regex - How to match exception with double character with Python regular expression? -

got string , regex findall:

txt = """   dx d_2,222.22     ,,   dy h..{3,333.33}  ,, dz b#(1,111.11) ,,   dx-ay relative 4,444.44 ,,  """ n in re.findall( r'([-\w]+){1}\w+([^,{2}]+)\s+,,\w+', txt ) :     axis, value = n     print "a:", axis      print "v:", value 

in second (value) group trying match except double commas, seems catch 1 ",". can got in example simple (.*?) reasons got except ",,". thank you.

edit: see want accomplish use r'([-\w]+){1}\w+(.*?)\s+,,\w+' instead. give such output:

a: dx v: d_2,222.22 a: dy v: h..{3,333.33} a: dz v: b#(1,111.11) a: dx-ay v: relative 4,444.44 

edit #2: please, answer did not include double comma exception not needed. there solution...should be. patern :

any whitespace - word possibly "-" - " " - , ",," except itself.

[^,{2}] character class matches character except: ',', '{', '2', '}'

with "character class", called "character set", can tell regex engine match only 1 out of several characters.

it should ([^,]{2})+

(                        group , capture \1   [^,]{2}                  character except: ',' (2 times) )+                       end of \1  

---

get matched group index 1 , 2

 ([-\w]+)\s+(.*?)\s+,, 

here online demo

sample code:

import re p = re.compile(ur'([-\w]+)\s+(.*?)\s+,,') test_str = u"..."  re.findall(p, test_str) 

note: use \s* instead of \s+ if spaces optional.