split string at commas respecting quotes when string not in csv format

[Tim Chase]

Paul McGuire wrote:
> On Mar 27, 5:19 am, Tim Chase <python.l... at tim.thechases.com> wrote:
>>>>  >>> import re
>>>>  >>> s = """a=1,b="0234,)#($)@", k="7" """
>>>>  >>> rx = re.compile(r'[ ]*(\w+)=([^",]+|"[^"]*")[ ]*(?:,|$)')
>>>>  >>> rx.findall(s)
>>>>  [('a', '1'), ('b', '"0234,)#($)@"'), ('k', '"7"')]
>>>>  >>> rx.findall('a=1, *DODGY*SYNTAX* b=2')
>>>>  [('a', '1'), ('b', '2')]
>>> I'm going to save this one and study it, too.  I'd like to learn
>>> to use regexes better, even if I do try to avoid them when possible :)
>> This regexp is fairly close to the one I used, but I employed the
>> re.VERBOSE flag to split it out for readability.  The above
>> breaks down as
>>
>>   [ ]*       # optional whitespace, traditionally "\s*"
>>   (\w+)      # tag the variable name as one or more "word" chars
>>   =          # the literal equals sign
>>   (          # tag the value
>>   [^",]+     # one or more non-[quote/comma] chars
>>   |          # or
>>   "[^"]*"    # quotes around a bunch of non-quote chars
>>   )          # end of the value being tagged
>>   [ ]*       # same as previously, optional whitespace  ("\s*")
>>   (?:        # a non-capturing group (why?)
>>   ,          # a literal comma
>>   |          # or
>>   $          # the end-of-line/string
>>   )          # end of the non-capturing group
> 
> Mightent there be whitespace on either side of the '=' sign?  And if
> you are using findall, why is the bit with the delimiting commas or
> end of line/string necessary?  I should think findall would just skip
> over this stuff, like it skips over *DODGY*SYNTAX* in your example.

Which would leave you with the solution(s) fairly close to what I 
original posited ;-)

(my comment about the "non-capturing group (why?)" was in 
relation to not needing to find the EOL/comma because findall() 
doesn't need it, as Paul points out, not the precedence of the 
"|" operator.)

-tkc