How to print number with commas as thousands separ

Question

I am trying to print an integer in Python 2.6.1 with commas as thousands separators. For example, I want to show the number 1234567 as 1,234,567. How would I go about doing this? I have seen many examples on Google, but I am looking for the simplest practical way.

It does not need to be locale-specific to decide between periods and commas. I would prefer something as simple as reasonably possible.

Answer 1

For inefficiency and unreadability it's hard to beat:

>>> import itertools
>>> s = '-1234567'
>>> ','.join(["%s%s%s" % (x[0], x[1] or '', x[2] or '') for x in itertools.izip_longest(s[::-1][::3], s[::-1][1::3], s[::-1][2::3])])[::-1].replace('-,','-')

Answer 2

Here's a one-line regex replacement:

re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%d" % val)

Works only for inegral outputs:

import re
val = 1234567890
re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%d" % val)
# Returns: '1,234,567,890'

val = 1234567890.1234567890
# Returns: '1,234,567,890'

Or for floats with less than 4 digits, change the format specifier to %.3f:

re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%.3f" % val)
# Returns: '1,234,567,890.123'

NB: Doesn't work correctly with more than three decimal digits as it will attempt to group the decimal part:

re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%.5f" % val)
# Returns: '1,234,567,890.12,346'

How it works

Let's break it down:

re.sub(pattern, repl, string)

pattern = \
    "(\d)           # Find one digit...
     (?=            # that is followed by...
         (\d{3})+   # one or more groups of three digits...
         (?!\d)     # which are not followed by any more digits.
     )",

repl = \
    r"\1,",         # Replace that one digit by itself, followed by a comma,
                    # and continue looking for more matches later in the string.
                    # (re.sub() replaces all matches it finds in the input)

string = \
    "%d" % val      # Format the string as a decimal to begin with

Answer 3

Slightly expanding the answer of Ian Schneider:

If you want to use a custom thousands separator, the simplest solution is:

'{:,}'.format(value).replace(',', your_custom_thousands_separator)

Examples

'{:,.2f}'.format(123456789.012345).replace(',', ' ')

If you want the German representation like this, it gets a bit more complicated:

('{:,.2f}'.format(123456789.012345)
          .replace(',', ' ')  # 'save' the thousands separators 
          .replace('.', ',')  # dot to comma
          .replace(' ', '.')) # thousand separators to dot

Answer 4

I prefer the locale-based solution for real projects, but I think the approach with the use of slice assignment should be mentioned here:

def format_money(f, delimiter=',', frac_digits=2):

    negative_fix = int(f < 0)

    s = '%.*f' % (frac_digits, f)
    if len(s) < 5 + frac_digits + negative_fix:
        return s

    l = list(s)
    l_fix = l[negative_fix:]
    p = len(l_fix) - frac_digits - 5
    l_fix[p::-3] = [i + delimiter for i in l_fix[p::-3]]

    return ''.join(l[:negative_fix] + l_fix)

Gist with doctests is here - https://gist.github.com/ei-grad/b290dc761ae253af69438bbb94d82683

Answer 5

I'm sure there must be a standard library function for this, but it was fun to try to write it myself using recursion so here's what I came up with:

def intToStringWithCommas(x):
    if type(x) is not int and type(x) is not long:
        raise TypeError("Not an integer!")
    if x < 0:
        return '-' + intToStringWithCommas(-x)
    elif x < 1000:
        return str(x)
    else:
        return intToStringWithCommas(x / 1000) + ',' + '%03d' % (x % 1000)

Having said that, if someone else does find a standard way to do it, you should use that instead.

Answer 6

from Python version 2.6 you can do this:

def format_builtin(n):
    return format(n, ',')

For Python versions < 2.6 and just for your information, here are 2 manual solutions, they turn floats to ints but negative numbers work correctly:

def format_number_using_lists(number):
    string = '%d' % number
    result_list = list(string)
    indexes = range(len(string))
    for index in indexes[::-3][1:]:
        if result_list[index] != '-':
            result_list.insert(index+1, ',')
    return ''.join(result_list)

few things to notice here:

• this line: string = '%d' % number beautifully converts a number to a string, it supports negatives and it drops fractions from floats, making them ints;
• this slice indexes[::-3] returns each third item starting from the end, so I used another slice [1:] to remove the very last item cuz I don't need a comma after the last number;
• this conditional if l[index] != '-' is being used to support negative numbers, do not insert a comma after the minus sign.

And a more hardcore version:

def format_number_using_generators_and_list_comprehensions(number):
    string = '%d' % number
    generator = reversed( 
        [
            value+',' if (index!=0 and value!='-' and index%3==0) else value
            for index,value in enumerate(reversed(string))
        ]
    )
    return ''.join(generator)