formatting long numbers as strings in python
PythonFormattingStringIntegerPython Problem Overview
What is an easy way in Python to format integers into strings representing thousands with K, and millions with M, and leaving just couple digits after comma?
I'd like to show 7436313 as 7.44M, and 2345 as 2,34K.
Is there some % string formatting operator available for that? Or that could be done only by actually dividing by 1000 in a loop and constructing result string step by step?
Python Solutions
Solution 1 - Python
I don't think there's a built-in function that does that. You'll have to roll your own, e.g.:
def human_format(num):
magnitude = 0
while abs(num) >= 1000:
magnitude += 1
num /= 1000.0
# add more suffixes if you need them
return '%.2f%s' % (num, ['', 'K', 'M', 'G', 'T', 'P'][magnitude])
print('the answer is %s' % human_format(7436313)) # prints 'the answer is 7.44M'
Solution 2 - Python
This version does not suffer from the bug in the previous answers where 999,999 gives you 1000.0K. It also only allows 3 significant figures and eliminates trailing 0's.
def human_format(num):
num = float('{:.3g}'.format(num))
magnitude = 0
while abs(num) >= 1000:
magnitude += 1
num /= 1000.0
return '{}{}'.format('{:f}'.format(num).rstrip('0').rstrip('.'), ['', 'K', 'M', 'B', 'T'][magnitude])
The output looks like:
>>> human_format(999999)
'1M'
>>> human_format(999499)
'999K'
>>> human_format(9994)
'9.99K'
>>> human_format(9900)
'9.9K'
>>> human_format(6543165413)
'6.54B'
Solution 3 - Python
A more "math-y" solution is to use math.log:
from math import log, floor
def human_format(number):
units = ['', 'K', 'M', 'G', 'T', 'P']
k = 1000.0
magnitude = int(floor(log(number, k)))
return '%.2f%s' % (number / k**magnitude, units[magnitude])
Tests:
>>> human_format(123456)
'123.46K'
>>> human_format(123456789)
'123.46M'
>>> human_format(1234567890)
'1.23G'
Solution 4 - Python
Variable precision and no 999999 bug:
def human_format(num, round_to=2):
magnitude = 0
while abs(num) >= 1000:
magnitude += 1
num = round(num / 1000.0, round_to)
return '{:.{}f}{}'.format(num, round_to, ['', 'K', 'M', 'G', 'T', 'P'][magnitude])
Solution 5 - Python
I needed this function today, refreshed the accepted answer a bit for people with Python >= 3.6:
def human_format(num, precision=2, suffixes=['', 'K', 'M', 'G', 'T', 'P']):
m = sum([abs(num/1000.0**x) >= 1 for x in range(1, len(suffixes))])
return f'{num/1000.0**m:.{precision}f}{suffixes[m]}'
print('the answer is %s' % human_format(7454538)) # prints 'the answer is 7.45M'
Edit: given the comments, you might want to change to round(num/1000.0)
Solution 6 - Python
I was kind of confused by some of the stuff that other people showed, so I made the below code. It rounds to the second decimal point, ex. '23.56 Billion', but you can change what decimal place it rounds to by replacing the two '100.0's in the last line with a larger or smaller number, ex. '10.0' rounds to one decimal point and '1000.0' rounds to three decimal points. Also, using this code, it always rounds down from what it actually is. You can change this if you like, by replacing 'floor' with 'ceil' or 'round'.
#make the dictionary to store what to put after the result (ex. 'Billion'). You can go further with this then I did, or to wherever you wish.
#import the desired rounding mechanism. You will not need to do this for round.
from math import floor
magnitudeDict={0:'', 1:'Thousand', 2:'Million', 3:'Billion', 4:'Trillion', 5:'Quadrillion', 6:'Quintillion', 7:'Sextillion', 8:'Septillion', 9:'Octillion', 10:'Nonillion', 11:'Decillion'}
def simplify(num):
num=floor(num)
magnitude=0
while num>=1000.0:
magnitude+=1
num=num/1000.0
return(f'{floor(num*100.0)/100.0} {magnitudeDict[magnitude]}')
The 'f' before the string in the last line is to let python know you are formatting it. The result from running print(simplify(34867123012.13)) is this:
34.86 Billion
Please let me know if you have questions! Thanks, Angus
Solution 7 - Python
Numerize library is good.
from numerize import numerize
a = numerize.numerize(1000)
print(a)
1k
Thanks @tdy for pointing this,
a = numerize.numerize(999999)
print(a) # 1000K
1000K
Solution 8 - Python
I had the same need. And if anyone comes on this topic, I found a lib to do so: https://github.com/azaitsev/millify
Hope it helps :)
Solution 9 - Python
No String Formatting Operator, according to the docs. I've never heard of such a thing, so you may have to roll your own, as you suggest.
Solution 10 - Python
I don't think there are format operators for that, but you can simply divide by 1000 until the result is between 1 and 999 and then use a format string for 2 digits after comma. Unit is a single character (or perhaps a small string) in most cases, which you can store in a string or array and iterate through it after each divide.
Solution 11 - Python
I don't know of any built-in capability like this, but here are a couple of list threads that may help:
http://coding.derkeiler.com/Archive/Python/comp.lang.python/2005-09/msg03327.html http://mail.python.org/pipermail/python-list/2008-August/503417.html
Solution 12 - Python
Based on the comments here, I made an improved code for that. It is a little bit longer but gives solutions for more cases including small numbers (m,u,n,p).
Hope it will be helpful for someone
# print number in a readable format.
# default is up to 3 decimal digits and can be changed
# works on numbers in the range of 1e-15 to 1e 1e15 include negatives numbers
# can force the number to a specific magnitude unit
def human_format(num:float, force=None, ndigits=3):
perfixes = ('p', 'n', 'u', 'm', '', 'K', 'M', 'G', 'T')
one_index = perfixes.index('')
if force:
if force in perfixes:
index = perfixes.index(force)
magnitude = 3*(index - one_index)
num = num/(10**magnitude)
else:
raise ValueError('force value not supported.')
else:
div_sum = 0
if(abs(num) >= 1000):
while abs(num) >= 1000:
div_sum += 1
num /= 1000
else:
while abs(num) <= 1:
div_sum -= 1
num *= 1000
temp = round(num, ndigits) if ndigits else num
if temp < 1000:
num = temp
else:
num = 1
div_sum += 1
index = one_index + div_sum
return str(num).rstrip('0').rstrip('.') + perfixes[index]
tests from here and some more
# some tests
print(human_format(999) ,' = ' , '999')
print(human_format(999.999) ,' = ' , '999.999')
print(human_format(999.9999) ,' = ' , '1K')
print(human_format(999999) ,' = ' , '999.999K')
print(human_format(999499) ,' = ' , '999.499K')
print(human_format(9994) ,' = ' , '9.994K')
print(human_format(9900) ,' = ' , '9.9K')
print(human_format(6543165413) ,' = ' , '6.543G')
print(human_format(46780.9) ,' = ' , '46.781K')
print(human_format(0.001) ,' = ' , '1m')
print(human_format(0.000000999999) ,' = ' , '999.999n')
print(human_format(1.00394200) ,' = ' , '1.004')
print(human_format(0.0999) ,' = ' , '99.9m')
print(human_format(0.00000000999999) ,' = ' , '10n')
print(human_format(0.0000000099995) ,' = ' , '9.999n')
print(human_format(0.000000009999) ,' = ' , '9.999n')
print(human_format(999999 ,ndigits=2) ,' = ' , '1M')
print(human_format(9994 ,force='') ,' = ' , '9994K')
print(human_format(6543165413 ,ndigits=5) ,' = ' , '6.54317G')
print(human_format(6543165413 ,ndigits=None) ,' = ' , '6.543165413G')
print(human_format(7436313 ,ndigits=2) ,' = ' , '7.44M')
print(human_format(2344 ,ndigits=2) ,' = ' , '2.34K')
print(human_format(34867123012.13 ,ndigits=2) ,' = ' , '34.87G')