how to print number with commas as thousands separators in Python 2.x

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

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 2

I got this to work:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US')
'en_US'
>>> locale.format("%d", 1255000, grouping=True)
'1,255,000'

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.

Answer 3

Just subclass long (or float, or whatever). This is highly practical, because this way you can still use your numbers in math ops (and therefore existing code), but they will all print nicely in your terminal.

>>> class number(long):

        def __init__(self, value):
            self = value

        def __repr__(self):
            s = str(self)
            l = [x for x in s if x in '1234567890']
            for x in reversed(range(len(s)-1)[::3]):
                l.insert(-x, ',')
            l = ''.join(l[1:])
            return ('-'+l if self < 0 else l) 

>>> number(-100000)
-100,000
>>> number(-100)
-100
>>> number(-12345)
-12,345
>>> number(928374)
928,374
>>> 345

Answer 4

Here is the locale grouping code after removing irrelevant parts and cleaning it up a little:

(The following only works for integers)

def group(number):
    s = '%d' % number
    groups = []
    while s and s[-1].isdigit():
        groups.append(s[-3:])
        s = s[:-3]
    return s + ','.join(reversed(groups))

>>> group(-23432432434.34)
'-23,432,432,434'

---

There are already some good answers in here. I just want to add this for future reference. In python 2.7 there is going to be a format specifier for thousands separator. According to python docs it works like this

>>> '{:20,.2}'.format(f)
'18,446,744,073,709,551,616.00'

In python3.1 you can do the same thing like this:

>>> format(1234567, ',d')
'1,234,567'

Answer 5

ok, but how to get number separated by dots?