How to print number with commas as thousands separators?


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.

6/29/2019 11:34:15 AM

Accepted Answer

Locale unaware

'{:,}'.format(value)  # For Python ≥2.7
f'{value:,}'  # For Python ≥3.7

Locale aware

import locale
locale.setlocale(locale.LC_ALL, '')  # Use '' for auto, or force e.g. to 'en_US.UTF-8'

'{:n}'.format(value)  # For Python ≥2.7
f'{value:n}'  # For Python ≥3.7


Per Format Specification Mini-Language,

The ',' option signals the use of a comma for a thousands separator. For a locale aware separator, use the 'n' integer presentation type instead.

7/14/2019 5:15:31 PM

I got this to work:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US')
>>> locale.format("%d", 1255000, grouping=True)

Sure, you don't need internationalization support, but it's clear, concise, and uses a built-in library.

P.S. That "%d" is the usual %-style formatter. You can have only one formatter, but it can be whatever you need in terms of field width and precision settings.

P.P.S. If you can't get locale to work, I'd suggest a modified version of Mark's answer:

def intWithCommas(x):
    if type(x) not in [type(0), type(0L)]:
        raise TypeError("Parameter must be an integer.")
    if x < 0:
        return '-' + intWithCommas(-x)
    result = ''
    while x >= 1000:
        x, r = divmod(x, 1000)
        result = ",%03d%s" % (r, result)
    return "%d%s" % (x, result)

Recursion is useful for the negative case, but one recursion per comma seems a bit excessive to me.

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