number_re(x)',
prep_base + prep_re_method), 'seconds')
print('Regex with integer strings', timeit.timeit('is_number_re(y)',
prep_base + prep_re_method), 'seconds')
print('Regex with float strings', timeit.timeit('is_number_re(z)',
prep_base + prep_re_method), 'seconds')
print()
print('fastnumbers with non-number strings', timeit.timeit('isfloat(x)',
prep_base + 'from fastnumbers import isfloat'), 'seconds')
print('fastnumbers with integer strings', timeit.timeit('isfloat(y)',
prep_base + 'from fastnumbers import isfloat'), 'seconds')
print('fastnumbers with float strings', timeit.timeit('isfloat(z)',
prep_base + 'from fastnumbers import isfloat'), 'seconds')
print()
Try with non-number strings 2.39108395576 seconds
Try with integer strings 0.375686168671 seconds
Try with float strings 0.369210958481 seconds
Regex with non-number strings 0.748660802841 seconds
Regex with integer strings 1.02021503448 seconds
Regex with float strings 1.08564686775 seconds
fastnumbers with non-number strings 0.174362897873 seconds
fastnumbers with integer strings 0.179651021957 seconds
fastnumbers with float strings 0.20222902298 seconds
As you can see
try: except: was fast for numeric input but very slow for an invalid input
regex is very efficient when the input is invalid
fastnumbers wins in both cases
I know this is particularly old but I would add an answer I believe covers the information missing from the highest voted answer that could be very valuable to any who find this:
For each of the following methods connect them with a count if you need any input to be accepted. (Assuming we are using vocal definitions of integers rather than 0-255, etc.)
x.isdigit()
works well for checking if x is an integer.
x.replace('-','').isdigit()
works well for checking if x is a negative.(Check - in first position)
x.replace('.','').isdigit()
works well for checking if x is a decimal.
x.replace(':','').isdigit()
works well for checking if x is a ratio.
x.replace('/','',1).isdigit()
works well for checking if x is a fraction.
Just Mimic C#
In C# there are two different functions that handle parsing of scalar values:
Float.Parse()
Float.TryParse()
float.parse():
def parse(string):
try:
return float(string)
except Exception:
throw TypeError
Note: If you're wondering why I changed the exception to a TypeError, here's the documentation.
float.try_parse():
def try_parse(string, fail=None):
try:
return float(string)
except Exception:
return fail;
Note: You don't want to return the boolean 'False' because that's still a value type. None is better because it indicates failure. Of course, if you want something different you can change the fail parameter to whatever you want.
To extend float to include the 'parse()' and 'try_parse()' you'll need to monkeypatch the 'float' class to add these methods.
If you want respect pre-existing functions the code should be something like:
def monkey_patch():
if(!hasattr(float, 'parse')):
float.parse = parse
if(!hasattr(float, 'try_parse')):
float.try_parse = try_parse
SideNote: I personally prefer to call it Monkey Punching because it feels like I'm abusing the language when I do this but YMMV.
Usage:
float.parse('giggity') // throws TypeException
float.parse('54.3') // returns the scalar value 54.3
float.tryParse('twank') // returns None
float.tryParse('32.2') // returns the scalar value 32.2
And the great Sage Pythonas said to the Holy See Sharpisus, "Anything you can do I can do better; I can do anything better than you."
Casting to float and catching ValueError is probably the fastest way, since float() is specifically meant for just that. Anything else that requires string parsing (regex, etc) will likely be slower due to the fact that it's not tuned for this operation. My $0.02.
You can use Unicode strings, they have a method to do just what you want:
>>> s = u"345"
>>> s.isnumeric()
True
Or:
>>> s = "345"
>>> u = unicode(s)
>>> u.isnumeric()
True
http://www.tutorialspoint.com/python/string_isnumeric.htm
http://docs.python.org/2/howto/unicode.html
So to put it all together, checking for Nan, infinity and complex numbers (it would seem they are specified with j, not i, i.e. 1+2j) it results in:
def is_number(s):
try:
n=str(float(s))
if n == "nan" or n=="inf" or n=="-inf" : return False
except ValueError:
try:
complex(s) # for complex
except ValueError:
return False
return True
I wanted to see which method is fastest. Overall the best and most consistent results were given by the check_replace function. The fastest results were given by the check_exception function, but only if there was no exception fired - meaning its code is the most efficient, but the overhead of throwing an exception is quite large.
Please note that checking for a successful cast is the only method which is accurate, for example, this works with check_exception but the other two test functions will return False for a valid float:
huge_number = float('1e+100')
Here is the benchmark code:
import time, re, random, string
ITERATIONS = 10000000
class Timer:
def __enter__(self):
self.start = time.clock()
return self
def __exit__(self, *args):
self.end = time.clock()
self.interval = self.end - self.start
def check_regexp(x):
return re.compile("^\d*\.?\d*$").match(x) is not None
def check_replace(x):
return x.replace('.','',1).isdigit()
def check_exception(s):
try:
float(s)
return True
except ValueError:
return False
to_check = [check_regexp, check_replace, check_exception]
print('preparing data...')
good_numbers = [
str(random.random() / random.random())
for x in range(ITERATIONS)]
bad_numbers = ['.' + x for x in good_numbers]
strings = [
''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(random.randint(1,10)))
for x in range(ITERATIONS)]
print('running test...')
for func in to_check:
with Timer() as t:
for x in good_numbers:
res = func(x)
print('%s with good floats: %s' % (func.__name__, t.interval))
with Timer() as t:
for x in bad_numbers:
res = func(x)
print('%s with bad floats: %s' % (func.__name__, t.interval))
with Timer() as t:
for x in strings:
res = func(x)
print('%s with strings: %s' % (func.__name__, t.interval))
Here are the results with Python 2.7.10 on a 2017 MacBook Pro 13:
check_regexp with good floats: 12.688639
check_regexp with bad floats: 11.624862
check_regexp with strings: 11.349414
check_replace with good floats: 4.419841
check_replace with bad floats: 4.294909
check_replace with strings: 4.086358
check_exception with good floats: 3.276668
check_exception with bad floats: 13.843092
check_exception with strings: 15.786169
Here are the results with Python 3.6.5 on a 2017 MacBook Pro 13:
check_regexp with good floats: 13.472906000000009
check_regexp with bad floats: 12.977665000000016
check_regexp with strings: 12.417542999999995
check_replace with good floats: 6.011045999999993
check_replace with bad floats: 4.849356
check_replace with strings: 4.282754000000011
check_exception with good floats: 6.039081999999979
check_exception with bad floats: 9.322753000000006
check_exception with strings: 9.952595000000002
Here are the results with PyPy 2.7.13 on a 2017 MacBook Pro 13:
check_regexp with good floats: 2.693217
check_regexp with bad floats: 2.744819
check_regexp with strings: 2.532414
check_replace with good floats: 0.604367
check_replace with bad floats: 0.538169
check_replace with strings: 0.598664
check_exception with good floats: 1.944103
check_exception with bad floats: 2.449182
check_exception with strings: 2.200056
The input may be as follows:
a="50"
b=50
c=50.1
d="50.1"
1-General input:
The input of this function can be everything!
Finds whether the given variable is numeric. Numeric strings consist of optional sign, any number of digits, optional decimal part and optional exponential part. Thus +0123.45e6 is a valid numeric value. Hexadecimal (e.g. 0xf4c3b00c) and binary (e.g. 0b10100111001) notation is not allowed.
is_numeric function
import ast
import numbers
def is_numeric(obj):
if isinstance(obj, numbers.Number):
return True
elif isinstance(obj, str):
nodes = list(ast.walk(ast.parse(obj)))[1:]
if not isinstance(nodes[0], ast.Expr):
return False
if not isinstance(nodes[-1], ast.Num):
return False
nodes = nodes[1:-1]
for i in range(len(nodes)):
#if used + or - in digit :
if i % 2 == 0:
if not isinstance(nodes[i], ast.UnaryOp):
return False
else:
if not isinstance(nodes[i], (ast.USub, ast.UAdd)):
return False
return True
else:
return False
test:
>>> is_numeric("54")
True
>>> is_numeric("54.545")
True
>>> is_numeric("0x45")
True
is_float function
Finds whether the given variable is float. float strings consist of optional sign, any number of digits, ...
import ast
def is_float(obj):
if isinstance(obj, float):
return True
if isinstance(obj, int):
return False
elif isinstance(obj, str):
nodes = list(ast.walk(ast.parse(obj)))[1:]
if not isinstance(nodes[0], ast.Expr):
return False
if not isinstance(nodes[-1], ast.Num):
return False
if not isinstance(nodes[-1].n, float):
return False
nodes = nodes[1:-1]
for i in range(len(nodes)):
if i % 2 == 0:
if not isinstance(nodes[i], ast.UnaryOp):
return False
else:
if not isinstance(nodes[i], (ast.USub, ast.UAdd)):
return False
return True
else:
return False
test:
>>> is_float("5.4")
True
>>> is_float("5")
False
>>> is_float(5)
False
>>> is_float("5")
False
>>> is_float("+5.4")
True
what is ast?
2- If you are confident that the variable content is String:
use str.isdigit() method
>>> a=454
>>> a.isdigit()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'int' object has no attribute 'isdigit'
>>> a="454"
>>> a.isdigit()
True
3-Numerical input:
detect int value:
>>> isinstance("54", int)
False
>>> isinstance(54, int)
True
>>>
detect float:
>>> isinstance("45.1", float)
False
>>> isinstance(45.1, float)
True
In a most general case for a float, one would like to take care of integers and decimals. Let's take the string "1.1" as an example.
I would try one of the following:
1.> isnumeric()
word = "1.1"
"".join(word.split(".")).isnumeric()
>>> True
2.> isdigit()
word = "1.1"
"".join(word.split(".")).isdigit()
>>> True
3.> isdecimal()
word = "1.1"
"".join(word.split(".")).isdecimal()
>>> True
Speed:
► All the aforementioned methods have similar speeds.
%timeit "".join(word.split(".")).isnumeric()
>>> 257 ns ± 12 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit "".join(word.split(".")).isdigit()
>>> 252 ns ± 11 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit "".join(word.split(".")).isdecimal()
>>> 244 ns ± 7.17 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
str.isnumeric()
Return True if all characters in the string are numeric characters,
and there is at least one character, False otherwise. Numeric
characters include digit characters, and all characters that have the
Unicode numeric value property, e.g. U+2155, VULGAR FRACTION ONE
FIFTH. Formally, numeric characters are those with the property value
Numeric_Type=Digit, Numeric_Type=Decimal or Numeric_Type=Numeric.
str.isdecimal()
Return True if all characters in the string are decimal characters and
there is at least one character, False otherwise. Decimal characters
are those that can be used to form numbers in base 10, e.g. U+0660,
ARABIC-INDIC DIGIT ZERO. Formally a decimal character is a character
in the Unicode General Category “Nd”.
Both available for string types from Python 3.0.
I needed to determine if a string cast into basic types (float,int,str,bool). After not finding anything on the internet I created this:
def str_to_type (s):
""" Get possible cast type for a string
Parameters
----------
s : string
Returns
-------
float,int,str,bool : type
Depending on what it can be cast to
"""
try:
f = float(s)
if "." not in s:
return int
return float
except ValueError:
value = s.upper()
if value == "TRUE" or value == "FALSE":
return bool
return type(s)
Example
str_to_type("true") # bool
str_to_type("6.0") # float
str_to_type("6") # int
str_to_type("6abc") # str
str_to_type(u"6abc") # unicode
You can capture the type and use it
s = "6.0"
type_ = str_to_type(s) # float
f = type_(s)
I think your solution is fine, but there is a correct regexp implementation.
There does seem to be a lot of regexp hate towards these answers which I think is unjustified, regexps can be reasonably clean and correct and fast. It really depends on what you're trying to do. The original question was how can you "check if a string can be represented as a number (float)" (as per your title). Presumably you would want to use the numeric/float value once you've checked that it's valid, in which case your try/except makes a lot of sense. But if, for some reason, you just want to validate that a string is a number then a regex also works fine, but it's hard to get correct. I think most of the regex answers so far, for example, do not properly parse strings without an integer part (such as ".7") which is a float as far as python is concerned. And that's slightly tricky to check for in a single regex where the fractional portion is not required. I've included two regex to show this.
It does raise the interesting question as to what a "number" is. Do you include "inf" which is valid as a float in python? Or do you include numbers that are "numbers" but maybe can't be represented in python (such as numbers that are larger than the float max).
There's also ambiguities in how you parse numbers. For example, what about "--20"? Is this a "number"? Is this a legal way to represent "20"? Python will let you do "var = --20" and set it to 20 (though really this is because it treats it as an expression), but float("--20") does not work.
Anyways, without more info, here's a regex that I believe covers all the ints and floats as python parses them.
# Doesn't properly handle floats missing the integer part, such as ".7"
SIMPLE_FLOAT_REGEXP = re.compile(r'^[-+]?[0-9]+\.?[0-9]+([eE][-+]?[0-9]+)?$')
# Example "-12.34E+56" # sign (-)
# integer (12)
# mantissa (34)
# exponent (E+56)
# Should handle all floats
FLOAT_REGEXP = re.compile(r'^[-+]?([0-9]+|[0-9]*\.[0-9]+)([eE][-+]?[0-9]+)?$')
# Example "-12.34E+56" # sign (-)
# integer (12)
# OR
# int/mantissa (12.34)
# exponent (E+56)
def is_float(str):
return True if FLOAT_REGEXP.match(str) else False
Some example test values:
True <- +42
True <- +42.42
False <- +42.42.22
True <- +42.42e22
True <- +42.42E-22
False <- +42.42e-22.8
True <- .42
False <- 42nope
Running the benchmarking code in #ron-reiter's answer shows that this regex is actually faster than the normal regex and is much faster at handling bad values than the exception, which makes some sense. Results:
check_regexp with good floats: 18.001921
check_regexp with bad floats: 17.861423
check_regexp with strings: 17.558862
check_correct_regexp with good floats: 11.04428
check_correct_regexp with bad floats: 8.71211
check_correct_regexp with strings: 8.144161
check_replace with good floats: 6.020597
check_replace with bad floats: 5.343049
check_replace with strings: 5.091642
check_exception with good floats: 5.201605
check_exception with bad floats: 23.921864
check_exception with strings: 23.755481
I did some speed test. Lets say that if the string is likely to be a number the try/except strategy is the fastest possible.If the string is not likely to be a number and you are interested in Integer check, it worths to do some test (isdigit plus heading '-').
If you are interested to check float number, you have to use the try/except code whitout escape.
RyanN suggests
If you want to return False for a NaN and Inf, change line to x = float(s); return (x == x) and (x - 1 != x). This should return True for all floats except Inf and NaN
But this doesn't quite work, because for sufficiently large floats, x-1 == x returns true. For example, 2.0**54 - 1 == 2.0**54
I was working on a problem that led me to this thread, namely how to convert a collection of data to strings and numbers in the most intuitive way. I realized after reading the original code that what I needed was different in two ways:
1 - I wanted an integer result if the string represented an integer
2 - I wanted a number or a string result to stick into a data structure
so I adapted the original code to produce this derivative:
def string_or_number(s):
try:
z = int(s)
return z
except ValueError:
try:
z = float(s)
return z
except ValueError:
return s
import re
def is_number(num):
pattern = re.compile(r'^[-+]?[-0-9]\d*\.\d*|[-+]?\.?[0-9]\d*$')
result = pattern.match(num)
if result:
return True
else:
return False
​>>>: is_number('1')
True
>>>: is_number('111')
True
>>>: is_number('11.1')
True
>>>: is_number('-11.1')
True
>>>: is_number('inf')
False
>>>: is_number('-inf')
False
This code handles the exponents, floats, and integers, wihtout using regex.
return True if str1.lstrip('-').replace('.','',1).isdigit() or float(str1) else False
Here's my simple way of doing it. Let's say that I'm looping through some strings and I want to add them to an array if they turn out to be numbers.
try:
myvar.append( float(string_to_check) )
except:
continue
Replace the myvar.apppend with whatever operation you want to do with the string if it turns out to be a number. The idea is to try to use a float() operation and use the returned error to determine whether or not the string is a number.
I also used the function you mentioned, but soon I notice that strings as "Nan", "Inf" and it's variation are considered as number. So I propose you improved version of your function, that will return false on those type of input and will not fail "1e3" variants:
def is_float(text):
try:
float(text)
# check for nan/infinity etc.
if text.isalpha():
return False
return True
except ValueError:
return False
User helper function:
def if_ok(fn, string):
try:
return fn(string)
except Exception as e:
return None
then
if_ok(int, my_str) or if_ok(float, my_str) or if_ok(complex, my_str)
is_number = lambda s: any([if_ok(fn, s) for fn in (int, float, complex)])
def is_float(s):
if s is None:
return False
if len(s) == 0:
return False
digits_count = 0
dots_count = 0
signs_count = 0
for c in s:
if '0' <= c <= '9':
digits_count += 1
elif c == '.':
dots_count += 1
elif c == '-' or c == '+':
signs_count += 1
else:
return False
if digits_count == 0:
return False
if dots_count > 1:
return False
if signs_count > 1:
return False
return True

How do I parse a string to a float or int?

How can I convert a str to float?
"545.2222" → 545.2222
How can I convert a str to int?
"31" → 31
For the reverse, see Convert integer to string in Python and Converting a float to a string without rounding it.
Please instead use How can I read inputs as numbers? to close duplicate questions where OP received a string from user input and immediately wants to convert it, or was hoping for input (in 3.x) to convert the type automatically.
>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545
Python2 method to check if a string is a float:
def is_float(value):
if value is None:
return False
try:
float(value)
return True
except:
return False
For the Python3 version of is_float see: Checking if a string can be converted to float in Python
A longer and more accurate name for this function could be: is_convertible_to_float(value)
What is, and is not a float in Python may surprise you:
The below unit tests were done using python2. Check it that Python3 has different behavior for what strings are convertable to float. One confounding difference is that any number of interior underscores are now allowed: (float("1_3.4") == float(13.4)) is True
val is_float(val) Note
-------------------- ---------- --------------------------------
"" False Blank string
"127" True Passed string
True True Pure sweet Truth
"True" False Vile contemptible lie
False True So false it becomes true
"123.456" True Decimal
" -127 " True Spaces trimmed
"\t\n12\r\n" True whitespace ignored
"NaN" True Not a number
"NaNanananaBATMAN" False I am Batman
"-iNF" True Negative infinity
"123.E4" True Exponential notation
".1" True mantissa only
"1_2_3.4" False Underscores not allowed
"12 34" False Spaces not allowed on interior
"1,234" False Commas gtfo
u'\x30' True Unicode is fine.
"NULL" False Null is not special
0x3fade True Hexadecimal
"6e7777777777777" True Shrunk to infinity
"1.797693e+308" True This is max value
"infinity" True Same as inf
"infinityandBEYOND" False Extra characters wreck it
"12.34.56" False Only one dot allowed
u'四' False Japanese '4' is not a float.
"#56" False Pound sign
"56%" False Percent of what?
"0E0" True Exponential, move dot 0 places
0**0 True 0___0 Exponentiation
"-5e-5" True Raise to a negative number
"+1e1" True Plus is OK with exponent
"+1e1^5" False Fancy exponent not interpreted
"+1e1.3" False No decimals in exponent
"-+1" False Make up your mind
"(1)" False Parenthesis is bad
You think you know what numbers are? You are not so good as you think! Not big surprise.
Don't use this code on life-critical software!
Catching broad exceptions this way, killing canaries and gobbling the exception creates a tiny chance that a valid float as string will return false. The float(...) line of code can failed for any of a thousand reasons that have nothing to do with the contents of the string. But if you're writing life-critical software in a duck-typing prototype language like Python, then you've got much larger problems.
def num(s):
try:
return int(s)
except ValueError:
return float(s)
This is another method which deserves to be mentioned here, ast.literal_eval:
This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself.
That is, a safe 'eval'
>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31
Localization and commas
You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.
Example 1 -- United States number conventions
In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:
>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>
Example 2 -- European number conventions
In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:
>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222
The method locale.atoi is also available, but the argument should be an integer.
float(x) if '.' in x else int(x)
If you aren't averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation:
>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int
Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int("545") without first casting to float.
If your strings are in a list, you could use the map function as well.
>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>
It is only good if they're all the same type.
In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31?
I just want to know how to parse a float string to a float, and (separately) an int string to an int.
It's good that you ask to do these separately. If you're mixing them, you may be setting yourself up for problems later. The simple answer is:
"545.2222" to float:
>>> float("545.2222")
545.2222
"31" to an integer:
>>> int("31")
31
Other conversions, ints to and from strings and literals:
Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:
>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31
If you don't know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:
>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31
Non-Decimal (i.e. Integer) Literals from other Bases
If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases - you can let Python do it for you automatically with the correct syntax.
You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:
Binary, prefix 0b
>>> 0b11111
31
Octal, prefix 0o
>>> 0o37
31
Hexadecimal, prefix 0x
>>> 0x1f
31
This can be useful when describing binary flags, file permissions in code, or hex values for colors - for example, note no quotes:
>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215
Making ambiguous Python 2 octals compatible with Python 3
If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.
>>> 037
31
It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:
>>> 037
File "<stdin>", line 1
037
^
SyntaxError: invalid token
Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:
>>> 0o37
31
The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:
>>> import string
>>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
... int(x) or x.isalnum() and x or \
... len(set(string.punctuation).intersection(x)) == 1 and \
... x.count('.') == 1 and float(x) or x
>>> parseStr('123')
123
>>> parseStr('123.3')
123.3
>>> parseStr('3HC1')
'3HC1'
>>> parseStr('12.e5')
1200000.0
>>> parseStr('12$5')
'12$5'
>>> parseStr('12.2.2')
'12.2.2'
float("545.2222") and int(float("545.2222"))
The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type:
>>> import yaml
>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>
>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>
>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>
I use this function for that
import ast
def parse_str(s):
try:
return ast.literal_eval(str(s))
except:
return
It will convert the string to its type
value = parse_str('1') # Returns Integer
value = parse_str('1.5') # Returns Float
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else number_as_float
You could use json.loads:
>>> import json
>>> json.loads('123.456')
123.456
>>> type(_)
<class 'float'>
>>>
As you can see it becomes a type of float.
def num(s):
"""num(s)
num(3),num(3.7)-->3
num('3')-->3, num('3.7')-->3.7
num('3,700')-->ValueError
num('3a'),num('a3'),-->ValueError
num('3e4') --> 30000.0
"""
try:
return int(s)
except ValueError:
try:
return float(s)
except ValueError:
raise ValueError('argument is not a string of number')
You need to take into account rounding to do this properly.
i.e. - int(5.1) => 5
int(5.6) => 5 -- wrong, should be 6 so we do int(5.6 + 0.5) => 6
def convert(n):
try:
return int(n)
except ValueError:
return float(n + 0.5)
To typecast in Python use the constructor functions of the type, passing the string (or whatever value you are trying to cast) as a parameter.
For example:
>>>float("23.333")
23.333
Behind the scenes, Python is calling the objects __float__ method, which should return a float representation of the parameter. This is especially powerful, as you can define your own types (using classes) with a __float__ method so that it can be casted into a float using float(myobject).
Handles hex, octal, binary, decimal, and float
This solution will handle all of the string conventions for numbers (all that I know about).
def to_number(n):
''' Convert any number representation to a number
This covers: float, decimal, hex, and octal numbers.
'''
try:
return int(str(n), 0)
except:
try:
# Python 3 doesn't accept "010" as a valid octal. You must use the
# '0o' prefix
return int('0o' + n, 0)
except:
return float(n)
This test case output illustrates what I'm talking about.
======================== CAPTURED OUTPUT =========================
to_number(3735928559) = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0") = 0 == 0
to_number(100) = 100 == 100
to_number("42") = 42 == 42
to_number(8) = 8 == 8
to_number("0o20") = 16 == 16
to_number("020") = 16 == 16
to_number(3.14) = 3.14 == 3.14
to_number("2.72") = 2.72 == 2.72
to_number("1e3") = 1000.0 == 1000
to_number(0.001) = 0.001 == 0.001
to_number("0xA") = 10 == 10
to_number("012") = 10 == 10
to_number("0o12") = 10 == 10
to_number("0b01010") = 10 == 10
to_number("10") = 10 == 10
to_number("10.0") = 10.0 == 10
to_number("1e1") = 10.0 == 10
Here is the test:
class test_to_number(unittest.TestCase):
def test_hex(self):
# All of the following should be converted to an integer
#
values = [
# HEX
# ----------------------
# Input | Expected
# ----------------------
(0xDEADBEEF , 3735928559), # Hex
("0xFEEDFACE", 4277009102), # Hex
("0x0" , 0), # Hex
# Decimals
# ----------------------
# Input | Expected
# ----------------------
(100 , 100), # Decimal
("42" , 42), # Decimal
]
values += [
# Octals
# ----------------------
# Input | Expected
# ----------------------
(0o10 , 8), # Octal
("0o20" , 16), # Octal
("020" , 16), # Octal
]
values += [
# Floats
# ----------------------
# Input | Expected
# ----------------------
(3.14 , 3.14), # Float
("2.72" , 2.72), # Float
("1e3" , 1000), # Float
(1e-3 , 0.001), # Float
]
values += [
# All ints
# ----------------------
# Input | Expected
# ----------------------
("0xA" , 10),
("012" , 10),
("0o12" , 10),
("0b01010" , 10),
("10" , 10),
("10.0" , 10),
("1e1" , 10),
]
for _input, expected in values:
value = to_number(_input)
if isinstance(_input, str):
cmd = 'to_number("{}")'.format(_input)
else:
cmd = 'to_number({})'.format(_input)
print("{:23} = {:10} == {:10}".format(cmd, value, expected))
self.assertEqual(value, expected)
Pass your string to this function:
def string_to_number(str):
if("." in str):
try:
res = float(str)
except:
res = str
elif(str.isdigit()):
res = int(str)
else:
res = str
return(res)
It will return int, float or string depending on what was passed.
String that is an int
print(type(string_to_number("124")))
<class 'int'>
String that is a float
print(type(string_to_number("12.4")))
<class 'float'>
String that is a string
print(type(string_to_number("hello")))
<class 'str'>
String that looks like a float
print(type(string_to_number("hel.lo")))
<class 'str'>
There is also regex, because sometimes string must be prepared and normalized before casting to a number:
import re
def parseNumber(value, as_int=False):
try:
number = float(re.sub('[^.\-\d]', '', value))
if as_int:
return int(number + 0.5)
else:
return number
except ValueError:
return float('nan') # or None if you wish
Usage:
parseNumber('13,345')
> 13345.0
parseNumber('- 123 000')
> -123000.0
parseNumber('99999\n')
> 99999.0
And by the way, something to verify you have a number:
import numbers
def is_number(value):
return isinstance(value, numbers.Number)
# Will work with int, float, long, Decimal
a = int(float(a)) if int(float(a)) == float(a) else float(a)
This is a corrected version of Totoro's answer.
This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else
number_as_float
If you are dealing with mixed integers and floats and want a consistent way to deal with your mixed data, here is my solution with the proper docstring:
def parse_num(candidate):
"""Parse string to number if possible
It work equally well with negative and positive numbers, integers and floats.
Args:
candidate (str): string to convert
Returns:
float | int | None: float or int if possible otherwise None
"""
try:
float_value = float(candidate)
except ValueError:
return None
# Optional part if you prefer int to float when decimal part is 0
if float_value.is_integer():
return int(float_value)
# end of the optional part
return float_value
# Test
candidates = ['34.77', '-13', 'jh', '8990', '76_3234_54']
res_list = list(map(parse_num, candidates))
print('Before:')
print(candidates)
print('After:')
print(res_list)
Output:
Before:
['34.77', '-13', 'jh', '8990', '76_3234_54']
After:
[34.77, -13, None, 8990, 76323454]
Use:
def num(s):
try:
for each in s:
yield int(each)
except ValueError:
yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()
This is the most Pythonic way I could come up with.
If you don't want to use third party modules the following might be the most robust solution:
def string_to_int_or_float(s):
try:
f = float(s) # replace s with str(s) if you are not sure that s is a string
except ValueError:
print("Provided string '" + s + "' is not interpretable as a literal number.")
raise
try:
i = int(str(f).rstrip('0').rstrip('.'))
except:
return f
return i
It might not be the fastest, but it handles correctly literal numbers where many other solutions fail, such as:
>>> string_to_int_or_float('789.')
789
>>> string_to_int_or_float('789.0')
789
>>> string_to_int_or_float('12.3e2')
1230
>>> string_to_int_or_float('12.3e-2')
0.123
>>> string_to_int_or_float('4560e-1')
456
>>> string_to_int_or_float('4560e-2')
45.6
You can simply do this by
s = '542.22'
f = float(s) # This converts string data to float data with a decimal point
print(f)
i = int(f) # This converts string data to integer data by just taking the whole number part of it
print(i)
For more information on parsing of data types check on python documentation!
This is a function which will convert any object (not just str) to int or float, based on if the actual string supplied looks like int or float. Further if it's an object which has both __float and __int__ methods, it defaults to using __float__
def conv_to_num(x, num_type='asis'):
'''Converts an object to a number if possible.
num_type: int, float, 'asis'
Defaults to floating point in case of ambiguity.
'''
import numbers
is_num, is_str, is_other = [False]*3
if isinstance(x, numbers.Number):
is_num = True
elif isinstance(x, str):
is_str = True
is_other = not any([is_num, is_str])
if is_num:
res = x
elif is_str:
is_float, is_int, is_char = [False]*3
try:
res = float(x)
if '.' in x:
is_float = True
else:
is_int = True
except ValueError:
res = x
is_char = True
else:
if num_type == 'asis':
funcs = [int, float]
else:
funcs = [num_type]
for func in funcs:
try:
res = func(x)
break
except TypeError:
continue
else:
res = x
By using int and float methods we can convert a string to integer and floats.
s="45.8"
print(float(s))
y='67'
print(int(y))
For numbers and characters together:
string_for_int = "498 results should get"
string_for_float = "498.45645765 results should get"
First import re:
import re
# For getting the integer part:
print(int(re.search(r'\d+', string_for_int).group())) #498
# For getting the float part:
print(float(re.search(r'\d+\.\d+', string_for_float).group())) #498.45645765
For easy model:
value1 = "10"
value2 = "10.2"
print(int(value1)) # 10
print(float(value2)) # 10.2

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