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I want to remove digits from a float to have a fixed number of digits after the dot, like:
1.923328437452 → 1.923
I need to output as a string to another function, not print.
Also I want to ignore the lost digits, not round them.
round(1.923328437452, 3)
See Python's documentation on the standard types. You'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.
First, the function, for those who just want some copy-and-paste code:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
This is valid in Python 2.7 and 3.1+. For older versions, it's not possible to get the same "intelligent rounding" effect (at least, not without a lot of complicated code), but rounding to 12 decimal places before truncation will work much of the time:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '%.12f' % f
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
Explanation
The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters. The latter step is easy; it can be done either with string manipulation
i, p, d = s.partition('.')
'.'.join([i, (d+'0'*n)[:n]])
or the decimal module
str(Decimal(s).quantize(Decimal((0, (1,), -n)), rounding=ROUND_DOWN))
The first step, converting to a string, is quite difficult because there are some pairs of floating point literals (i.e. what you write in the source code) which both produce the same binary representation and yet should be truncated differently. For example, consider 0.3 and 0.29999999999999998. If you write 0.3 in a Python program, the compiler encodes it using the IEEE floating-point format into the sequence of bits (assuming a 64-bit float)
0011111111010011001100110011001100110011001100110011001100110011
This is the closest value to 0.3 that can accurately be represented as an IEEE float. But if you write 0.29999999999999998 in a Python program, the compiler translates it into exactly the same value. In one case, you meant it to be truncated (to one digit) as 0.3, whereas in the other case you meant it to be truncated as 0.2, but Python can only give one answer. This is a fundamental limitation of Python, or indeed any programming language without lazy evaluation. The truncation function only has access to the binary value stored in the computer's memory, not the string you actually typed into the source code.1
If you decode the sequence of bits back into a decimal number, again using the IEEE 64-bit floating-point format, you get
0.2999999999999999888977697537484345957637...
so a naive implementation would come up with 0.2 even though that's probably not what you want. For more on floating-point representation error, see the Python tutorial.
It's very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number. So when truncating, it probably makes sense to choose the "nicest" decimal representation out of all that could correspond to the value in memory. Python 2.7 and up (but not 3.0) includes a sophisticated algorithm to do just that, which we can access through the default string formatting operation.
'{}'.format(f)
The only caveat is that this acts like a g format specification, in the sense that it uses exponential notation (1.23e+4) if the number is large or small enough. So the method has to catch this case and handle it differently. There are a few cases where using an f format specification instead causes a problem, such as trying to truncate 3e-10 to 28 digits of precision (it produces 0.0000000002999999999999999980), and I'm not yet sure how best to handle those.
If you actually are working with floats that are very close to round numbers but intentionally not equal to them (like 0.29999999999999998 or 99.959999999999994), this will produce some false positives, i.e. it'll round numbers that you didn't want rounded. In that case the solution is to specify a fixed precision.
'{0:.{1}f}'.format(f, sys.float_info.dig + n + 2)
The number of digits of precision to use here doesn't really matter, it only needs to be large enough to ensure that any rounding performed in the string conversion doesn't "bump up" the value to its nice decimal representation. I think sys.float_info.dig + n + 2 may be enough in all cases, but if not that 2 might have to be increased, and it doesn't hurt to do so.
In earlier versions of Python (up to 2.6, or 3.0), the floating point number formatting was a lot more crude, and would regularly produce things like
>>> 1.1
1.1000000000000001
If this is your situation, if you do want to use "nice" decimal representations for truncation, all you can do (as far as I know) is pick some number of digits, less than the full precision representable by a float, and round the number to that many digits before truncating it. A typical choice is 12,
'%.12f' % f
but you can adjust this to suit the numbers you're using.
1Well... I lied. Technically, you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function. If that argument is a floating-point literal, you can just cut it off a certain number of places after the decimal point and return that. However this strategy doesn't work if the argument is a variable, which makes it fairly useless. The following is presented for entertainment value only:
def trunc_introspect(f, n):
'''Truncates/pads the float f to n decimal places by looking at the caller's source code'''
current_frame = None
caller_frame = None
s = inspect.stack()
try:
current_frame = s[0]
caller_frame = s[1]
gen = tokenize.tokenize(io.BytesIO(caller_frame[4][caller_frame[5]].encode('utf-8')).readline)
for token_type, token_string, _, _, _ in gen:
if token_type == tokenize.NAME and token_string == current_frame[3]:
next(gen) # left parenthesis
token_type, token_string, _, _, _ = next(gen) # float literal
if token_type == tokenize.NUMBER:
try:
cut_point = token_string.index('.') + n + 1
except ValueError: # no decimal in string
return token_string + '.' + '0' * n
else:
if len(token_string) < cut_point:
token_string += '0' * (cut_point - len(token_string))
return token_string[:cut_point]
else:
raise ValueError('Unable to find floating-point literal (this probably means you called {} with a variable)'.format(current_frame[3]))
break
finally:
del s, current_frame, caller_frame
Generalizing this to handle the case where you pass in a variable seems like a lost cause, since you'd have to trace backwards through the program's execution until you find the floating-point literal which gave the variable its value. If there even is one. Most variables will be initialized from user input or mathematical expressions, in which case the binary representation is all there is.
The result of round is a float, so watch out (example is from Python 2.6):
>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001
You will be better off when using a formatted string:
>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'
n = 1.923328437452
str(n)[:4]
At my Python 2.7 prompt:
>>> int(1.923328437452 * 1000)/1000.0
1.923
The truely pythonic way of doing it is
from decimal import *
with localcontext() as ctx:
ctx.rounding = ROUND_DOWN
print Decimal('1.923328437452').quantize(Decimal('0.001'))
or shorter:
from decimal import Decimal as D, ROUND_DOWN
D('1.923328437452').quantize(D('0.001'), rounding=ROUND_DOWN)
Update
Usually the problem is not in truncating floats itself, but in the improper usage of float numbers before rounding.
For example: int(0.7*3*100)/100 == 2.09.
If you are forced to use floats (say, you're accelerating your code with numba), it's better to use cents as "internal representation" of prices: (70*3 == 210) and multiply/divide the inputs/outputs.
Simple python script -
n = 1.923328437452
n = float(int(n * 1000))
n /=1000
def trunc(num, digits):
sp = str(num).split('.')
return '.'.join([sp[0], sp[1][:digits]])
This should work. It should give you the truncation you are looking for.
So many of the answers given for this question are just completely wrong. They either round up floats (rather than truncate) or do not work for all cases.
This is the top Google result when I search for 'Python truncate float', a concept which is really straightforward, and which deserves better answers. I agree with Hatchkins that using the decimal module is the pythonic way of doing this, so I give here a function which I think answers the question correctly, and which works as expected for all cases.
As a side-note, fractional values, in general, cannot be represented exactly by binary floating point variables (see here for a discussion of this), which is why my function returns a string.
from decimal import Decimal, localcontext, ROUND_DOWN
def truncate(number, places):
if not isinstance(places, int):
raise ValueError("Decimal places must be an integer.")
if places < 1:
raise ValueError("Decimal places must be at least 1.")
# If you want to truncate to 0 decimal places, just do int(number).
with localcontext() as context:
context.rounding = ROUND_DOWN
exponent = Decimal(str(10 ** - places))
return Decimal(str(number)).quantize(exponent).to_eng_string()
>>> from math import floor
>>> floor((1.23658945) * 10**4) / 10**4
1.2365
# divide and multiply by 10**number of desired digits
If you fancy some mathemagic, this works for +ve numbers:
>>> v = 1.923328437452
>>> v - v % 1e-3
1.923
I did something like this:
from math import trunc
def truncate(number, decimals=0):
if decimals < 0:
raise ValueError('truncate received an invalid value of decimals ({})'.format(decimals))
elif decimals == 0:
return trunc(number)
else:
factor = float(10**decimals)
return trunc(number*factor)/factor
You can do:
def truncate(f, n):
return math.floor(f * 10 ** n) / 10 ** n
testing:
>>> f=1.923328437452
>>> [truncate(f, n) for n in range(5)]
[1.0, 1.9, 1.92, 1.923, 1.9233]
Just wanted to mention that the old "make round() with floor()" trick of
round(f) = floor(f+0.5)
can be turned around to make floor() from round()
floor(f) = round(f-0.5)
Although both these rules break around negative numbers, so using it is less than ideal:
def trunc(f, n):
if f > 0:
return "%.*f" % (n, (f - 0.5*10**-n))
elif f == 0:
return "%.*f" % (n, f)
elif f < 0:
return "%.*f" % (n, (f + 0.5*10**-n))
def precision(value, precision):
"""
param: value: takes a float
param: precision: int, number of decimal places
returns a float
"""
x = 10.0**precision
num = int(value * x)/ x
return num
precision(1.923328437452, 3)
1.923
Short and easy variant
def truncate_float(value, digits_after_point=2):
pow_10 = 10 ** digits_after_point
return (float(int(value * pow_10))) / pow_10
>>> truncate_float(1.14333, 2)
>>> 1.14
>>> truncate_float(1.14777, 2)
>>> 1.14
>>> truncate_float(1.14777, 4)
>>> 1.1477
When using a pandas df this worked for me
import math
def truncate(number, digits) -> float:
stepper = 10.0 ** digits
return math.trunc(stepper * number) / stepper
df['trunc'] = df['float_val'].apply(lambda x: truncate(x,1))
df['trunc']=df['trunc'].map('{:.1f}'.format)
int(16.5);
this will give an integer value of 16, i.e. trunc, won't be able to specify decimals, but guess you can do that by
import math;
def trunc(invalue, digits):
return int(invalue*math.pow(10,digits))/math.pow(10,digits);
Here is an easy way:
def truncate(num, res=3):
return (floor(num*pow(10, res)+0.5))/pow(10, res)
for num = 1.923328437452, this outputs 1.923
def trunc(f,n):
return ('%.16f' % f)[:(n-16)]
A general and simple function to use:
def truncate_float(number, length):
"""Truncate float numbers, up to the number specified
in length that must be an integer"""
number = number * pow(10, length)
number = int(number)
number = float(number)
number /= pow(10, length)
return number
There is an easy workaround in python 3. Where to cut I defined with an help variable decPlace to make it easy to adapt.
f = 1.12345
decPlace= 4
f_cut = int(f * 10**decPlace) /10**decPlace
Output:
f = 1.1234
Hope it helps.
Most answers are way too complicated in my opinion, how about this?
digits = 2 # Specify how many digits you want
fnum = '122.485221'
truncated_float = float(fnum[:fnum.find('.') + digits + 1])
>>> 122.48
Simply scanning for the index of '.' and truncate as desired (no rounding).
Convert string to float as final step.
Or in your case if you get a float as input and want a string as output:
fnum = str(122.485221) # convert float to string first
truncated_float = fnum[:fnum.find('.') + digits + 1] # string output
I think a better version would be just to find the index of decimal point . and then to take the string slice accordingly:
def truncate(number, n_digits:int=1)->float:
'''
:param number: real number ℝ
:param n_digits: Maximum number of digits after the decimal point after truncation
:return: truncated floating point number with at least one digit after decimal point
'''
decimalIndex = str(number).find('.')
if decimalIndex == -1:
return float(number)
else:
return float(str(number)[:decimalIndex+n_digits+1])
int(1.923328437452 * 1000) / 1000
>>> 1.923
int(1.9239 * 1000) / 1000
>>> 1.923
By multiplying the number by 1000 (10 ^ 3 for 3 digits) we shift the decimal point 3 places to the right and get 1923.3284374520001. When we convert that to an int the fractional part 3284374520001 will be discarded. Then we undo the shifting of the decimal point again by dividing by 1000 which returns 1.923.
use numpy.round
import numpy as np
precision = 3
floats = [1.123123123, 2.321321321321]
new_float = np.round(floats, precision)
Something simple enough to fit in a list-comprehension, with no libraries or other external dependencies. For Python >=3.6, it's very simple to write with f-strings.
The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit.
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1] for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.800', '0.888', '1.125', '1.250', '1.500']
Of course, there is rounding happening here (namely for the fourth digit), but rounding at some point is unvoidable. In case the transition between truncation and rounding is relevant, here's a slightly better example:
>>> nacc = 6 # desired accuracy (maximum 15!)
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nacc}f}'[:-(nacc-nout)] for x in [2.9999, 2.99999, 2.999999, 2.9999999]]
>>> ['2.999', '2.999', '2.999', '3.000']
Bonus: removing zeros on the right
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1].rstrip('0') for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.8', '0.888', '1.125', '1.25', '1.5']
The core idea given here seems to me to be the best approach for this problem.
Unfortunately, it has received less votes while the later answer that has more votes is not complete (as observed in the comments). Hopefully, the implementation below provides a short and complete solution for truncation.
def trunc(num, digits):
l = str(float(num)).split('.')
digits = min(len(l[1]), digits)
return l[0] + '.' + l[1][:digits]
which should take care of all corner cases found here and here.
Am also a python newbie and after making use of some bits and pieces here, I offer my two cents
print str(int(time.time()))+str(datetime.now().microsecond)[:3]
str(int(time.time())) will take the time epoch as int and convert it to string and join with...
str(datetime.now().microsecond)[:3] which returns the microseconds only, convert to string and truncate to first 3 chars
# value value to be truncated
# n number of values after decimal
value = 0.999782
n = 3
float(int(value*1en))*1e-n
I'd simply like to convert a base-2 binary number string into an int, something like this:
>>> '11111111'.fromBinaryToInt()
255
Is there a way to do this in Python?
You use the built-in int() function, and pass it the base of the input number, i.e. 2 for a binary number:
>>> int('11111111', 2)
255
Here is documentation for Python 2, and for Python 3.
Just type 0b11111111 in python interactive interface:
>>> 0b11111111
255
Another way to do this is by using the bitstring module:
>>> from bitstring import BitArray
>>> b = BitArray(bin='11111111')
>>> b.uint
255
Note that the unsigned integer (uint) is different from the signed integer (int):
>>> b.int
-1
Your question is really asking for the unsigned integer representation; this is an important distinction.
The bitstring module isn't a requirement, but it has lots of performant methods for turning input into and from bits into other forms, as well as manipulating them.
Using int with base is the right way to go. I used to do this before I found int takes base also. It is basically a reduce applied on a list comprehension of the primitive way of converting binary to decimal ( e.g. 110 = 2**0 * 0 + 2 ** 1 * 1 + 2 ** 2 * 1)
add = lambda x,y : x + y
reduce(add, [int(x) * 2 ** y for x, y in zip(list(binstr), range(len(binstr) - 1, -1, -1))])
If you wanna know what is happening behind the scene, then here you go.
class Binary():
def __init__(self, binNumber):
self._binNumber = binNumber
self._binNumber = self._binNumber[::-1]
self._binNumber = list(self._binNumber)
self._x = [1]
self._count = 1
self._change = 2
self._amount = 0
print(self._ToNumber(self._binNumber))
def _ToNumber(self, number):
self._number = number
for i in range (1, len (self._number)):
self._total = self._count * self._change
self._count = self._total
self._x.append(self._count)
self._deep = zip(self._number, self._x)
for self._k, self._v in self._deep:
if self._k == '1':
self._amount += self._v
return self._amount
mo = Binary('101111110')
Here's another concise way to do it not mentioned in any of the above answers:
>>> eval('0b' + '11111111')
255
Admittedly, it's probably not very fast, and it's a very very bad idea if the string is coming from something you don't have control over that could be malicious (such as user input), but for completeness' sake, it does work.
A recursive Python implementation:
def int2bin(n):
return int2bin(n >> 1) + [n & 1] if n > 1 else [1]
If you are using python3.6 or later you can use f-string to do the
conversion:
Binary to decimal:
>>> print(f'{0b1011010:#0}')
90
>>> bin_2_decimal = int(f'{0b1011010:#0}')
>>> bin_2_decimal
90
binary to octal hexa and etc.
>>> f'{0b1011010:#o}'
'0o132' # octal
>>> f'{0b1011010:#x}'
'0x5a' # hexadecimal
>>> f'{0b1011010:#0}'
'90' # decimal
Pay attention to 2 piece of information separated by colon.
In this way, you can convert between {binary, octal, hexadecimal, decimal} to {binary, octal, hexadecimal, decimal} by changing right side of colon[:]
:#b -> converts to binary
:#o -> converts to octal
:#x -> converts to hexadecimal
:#0 -> converts to decimal as above example
Try changing left side of colon to have octal/hexadecimal/decimal.
For large matrix (10**5 rows and up) it is better to use a vectorized matmult. Pass in all rows and cols in one shot. It is extremely fast. There is no looping in python here. I originally designed it for converting many binary columns like 0/1 for like 10 different genre columns in MovieLens into a single integer for each example row.
def BitsToIntAFast(bits):
m,n = bits.shape
a = 2**np.arange(n)[::-1] # -1 reverses array of powers of 2 of same length as bits
return bits # a
For the record to go back and forth in basic python3:
a = 10
bin(a)
# '0b1010'
int(bin(a), 2)
# 10
eval(bin(a))
# 10
I have the following code of two binary lists and I want to obtain a resulting list in which the element i will be the OR operation applied to the i elements of the two lists:
from operator import ior
l_0 = [01100]
l_1 = [11000]
print map(ior, l_0, l_1)
And I was expecting a result of [11100], but the result is:
[11000]
I have checked ior operator and the documentation says that it performs the operation:
a = ior(a, b) is equivalent to a |= b
So I tried the following to check as well:
print ior(0,0)
print ior(1,0)
print ior(0,1)
print ior(1,1)
Getting as results:
0
1
1
1
Which makes sense, but doesn't coincide with the result obtained in the 3rd position of the lists. I don't understand why the result of the map operation of above is not [11100]. I am missing something here and I hope that you can throw some light on it.
[11000] (for example) is not a list of five binary digits, it's a list of one decimal number, 11000.
Similarly, 01100 is a single octal number equal to 576 decimal, so:
11000d = 0010101011111000b
01100o = 576d = 0000001001000000b
-----------------
perform or: 0010101011111000b = 11000d
That's why you're getting 11000 as the answer, exactly the same as if you'd done:
[l_0[i] | l_1[i] for i in range(len(l_0))]
If you want to process a list of five binary digits, that would be something like:
>>> l_0 = [0,1,1,0,0]
>>> l_1 = [1,1,0,0,0]
>>> [l_0[i] | l_1[i] for i in range(len(l_0))]
[1, 1, 1, 0, 0]
01100 is octal representation (576 in decimal). 11000 is decimal representation. They are not binary representations.
To represent binary, prefix them with 0b:
[`01100`, `11000`]
And to get binary representation from the number use bin:
bin(num)
or
'{:b}'.format(num)
How can I add, subtract, and compare binary numbers in Python without converting to decimal?
You can convert between a string representation of the binary using bin() and int()
>>> bin(88)
'0b1011000'
>>> int('0b1011000', 2)
88
>>>
>>> a=int('01100000', 2)
>>> b=int('00100110', 2)
>>> bin(a & b)
'0b100000'
>>> bin(a | b)
'0b1100110'
>>> bin(a ^ b)
'0b1000110'
I think you're confused about what binary is. Binary and decimal are just different representations of a number - e.g. 101 base 2 and 5 base 10 are the same number. The operations add, subtract, and compare operate on numbers - 101 base 2 == 5 base 10 and addition is the same logical operation no matter what base you're working in. The fact that your python interpreter may store things as binary internally doesn't affect how you work with it - if you have an integer type, just use +, -, etc.
If you have strings of binary digits, you'll have to either write your own implementation or convert them using the int(binaryString, 2) function.
If you're talking about bitwise operators, then you're after:
~ Not
^ XOR
| Or
& And
Otherwise, binary numbers work exactly the same as decimal numbers, because numbers are numbers, no matter how you look at them. The only difference between decimal and binary is how we represent that data when we are looking at it.
Binary, decimal, hexadecimal... the base only matters when reading or outputting numbers, adding binary numbers is just the same as adding decimal number : it is just a matter of representation.
Below is a re-write of a previously posted function:
def addBinary(a, b): # Example: a = '11' + b =' 100' returns as '111'.
for ch in a: assert ch in {'0','1'}, 'bad digit: ' + ch
for ch in b: assert ch in {'0','1'}, 'bad digit: ' + ch
sumx = int(a, 2) + int(b, 2)
return bin(sumx)[2:]
'''
I expect the intent behind this assignment was to work in binary string format.
This is absolutely doable.
'''
def compare(bin1, bin2):
return bin1.lstrip('0') == bin2.lstrip('0')
def add(bin1, bin2):
result = ''
blen = max((len(bin1), len(bin2))) + 1
bin1, bin2 = bin1.zfill(blen), bin2.zfill(blen)
carry_s = '0'
for b1, b2 in list(zip(bin1, bin2))[::-1]:
count = (carry_s, b1, b2).count('1')
carry_s = '1' if count >= 2 else '0'
result += '1' if count % 2 else '0'
return result[::-1]
if __name__ == '__main__':
print(add('101', '100'))
I leave the subtraction func as an exercise for the reader.
For example, 00000011 - 00000001 = 00000010
You can remove the zeroes and then add them again after you do your calculation! This works very easy.
If your binary is stored as a string then you can convert to int which will automatically strip the zeroes from the start. After you have your answer you can turn it back into a string and add the zeroes to the start.
Not sure if helpful, but I leave my solution here:
class Solution:
# #param A : string
# #param B : string
# #return a strings
def addBinary(self, A, B):
num1 = bin(int(A, 2))
num2 = bin(int(B, 2))
bin_str = bin(int(num1, 2)+int(num2, 2))
b_index = bin_str.index('b')
return bin_str[b_index+1:]
s = Solution()
print(s.addBinary("11", "100"))
x = x + 1
print(x)
a = x + 5
print(a)
I think you're confused about what binary is. Binary and decimal are just different representations of a number - e.g. 101 base 2 and 5 base 10 are the same number. The operations add, subtract, and compare operate on numbers - 101 base 2 == 5 base 10 and addition is the same logical operation no matter what base you're working in.
I want to remove digits from a float to have a fixed number of digits after the dot, like:
1.923328437452 → 1.923
I need to output as a string to another function, not print.
Also I want to ignore the lost digits, not round them.
round(1.923328437452, 3)
See Python's documentation on the standard types. You'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.
First, the function, for those who just want some copy-and-paste code:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
This is valid in Python 2.7 and 3.1+. For older versions, it's not possible to get the same "intelligent rounding" effect (at least, not without a lot of complicated code), but rounding to 12 decimal places before truncation will work much of the time:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '%.12f' % f
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
Explanation
The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters. The latter step is easy; it can be done either with string manipulation
i, p, d = s.partition('.')
'.'.join([i, (d+'0'*n)[:n]])
or the decimal module
str(Decimal(s).quantize(Decimal((0, (1,), -n)), rounding=ROUND_DOWN))
The first step, converting to a string, is quite difficult because there are some pairs of floating point literals (i.e. what you write in the source code) which both produce the same binary representation and yet should be truncated differently. For example, consider 0.3 and 0.29999999999999998. If you write 0.3 in a Python program, the compiler encodes it using the IEEE floating-point format into the sequence of bits (assuming a 64-bit float)
0011111111010011001100110011001100110011001100110011001100110011
This is the closest value to 0.3 that can accurately be represented as an IEEE float. But if you write 0.29999999999999998 in a Python program, the compiler translates it into exactly the same value. In one case, you meant it to be truncated (to one digit) as 0.3, whereas in the other case you meant it to be truncated as 0.2, but Python can only give one answer. This is a fundamental limitation of Python, or indeed any programming language without lazy evaluation. The truncation function only has access to the binary value stored in the computer's memory, not the string you actually typed into the source code.1
If you decode the sequence of bits back into a decimal number, again using the IEEE 64-bit floating-point format, you get
0.2999999999999999888977697537484345957637...
so a naive implementation would come up with 0.2 even though that's probably not what you want. For more on floating-point representation error, see the Python tutorial.
It's very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number. So when truncating, it probably makes sense to choose the "nicest" decimal representation out of all that could correspond to the value in memory. Python 2.7 and up (but not 3.0) includes a sophisticated algorithm to do just that, which we can access through the default string formatting operation.
'{}'.format(f)
The only caveat is that this acts like a g format specification, in the sense that it uses exponential notation (1.23e+4) if the number is large or small enough. So the method has to catch this case and handle it differently. There are a few cases where using an f format specification instead causes a problem, such as trying to truncate 3e-10 to 28 digits of precision (it produces 0.0000000002999999999999999980), and I'm not yet sure how best to handle those.
If you actually are working with floats that are very close to round numbers but intentionally not equal to them (like 0.29999999999999998 or 99.959999999999994), this will produce some false positives, i.e. it'll round numbers that you didn't want rounded. In that case the solution is to specify a fixed precision.
'{0:.{1}f}'.format(f, sys.float_info.dig + n + 2)
The number of digits of precision to use here doesn't really matter, it only needs to be large enough to ensure that any rounding performed in the string conversion doesn't "bump up" the value to its nice decimal representation. I think sys.float_info.dig + n + 2 may be enough in all cases, but if not that 2 might have to be increased, and it doesn't hurt to do so.
In earlier versions of Python (up to 2.6, or 3.0), the floating point number formatting was a lot more crude, and would regularly produce things like
>>> 1.1
1.1000000000000001
If this is your situation, if you do want to use "nice" decimal representations for truncation, all you can do (as far as I know) is pick some number of digits, less than the full precision representable by a float, and round the number to that many digits before truncating it. A typical choice is 12,
'%.12f' % f
but you can adjust this to suit the numbers you're using.
1Well... I lied. Technically, you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function. If that argument is a floating-point literal, you can just cut it off a certain number of places after the decimal point and return that. However this strategy doesn't work if the argument is a variable, which makes it fairly useless. The following is presented for entertainment value only:
def trunc_introspect(f, n):
'''Truncates/pads the float f to n decimal places by looking at the caller's source code'''
current_frame = None
caller_frame = None
s = inspect.stack()
try:
current_frame = s[0]
caller_frame = s[1]
gen = tokenize.tokenize(io.BytesIO(caller_frame[4][caller_frame[5]].encode('utf-8')).readline)
for token_type, token_string, _, _, _ in gen:
if token_type == tokenize.NAME and token_string == current_frame[3]:
next(gen) # left parenthesis
token_type, token_string, _, _, _ = next(gen) # float literal
if token_type == tokenize.NUMBER:
try:
cut_point = token_string.index('.') + n + 1
except ValueError: # no decimal in string
return token_string + '.' + '0' * n
else:
if len(token_string) < cut_point:
token_string += '0' * (cut_point - len(token_string))
return token_string[:cut_point]
else:
raise ValueError('Unable to find floating-point literal (this probably means you called {} with a variable)'.format(current_frame[3]))
break
finally:
del s, current_frame, caller_frame
Generalizing this to handle the case where you pass in a variable seems like a lost cause, since you'd have to trace backwards through the program's execution until you find the floating-point literal which gave the variable its value. If there even is one. Most variables will be initialized from user input or mathematical expressions, in which case the binary representation is all there is.
The result of round is a float, so watch out (example is from Python 2.6):
>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001
You will be better off when using a formatted string:
>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'
n = 1.923328437452
str(n)[:4]
At my Python 2.7 prompt:
>>> int(1.923328437452 * 1000)/1000.0
1.923
The truely pythonic way of doing it is
from decimal import *
with localcontext() as ctx:
ctx.rounding = ROUND_DOWN
print Decimal('1.923328437452').quantize(Decimal('0.001'))
or shorter:
from decimal import Decimal as D, ROUND_DOWN
D('1.923328437452').quantize(D('0.001'), rounding=ROUND_DOWN)
Update
Usually the problem is not in truncating floats itself, but in the improper usage of float numbers before rounding.
For example: int(0.7*3*100)/100 == 2.09.
If you are forced to use floats (say, you're accelerating your code with numba), it's better to use cents as "internal representation" of prices: (70*3 == 210) and multiply/divide the inputs/outputs.
Simple python script -
n = 1.923328437452
n = float(int(n * 1000))
n /=1000
def trunc(num, digits):
sp = str(num).split('.')
return '.'.join([sp[0], sp[1][:digits]])
This should work. It should give you the truncation you are looking for.
So many of the answers given for this question are just completely wrong. They either round up floats (rather than truncate) or do not work for all cases.
This is the top Google result when I search for 'Python truncate float', a concept which is really straightforward, and which deserves better answers. I agree with Hatchkins that using the decimal module is the pythonic way of doing this, so I give here a function which I think answers the question correctly, and which works as expected for all cases.
As a side-note, fractional values, in general, cannot be represented exactly by binary floating point variables (see here for a discussion of this), which is why my function returns a string.
from decimal import Decimal, localcontext, ROUND_DOWN
def truncate(number, places):
if not isinstance(places, int):
raise ValueError("Decimal places must be an integer.")
if places < 1:
raise ValueError("Decimal places must be at least 1.")
# If you want to truncate to 0 decimal places, just do int(number).
with localcontext() as context:
context.rounding = ROUND_DOWN
exponent = Decimal(str(10 ** - places))
return Decimal(str(number)).quantize(exponent).to_eng_string()
>>> from math import floor
>>> floor((1.23658945) * 10**4) / 10**4
1.2365
# divide and multiply by 10**number of desired digits
If you fancy some mathemagic, this works for +ve numbers:
>>> v = 1.923328437452
>>> v - v % 1e-3
1.923
I did something like this:
from math import trunc
def truncate(number, decimals=0):
if decimals < 0:
raise ValueError('truncate received an invalid value of decimals ({})'.format(decimals))
elif decimals == 0:
return trunc(number)
else:
factor = float(10**decimals)
return trunc(number*factor)/factor
You can do:
def truncate(f, n):
return math.floor(f * 10 ** n) / 10 ** n
testing:
>>> f=1.923328437452
>>> [truncate(f, n) for n in range(5)]
[1.0, 1.9, 1.92, 1.923, 1.9233]
Just wanted to mention that the old "make round() with floor()" trick of
round(f) = floor(f+0.5)
can be turned around to make floor() from round()
floor(f) = round(f-0.5)
Although both these rules break around negative numbers, so using it is less than ideal:
def trunc(f, n):
if f > 0:
return "%.*f" % (n, (f - 0.5*10**-n))
elif f == 0:
return "%.*f" % (n, f)
elif f < 0:
return "%.*f" % (n, (f + 0.5*10**-n))
def precision(value, precision):
"""
param: value: takes a float
param: precision: int, number of decimal places
returns a float
"""
x = 10.0**precision
num = int(value * x)/ x
return num
precision(1.923328437452, 3)
1.923
Short and easy variant
def truncate_float(value, digits_after_point=2):
pow_10 = 10 ** digits_after_point
return (float(int(value * pow_10))) / pow_10
>>> truncate_float(1.14333, 2)
>>> 1.14
>>> truncate_float(1.14777, 2)
>>> 1.14
>>> truncate_float(1.14777, 4)
>>> 1.1477
When using a pandas df this worked for me
import math
def truncate(number, digits) -> float:
stepper = 10.0 ** digits
return math.trunc(stepper * number) / stepper
df['trunc'] = df['float_val'].apply(lambda x: truncate(x,1))
df['trunc']=df['trunc'].map('{:.1f}'.format)
int(16.5);
this will give an integer value of 16, i.e. trunc, won't be able to specify decimals, but guess you can do that by
import math;
def trunc(invalue, digits):
return int(invalue*math.pow(10,digits))/math.pow(10,digits);
Here is an easy way:
def truncate(num, res=3):
return (floor(num*pow(10, res)+0.5))/pow(10, res)
for num = 1.923328437452, this outputs 1.923
def trunc(f,n):
return ('%.16f' % f)[:(n-16)]
A general and simple function to use:
def truncate_float(number, length):
"""Truncate float numbers, up to the number specified
in length that must be an integer"""
number = number * pow(10, length)
number = int(number)
number = float(number)
number /= pow(10, length)
return number
There is an easy workaround in python 3. Where to cut I defined with an help variable decPlace to make it easy to adapt.
f = 1.12345
decPlace= 4
f_cut = int(f * 10**decPlace) /10**decPlace
Output:
f = 1.1234
Hope it helps.
Most answers are way too complicated in my opinion, how about this?
digits = 2 # Specify how many digits you want
fnum = '122.485221'
truncated_float = float(fnum[:fnum.find('.') + digits + 1])
>>> 122.48
Simply scanning for the index of '.' and truncate as desired (no rounding).
Convert string to float as final step.
Or in your case if you get a float as input and want a string as output:
fnum = str(122.485221) # convert float to string first
truncated_float = fnum[:fnum.find('.') + digits + 1] # string output
I think a better version would be just to find the index of decimal point . and then to take the string slice accordingly:
def truncate(number, n_digits:int=1)->float:
'''
:param number: real number ℝ
:param n_digits: Maximum number of digits after the decimal point after truncation
:return: truncated floating point number with at least one digit after decimal point
'''
decimalIndex = str(number).find('.')
if decimalIndex == -1:
return float(number)
else:
return float(str(number)[:decimalIndex+n_digits+1])
int(1.923328437452 * 1000) / 1000
>>> 1.923
int(1.9239 * 1000) / 1000
>>> 1.923
By multiplying the number by 1000 (10 ^ 3 for 3 digits) we shift the decimal point 3 places to the right and get 1923.3284374520001. When we convert that to an int the fractional part 3284374520001 will be discarded. Then we undo the shifting of the decimal point again by dividing by 1000 which returns 1.923.
use numpy.round
import numpy as np
precision = 3
floats = [1.123123123, 2.321321321321]
new_float = np.round(floats, precision)
Something simple enough to fit in a list-comprehension, with no libraries or other external dependencies. For Python >=3.6, it's very simple to write with f-strings.
The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit.
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1] for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.800', '0.888', '1.125', '1.250', '1.500']
Of course, there is rounding happening here (namely for the fourth digit), but rounding at some point is unvoidable. In case the transition between truncation and rounding is relevant, here's a slightly better example:
>>> nacc = 6 # desired accuracy (maximum 15!)
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nacc}f}'[:-(nacc-nout)] for x in [2.9999, 2.99999, 2.999999, 2.9999999]]
>>> ['2.999', '2.999', '2.999', '3.000']
Bonus: removing zeros on the right
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1].rstrip('0') for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.8', '0.888', '1.125', '1.25', '1.5']
The core idea given here seems to me to be the best approach for this problem.
Unfortunately, it has received less votes while the later answer that has more votes is not complete (as observed in the comments). Hopefully, the implementation below provides a short and complete solution for truncation.
def trunc(num, digits):
l = str(float(num)).split('.')
digits = min(len(l[1]), digits)
return l[0] + '.' + l[1][:digits]
which should take care of all corner cases found here and here.
Am also a python newbie and after making use of some bits and pieces here, I offer my two cents
print str(int(time.time()))+str(datetime.now().microsecond)[:3]
str(int(time.time())) will take the time epoch as int and convert it to string and join with...
str(datetime.now().microsecond)[:3] which returns the microseconds only, convert to string and truncate to first 3 chars
# value value to be truncated
# n number of values after decimal
value = 0.999782
n = 3
float(int(value*1en))*1e-n