2011/6/16 TheSaint <nob...@nowhere.net.no>: > Hello, > Is it possible to compile a regex by supplying a list? > > lst= ['good', 'brilliant'. 'solid'] > re.compile(r'^'(any_of_lst)) > > without to go into a *for* cicle? >
In simple cases, you can just join the list of alternatives on "|" and incorporate it in the pattern - e.g. in non capturing parentheses: (?: ...) cf.: >>> >>> lst= ['good', 'brilliant', 'solid'] >>> import re >>> re.findall(r"^(?:"+"|".join(lst)+")", u"solid sample text; brilliant QWERT") [u'solid'] >>> [using findall just to show the result directly, it is not that usual with starting ^ ...] However, if there can be metacharacters like [ ] | . ? * + ... in the alternative "words", you have to use re.escape(...) on each of these before. Or you can use a newer regex implementation with more features http://pypi.python.org/pypi/regex which was just provisionally enhanced with an option for exactly this usecase: cf. Additional features: Named lists on the above page; in this case: >>> import regex # http://pypi.python.org/pypi/regex >>> regex.findall(r"^\L<options>", u"solid sample text; brilliant QWERT", >>> options=lst) [u'solid'] >>> hth, vbr -- http://mail.python.org/mailman/listinfo/python-list
