Formatting question : printing numbers with thousands separators ?
Fred Pacquier
fredp at mygale.org.nospam
Mon Sep 3 04:37:36 EDT 2001
juneaftn at orgio.net (June Kim) said :
> * and the most(or more or less) obfuscated way as a one-liner is:
>
> def sep(s): return
> re.sub(r"([-+]?\d{1,3}(\.\d*)?)(?=(\d{3})*(\.|$))",r",\1",s)[1:]
tanzer at swing.co.at (Christian Tanzer) said :
> How about:
>>>> import re
>>>> sep_1000_pat = re.compile("(\d{1,3}) (?= (?: \d\d\d)+ (?! \d) )",
>>>> re.X) sep_1000_pat.sub(r"\g<1>,", "2342342342344567")
> '2,342,342,342,344,567'
> Short, but then you need Friedl's masterpiece on regular expression to
> understand it...
Ouch. Didn't know you could also write Perl in Python... :-))))))
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