formatting long numbers as strings in 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'
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'
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'
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])
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)