I want a to be rounded to 13.95. I tried using round, but I get:
>>> a
13.949999999999999
>>> round(a, 2)
13.949999999999999
For the analogous issue with the standard library Decimal class, see How can I format a decimal to always show 2 decimal places?.
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True
The built-in round() works just fine in Python 2.7 or later.
Example:
>>> round(14.22222223, 2)
14.22
Check out the documentation.
Let me give an example in Python 3.6's f-string/template-string format, which I think is beautifully neat:
>>> f'{a:.2f}'
It works well with longer examples too, with operators and not needing parentheses:
>>> print(f'Completed in {time.time() - start:.2f}s')
I feel that the simplest approach is to use the format() function.
For example:
a = 13.949999999999999
format(a, '.2f')
13.95
This produces a float number as a string rounded to two decimal points.
Most numbers cannot be exactly represented in floats. If you want to round the number because that's what your mathematical formula or algorithm requires, then you want to use round. If you just want to restrict the display to a certain precision, then don't even use round and just format it as that string. (If you want to display it with some alternate rounding method, and there are tons, then you need to mix the two approaches.)
>>> "%.2f" % 3.14159
'3.14'
>>> "%.2f" % 13.9499999
'13.95'
And lastly, though perhaps most importantly, if you want exact math then you don't want floats at all. The usual example is dealing with money and to store 'cents' as an integer.
Use
print"{:.2f}".format(a)
instead of
print"{0:.2f}".format(a)
Because the latter may lead to output errors when trying to output multiple variables (see comments).
Try the code below:
>>> a = 0.99334
>>> a = int((a * 100) + 0.5) / 100.0 # Adding 0.5 rounds it up
>>> print a
0.99
TLDR ;)
The rounding problem of input and output has been solved definitively by Python 3.1 and the fix is backported also to Python 2.7.0.
Rounded numbers can be reversibly converted between float and string back and forth:
str -> float() -> repr() -> float() ... or Decimal -> float -> str -> Decimal
>>> 0.3
0.3
>>> float(repr(0.3)) == 0.3
True
A Decimal type is not necessary for storage anymore.
Results of arithmetic operations must be rounded again because rounding errors could accumulate more inaccuracy than that is possible after parsing one number. That is not fixed by the improved repr() algorithm (Python >= 3.1, >= 2.7.0):
>>> 0.1 + 0.2
0.30000000000000004
>>> 0.1, 0.2, 0.3
(0.1, 0.2, 0.3)
The output string function str(float(...)) was rounded to 12 valid digits in Python < 2.7x and < 3.1, to prevent excessive invalid digits similar to unfixed repr() output. That was still insufficientl after subtraction of very similar numbers and it was too much rounded after other operations. Python 2.7 and 3.1 use the same length of str() although the repr() is fixed. Some old versions of Numpy had also excessive invalid digits, even with fixed Python. The current Numpy is fixed. Python versions >= 3.2 have the same results of str() and repr() function and also output of similar functions in Numpy.
Test
import random
from decimal import Decimal
for _ in range(1000000):
x = random.random()
assert x == float(repr(x)) == float(Decimal(repr(x))) # Reversible repr()
assert str(x) == repr(x)
assert len(repr(round(x, 12))) <= 14 # no excessive decimal places.
Documentation
See the Release notes Python 2.7 - Other Language Changes the fourth paragraph:
Conversions between floating-point numbers and strings are now correctly rounded on most platforms. These conversions occur in many different places: str() on floats and complex numbers; the float and complex constructors; numeric formatting; serializing and de-serializing floats and complex numbers using the marshal, pickle and json modules; parsing of float and imaginary literals in Python code; and Decimal-to-float conversion.
Related to this, the repr() of a floating-point number x now returns a result based on the shortest decimal string that’s guaranteed to round back to x under correct rounding (with round-half-to-even rounding mode). Previously it gave a string based on rounding x to 17 decimal digits.
The related issue
More information: The formatting of float before Python 2.7 was similar to the current numpy.float64. Both types use the same 64 bit IEEE 754 double precision with 52 bit mantissa. A big difference is that np.float64.__repr__ is formatted frequently with an excessive decimal number so that no bit can be lost, but no valid IEEE 754 number exists between 13.949999999999999 and 13.950000000000001. The result is not nice and the conversion repr(float(number_as_string)) is not reversible with numpy. On the other hand: float.__repr__ is formatted so that every digit is important; the sequence is without gaps and the conversion is reversible. Simply: If you perhaps have a numpy.float64 number, convert it to normal float in order to be formatted for humans, not for numeric processors, otherwise nothing more is necessary with Python 2.7+.
Use:
float_number = 12.234325335563
round(float_number, 2)
This will return;
12.23
Explanation:
The round function takes two arguments;
The number to be rounded and the number of decimal places to be returned. Here I returned two decimal places.
You can modify the output format:
>>> a = 13.95
>>> a
13.949999999999999
>>> print "%.2f" % a
13.95
With Python < 3 (e.g. 2.6 or 2.7), there are two ways to do so.
# Option one
older_method_string = "%.9f" % numvar
# Option two (note ':' before the '.9f')
newer_method_string = "{:.9f}".format(numvar)
But note that for Python versions above 3 (e.g. 3.2 or 3.3), option two is preferred.
For more information on option two, I suggest this link on string formatting from the Python documentation.
And for more information on option one, this link will suffice and has information on the various flags.
Reference: Convert floating point number to a certain precision, and then copy to string
You can use format operator for rounding the value up to two decimal places in Python:
print(format(14.4499923, '.2f')) // The output is 14.45
As Matt pointed out, Python 3.6 provides f-strings, and they can also use nested parameters:
value = 2.34558
precision = 2
width = 4
print(f'result: {value:{width}.{precision}f}')
which will display result: 2.35
In Python 2.7:
a = 13.949999999999999
output = float("%0.2f"%a)
print output
We multiple options to do that:
Option 1:
x = 1.090675765757
g = float("{:.2f}".format(x))
print(g)
Option 2:
The built-in round() supports Python 2.7 or later.
x = 1.090675765757
g = round(x, 2)
print(g)
The Python tutorial has an appendix called Floating Point Arithmetic: Issues and Limitations. Read it. It explains what is happening and why Python is doing its best. It has even an example that matches yours. Let me quote a bit:
>>> 0.1
0.10000000000000001
you may be tempted to use the round()
function to chop it back to the single
digit you expect. But that makes no
difference:
>>> round(0.1, 1)
0.10000000000000001
The problem is that the binary
floating-point value stored for “0.1”
was already the best possible binary
approximation to 1/10, so trying to
round it again can’t make it better:
it was already as good as it gets.
Another consequence is that since 0.1
is not exactly 1/10, summing ten
values of 0.1 may not yield exactly
1.0, either:
>>> sum = 0.0
>>> for i in range(10):
... sum += 0.1
...
>>> sum
0.99999999999999989
One alternative and solution to your problems would be using the decimal module.
Use combination of Decimal object and round() method.
Python 3.7.3
>>> from decimal import Decimal
>>> d1 = Decimal (13.949999999999999) # define a Decimal
>>> d1
Decimal('13.949999999999999289457264239899814128875732421875')
>>> d2 = round(d1, 2) # round to 2 decimals
>>> d2
Decimal('13.95')
It's doing exactly what you told it to do and is working correctly. Read more about floating point confusion and maybe try decimal objects instead.
from decimal import Decimal
def round_float(v, ndigits=2, rt_str=False):
d = Decimal(v)
v_str = ("{0:.%sf}" % ndigits).format(round(d, ndigits))
if rt_str:
return v_str
return Decimal(v_str)
Results:
Python 3.6.1 (default, Dec 11 2018, 17:41:10)
>>> round_float(3.1415926)
Decimal('3.14')
>>> round_float(3.1445926)
Decimal('3.14')
>>> round_float(3.1455926)
Decimal('3.15')
>>> round_float(3.1455926, rt_str=True)
'3.15'
>>> str(round_float(3.1455926))
'3.15'
The simple solution is here
value = 5.34343
rounded_value = round(value, 2) # 5.34
Use a lambda function like this:
arred = lambda x,n : x*(10**n)//1/(10**n)
This way you could just do:
arred(3.141591657, 2)
and get
3.14
orig_float = 232569 / 16000.0
14.5355625
short_float = float("{:.2f}".format(orig_float))
14.54
For fixing the floating point in type-dynamic languages such as Python and JavaScript, I use this technique
# For example:
a = 70000
b = 0.14
c = a * b
print c # Prints 980.0000000002
# Try to fix
c = int(c * 10000)/100000
print c # Prints 980
You can also use Decimal as following:
from decimal import *
getcontext().prec = 6
Decimal(1) / Decimal(7)
# Results in 6 precision -> Decimal('0.142857')
getcontext().prec = 28
Decimal(1) / Decimal(7)
# Results in 28 precision -> Decimal('0.1428571428571428571428571429')
It's simple like:
use decimal module for fast correctly-rounded decimal floating point arithmetic:
d = Decimal(10000000.0000009)
to achieve rounding:
d.quantize(Decimal('0.01'))
will result with Decimal('10000000.00')
make the above DRY:
def round_decimal(number, exponent='0.01'):
decimal_value = Decimal(number)
return decimal_value.quantize(Decimal(exponent))
or
def round_decimal(number, decimal_places=2):
decimal_value = Decimal(number)
return decimal_value.quantize(Decimal(10) ** -decimal_places)
PS: critique of others: formatting is not rounding.
Here is the simple solution using the format function.
float(format(num, '.2f'))
Note: We are converting numbers to float, because the format method is returning a string.
If you want to handle money, use the Python decimal module:
from decimal import Decimal, ROUND_HALF_UP
# 'amount' can be integer, string, tuple, float, or another Decimal object
def to_money(amount) -> Decimal:
money = Decimal(amount).quantize(Decimal('.00'), rounding=ROUND_HALF_UP)
return money
lambda x, n:int(x*10^n + 0.5)/10^n
has worked for me for many years in many languages.
To round a number to a resolution, the best way is the following one, which can work with any resolution (0.01 for two decimals or even other steps):
>>> import numpy as np
>>> value = 13.949999999999999
>>> resolution = 0.01
>>> newValue = int(np.round(value/resolution))*resolution
>>> print newValue
13.95
>>> resolution = 0.5
>>> newValue = int(np.round(value/resolution))*resolution
>>> print newValue
14.0
The answers I saw didn't work with the float(52.15) case. After some tests, there is the solution that I'm using:
import decimal
def value_to_decimal(value, decimal_places):
decimal.getcontext().rounding = decimal.ROUND_HALF_UP # define rounding method
return decimal.Decimal(str(float(value))).quantize(decimal.Decimal('1e-{}'.format(decimal_places)))
(The conversion of the 'value' to float and then string is very important, that way, 'value' can be of the type float, decimal, integer or string!)
Hope this helps anyone.
I have some number 0.0000002345E^-60. I want to print the floating point value as it is.
What is the way to do it?
print %f truncates it to 6 digits. Also %n.nf gives fixed numbers. What is the way to print without truncation.
Like this?
>>> print('{:.100f}'.format(0.0000002345E-60))
0.0000000000000000000000000000000000000000000000000000000000000000002344999999999999860343602938602754
As you might notice from the output, it’s not really that clear how you want to do it. Due to the float representation you lose precision and can’t really represent the number precisely. As such it’s not really clear where you want the number to stop displaying.
Also note that the exponential representation is often used to more explicitly show the number of significant digits the number has.
You could also use decimal to not lose the precision due to binary float truncation:
>>> from decimal import Decimal
>>> d = Decimal('0.0000002345E-60')
>>> p = abs(d.as_tuple().exponent)
>>> print(('{:.%df}' % p).format(d))
0.0000000000000000000000000000000000000000000000000000000000000000002345
You can use decimal.Decimal:
>>> from decimal import Decimal
>>> str(Decimal(0.0000002345e-60))
'2.344999999999999860343602938602754401109865640550232148836753621775217856801120686600683401464097113374472942165409862789978024748827516129306833728589548440037314681709534891496105046826414763927459716796875E-67'
This is the actual value of float created by literal 0.0000002345e-60. Its value is a number representable as python float which is closest to actual 0.0000002345 * 10**-60.
float should be generally used for approximate calculations. If you want accurate results you should use something else, like mentioned Decimal.
If I understand, you want to print a float?
The problem is, you cannot print a float.
You can only print a string representation of a float. So, in short, you cannot print a float, that is your answer.
If you accept that you need to print a string representation of a float, and your question is how specify your preferred format for the string representations of your floats, then judging by the comments you have been very unclear in your question.
If you would like to print the string representations of your floats in exponent notation, then the format specification language allows this:
{:g} or {:G}, depending whether or not you want the E in the output to be capitalized). This gets around the default precision for e and E types, which leads to unwanted trailing 0s in the part before the exponent symbol.
Assuming your value is my_float, "{:G}".format(my_float) would print the output the way that the Python interpreter prints it. You could probably just print the number without any formatting and get the same exact result.
If your goal is to print the string representation of the float with its current precision, in non-exponentiated form, User poke describes a good way to do this by casting the float to a Decimal object.
If, for some reason, you do not want to do this, you can do something like is mentioned in this answer. However, you should set 'max_digits' to sys.float_info.max_10_exp, instead of 14 used in the answer. This requires you to import sys at some point prior in the code.
A full example of this would be:
import math
import sys
def precision_and_scale(x):
max_digits = sys.float_info.max_10_exp
int_part = int(abs(x))
magnitude = 1 if int_part == 0 else int(math.log10(int_part)) + 1
if magnitude >= max_digits:
return (magnitude, 0)
frac_part = abs(x) - int_part
multiplier = 10 ** (max_digits - magnitude)
frac_digits = multiplier + int(multiplier * frac_part + 0.5)
while frac_digits % 10 == 0:
frac_digits /= 10
scale = int(math.log10(frac_digits))
return (magnitude + scale, scale)
f = 0.0000002345E^-60
p, s = precision_and_scale(f)
print "{:.{p}f}".format(f, p=p)
But I think the method involving casting to Decimal is probably better, overall.
How to correctly add or subtract using floats?
For example how to perform:
2.4e-07 - 1e-8
so that it returns 2.3e-7 instead of 2.2999999999999997e-07.
Converting to int first yields unexpected results, the below returns 2.2e-07:
int(2.4e-07 * 1e8 - 1) * 1e-8
Similarly,
(2.4e-07 * 1e8 - 1) * 1e-8
returns 2.2999999999999997e-07.
How to perform subtraction and addition of numbers with 8 decimal point precision?
2.2999999999999997e-07 is not sufficient as the number is used as a lookup in a dictionary, and the key is 2.3e-7. This means that any value other than 2.3e-7 results in an incorrect lookup.
I suggest using the decimal data type (it is present in the stardard installation of Python), because it uses fixed precision to avoid just the differences you are talking about.
>>> from decimal import Decimal
>>> x = Decimal('2.4e-7')
>>> x
Decimal('2.4E-7')
>>> y = Decimal('1e-8')
>>> y
Decimal('1E-8')
>>> x - y
Decimal('2.3E-7')
It's really just a way of skirting around the issue of floating point arithmetic, but I suggest using the decimal package from the standard library. It lets you do exact floating point math.
Using your example,
$ from decimal import Decimal
$ x = Decimal('2.4e-7')
$ y = Decimal('1e-8')
$ x-y
Decimal('2.3E-7')
It's worth noting that Decimal objects are different than the float built-in, but they are mostly interchangeable.
I do not know if it is what you are looking for but you can try that kind of thing:
a = 0.555555555
a = float("{0:.2f}".format(a))
>>> 0.56
I hope it will help you!
Adrien
I'm working on the math program and I have a quite big problem with round. So after my program did some math, it rounds the result.
Everything works fine but if the result == 2.49999999999999992 , round function return 3.0 instead of 2.0.
How can I fix that?
Thanks.
As #Pavel Anossov says in his comment there's no such thing as 2.49999999999999992 in IEEE 754, 2.49999999999999992 == 2.5.. Float might always be critical for your calculations, because in any case (32/64/128 bit float), you have a precision limit. This is obviously also limited for Python floats.
There are different options to deal with that, you could e.g. use the decimal library:
>>> from decimal import *
>>> getcontext().prec = 6
>>> Decimal(1) / Decimal(7)
Decimal('0.142857')
>>> getcontext().prec = 28
>>> Decimal(1) / Decimal(7)
Decimal('0.1428571428571428571428571429')
It's possible to set the precision yourself in that case. decimal is in the standard library.
There are also third party libraries like bigfloat, that you could use (I have no experience with it):
>>> from bigfloat import *
>>> sqrt(2, precision(100)) # compute sqrt(2) with 100 bits of precision
But as you can see, you always have to choose a precision. If you really don't want to lose any kind of precision, use fractions (also in the standard library):
>>> from fractions import Fraction
>>> a = Fraction(16, -10)
>>> a
Fraction(-8, 5)
>>> a / 23
Fraction(-8, 115)
>>> float(a/23)
-0.06956521739130435
The reason is that Python's float type (typically IEEE 754 double precision floating point numbers) does not have such a value as 2.49999999999999992. Floating point numbers are generally on the form mantissa*base**exponent, and in Python you can find the limits for float in particular within sys.float_info. For starters, let's calculate how many digits the mantissa itself can hold:
>>> from sys import float_info
>>> print float_info.radix**float_info.mant_dig # How big can the mantissa get?
9007199254740992
>>> print "2.49999999999999992"
2.49999999999999992
>>> 2.49999999999999992
2.5
Clearly the number we've entered is wider. Just how close can we go before things go wrong?
>>> print 2.5*float_info.epsilon
5.55111512313e-16
e-16 here means *10**-16, so let's reformat that for comparison:
>>> print "%.17f"%(2.5*float_info.epsilon); print "2.49999999999999992"
0.00000000000000056
2.49999999999999992
This indicates that at a magnitude around 2.5, differences lower than about 5.6e-16 (including this 8e-17) will be lost to the storage itself. Therefore this value is 2.5, which rounds up.
We can also calculate an estimate of how many significant digits we can use:
>>> import math, sys
>>> print math.log10(sys.float_info.radix**sys.float_info.mant_dig)
15.9545897702
Very nearly, but not quite, 16. In binary the first digit will always be 1, so we can have a known number of significant digits (mant_dig), but in decimal the first digit will consume between one and four bits. This means the last digit may be off by more than one. Usually we hide this by printing only with a limited precision, but it actually occurs to lots of numbers:
>>> print '%f = %.17f'%(1.1, 1.1)
1.100000 = 1.10000000000000009
>>> print '%f = %.17f'%(10.1, 10.1)
10.100000 = 10.09999999999999964
Such is the inherent imprecision of floating point numbers. Types like bigfloat, decimal and fractions (thanks to David Halter for these examples) can push the limits around, but if you start looking at many digits you need to be aware of them. Also note that this is not unique to computers; an irrational number, such as pi or sqrt(2), cannot be exactly written in any integer base.