John Machin <[EMAIL PROTECTED]> wrote:
> On 2/06/2006 9:08 AM, A.M wrote:
>> Hi,
>>
>> Is there any built in feature in Python that can format long integer
>> 123456789 to 12,3456,789 ?
>>
>> Thank you,
>> Alan
>
> Not that I know of, but this little kludge may help:
> 8<---
> import re
>
> subber = re.compile(r'^(-?\d+)(\d{3})').sub
>
> def fmt_thousands(amt, sep):
> if amt in ('', '-'):
> return ''
> repl = r'\1' + sep + r'\2'
> while True:
> new_amt = subber(repl, amt)
> if new_amt == amt:
> return amt
> amt = new_amt
>
> if __name__ == "__main__":
> for prefix in ['', '-']:
> for k in range(11):
> arg = prefix + "1234567890.1234"[k:]
> print "<%s> <%s>" % (arg, fmt_thousands(arg, ","))
> 8<---
Why not just port the Perl "commify" code? You're close to it, at least
for the regex:
# From perldoc perlfaq5
# 1 while s/^([-+]?\d+)(\d{3})/$1,$2/;
# Python version
import re
def commify(n):
while True:
(n, count) = re.subn(r'^([-+]?\d+)(\d{3})', r'\1,\2', n)
if count == 0: break
return n
--
Warren Block * Rapid City, South Dakota * USA
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