Paul McGuire wrote:
On Mar 27, 5:19 am, Tim Chase <python.l...@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
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