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Re: how to get all repeated group with regular expression

P: n/a
Steve Holden wrote:
Please keep this on the list.

scsoce wrote:
>Steve Holden wrote:
>>scsoce wrote:

say, when I try to search and match every char from variable length
string, such as string '123456', i tried re.findall( r'(\d)*, '12346' )

I think you will find you missed a quote out there. Always better to
copy and paste ...
, but only get '6' and Python doc indeed say: "If a group is contained
in a part of the pattern that matched multiple times, the last match is
returned."

So use

r'(\d*)'

instead and then the group includes all the digits you match.
cause the regx engine cannot remember all the past history then ? is it
nature to all regx engine or only to Python ?

Different regex engines have different capabilities, so I can't speak to
them all. If you wanted *all* the matches of *all* groups, how would you
have them returned? As a list? That would make the case where there was
only one match much tricker to handle. And what would you do with

r'((\w)*\d)*)'

Also, what about named groups? I can see enough potential implementation
issues that I can perfectly understand why Python works the way it does,
so I'd be interested to know why it doesn't makes sense to you, and what
you would prefer it to do.

regards
Steve
maybe my expression was not clear. I want to capture every matched part
in a repeated pattern, not only the last, say, for string '123456', I
want to back reference any one char, not only the '6'. and i know the
example is very simple, so we can got the whole string using regx and
get every char using other python statements, but if the pattern in
group is complex?
and I test in VIM, it can do the 'back reference':
==you text in vim:
123456
== pattern:
:%s/\(\d\)*/$2
text will turn to be:
2
'Fraid the Python re implementers just decided not to do it that way.
Nor Perl.

Probably what you want is re.findall(r"(\d)", "123456"), which returns a
list of what it captured.

Nov 21 '08 #1
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