How can I print a float with thousands separators?

问题

How can I format a decimal number so that 32757121.33 will display as 32.757.121,33?

回答1:

Use locale.format():

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'German')
'German_Germany.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32.757.121,33

You can restrict the locale changes to the display of numeric values (when using locale.format(), locale.str() etc.) and leave other locale settings unaffected:

>>> locale.setlocale(locale.LC_NUMERIC, 'English')
'English_United States.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32,757,121.33
>>> locale.setlocale(locale.LC_NUMERIC, 'German')
'German_Germany.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32.757.121,33

回答2:

I have found another solution:

'{:,.2f}'.format(num).replace(".","%").replace(",",".").replace("%",",")

回答3:

If you can't or don't want to use locale for some reason, you can also do it with a regular expression:

import re
def sep(s, thou=",", dec="."):
    integer, decimal = s.split(".")
    integer = re.sub(r"\B(?=(?:\d{3})+$)", thou, integer)
    return integer + dec + decimal

sep() takes the string representation of a standard Python float and returns it with custom thousands and decimal separators.

>>> s = "%.2f" % 32757121.33
>>> sep(s)
'32,757,121.33'
>>> sep(s, thou=".", dec=",")
'32.757.121,33'

Explanation:

\B      # Assert that we're not at the start of the number
(?=     # Match at a position where it's possible to match...
 (?:    #  the following regex:
  \d{3} #   3 digits
 )+     #  repeated at least once
 $      #  until the end of the string
)       # (thereby ensuring a number of digits divisible by 3

来源：https://stackoverflow.com/questions/13082620/how-can-i-print-a-float-with-thousands-separators

标签

python

format

decimal