Qiangning Hong wrote:
> faulkner wrote:
> > re.findall('\([^\)]*\)|\[[^\]]*|\S+', s)
>
> sorry i forgot to give a limitation: if a letter is next to a bracket,
> they should be considered as one word. i.e.:
> "a(b c) d" becomes ["a(b c)", "d"]
> because there is no blank between "a" and "(".
This variation seems to do it:
import re
s = "a (b c) d [e f g] h i(j k) l [m n o]p q"
def splitup(s):
return re.findall('''
\S*\( [^\)]* \)\S* |
\S*\[ [^\]]* \]\S* |
\S+
''', s, re.VERBOSE)
print splitup(s)
# Prints
['a', '(b c)', 'd', '[e f g]', 'h', 'i(j k)', 'l', '[m n o]p', 'q']
Peace,
~Simon
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