浮点数:

float(filter(lambda x: x!=',', '1,234.52'))
# returns 1234.52

对于整数：

int(filter(lambda x: x!=',', '1,234'))
# returns 1234

下面是移除不相关部分并稍微清理后的地区分组代码:

(以下仅适用于整数)

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'

这里已经有一些很好的答案了。我只是想补充一下，以备将来参考。在python 2.7中，将有一个用于千位分隔符的格式说明符。根据python文档，它是这样工作的

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

在python3.1中，你可以这样做:

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

稍微扩展一下Ian Schneider的回答:

如果你想使用自定义千位分隔符，最简单的解决方案是:

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

例子

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

如果你想要像这样的德语表示，它会变得有点复杂:

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

下面是一行正则表达式替换:

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

仅适用于积分输出:

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'

或者对于小于4位的浮点数，将格式说明符更改为%.3f:

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

注意:不能正确工作与超过三个十进制数字，因为它将尝试分组小数部分:

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

它是如何工作的

让我们来分析一下:

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

下面是另一个使用生成器函数处理整数的变体:

def ncomma(num):
    def _helper(num):
        # assert isinstance(numstr, basestring)
        numstr = '%d' % num
        for ii, digit in enumerate(reversed(numstr)):
            if ii and ii % 3 == 0 and digit.isdigit():
                yield ','
            yield digit

    return ''.join(reversed([n for n in _helper(num)]))

下面是一个测试:

>>> for i in (0, 99, 999, 9999, 999999, 1000000, -1, -111, -1111, -111111, -1000000):
...     print i, ncomma(i)
... 
0 0
99 99
999 999
9999 9,999
999999 999,999
1000000 1,000,000
-1 -1
-111 -111
-1111 -1,111
-111111 -111,111
-1000000 -1,000,000

从评论到activestate recipe 498181，我重做了这个:

import re
def thous(x, sep=',', dot='.'):
    num, _, frac = str(x).partition(dot)
    num = re.sub(r'(\d{3})(?=\d)', r'\1'+sep, num[::-1])[::-1]
    if frac:
        num += dot + frac
    return num

它使用正则表达式特性:超前，即(?=\d)来确保只有三个数字组成的“后面”有一个数字的组才会有逗号。我说'after'是因为在这一点上字符串是反向的。

[::-1]只是反转一个字符串。