formatting long numbers as strings in python

51
votes

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?

10
pythonformattingstringinteger

10 Answers

70
votes

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'
56
votes

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'
10
votes

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'
6
votes

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)

6
votes

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(round(num, round_to), round_to, ['', 'K', 'M', 'G', 'T', 'P'][magnitude])
2
votes

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

1
votes

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 :)

0
votes

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.

0
votes

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.