python 千分位輸出 1,233,232格式

摘抄自郵件列表，我的方法比較土，小數點也沒有加進去，locale未盡測試。
郵件收錄在：http://wiki.woodpecker.org.cn/moin/MiscItems/2009-01-13

最新方法：
>>> while True:
...     (s,count) = re.subn(r"(/d)(/d{3})((:?,/d/d/d)*)$",r"/1,/2/3",s)
...     if count == 0 : break

locale
國外同志們有遇到同樣的問題。看這裏：
http://bytes.com/groups/python/454763-number-format-function
其中有個回帖是這樣的。
This is a little faster:
def number_format(num, places=0):

"""Format a number according to locality and given places"""
locale.setlocale(locale.LC_ALL, "")
return locale.format("%.*f", (places, num), True)

I tested this ok with my test

再一例
eric <[email protected]>
reply-to        [email protected]
to      python-cn`CPyUG`華蟒用戶組 <[email protected]>
date    Tue, Jan 13, 2009 at 16:53
subject [CPyUG:76807] Re: python怎麼輸出1,233,232這種形式？

http://www.jaharmi.com/2008/05/26/format_numbers_with_the_python_locale_module

>>> import locale
>>> a = {'size': 123456789, 'unit': 'bytes'}
>>> print(locale.format("%(size).2f", a, 1))
123456789.00
>>> locale.setlocale(locale.LC_ALL, '') # Set the locale for your system
'en_US.UTF-8'
>>> print(locale.format("%(size).2f", a, 1))
123,456,789.00

DIY
smallfish <[email protected]>
reply-to        [email protected]
to      [email protected]
date    Tue, Jan 13, 2009 at 16:53
subject [CPyUG:76806] Re: python怎麼輸出1,233,232這種形式？

我試了一個土方法：

>>> s = "1234567890"
>>> s = s[::-1]
>>> a = [s[i:i+3] for i in range(0,len(s),3)]
>>> print (",".join(a))[::-1]

其實用perl實現感覺更簡單了些：
$size = "1234567890";
1 while $size =~ s/(/d)(/d{3})((:?,/d/d/d)*)$/$1,$2$3/;
print $size, "/n";