>Please keep this on the list.
Quote:
>>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.
