Le Tue, 19 May 2009 10:49:15 -0700, Emile van Sebille <em...@fenx.com> s'exprima ainsi:
> On 5/19/2009 10:19 AM spir said... > > Le Tue, 19 May 2009 11:36:17 +0200, > > spir <denis.s...@free.fr> s'exprima ainsi: > > > > [...] > > > > Thank you Albert, Kent, Sanders, Lie, Malcolm. > > > > This time regex wins! Thought it wouldn't because of the additional func > > call (too bad we cannot pass a mapping to re.sub). Actually the diff. is > > very small ;-) The relevant change is indeed using a dict. Replacing > > string concat with ''.join() is slower (tested with 10 times and 100 > > times bigger strings too). Strange... Membership test in a set is only > > very slightly faster than in dict keys. > > Hmm... this seems faster assuming it does the same thing... > > xlate = dict( (chr(c),chr(c)) for c in range(256)) > xlate.update(control_char_map) > > def cleanRepr5(text): > return "".join([ xlate[c] for c in text ]) > > > Emile Thank you, Emile. I thought at this solution (having a dict for all chars). But I cannot use it because later I will extend the app to cope with unicode (~ 100_000 chars). So that I really need to filter which chars have to be converted. A useful help I guess would be to have a builtin func that returns conventional char/string repr without "'...'" around. Denis PS By the way, you don't need (anymore) to build a list comprehension for an outer func that walks through a sequence: "".join( xlate[c] for c in text ) is a shortcut for "".join( (xlate[c] for c in text) ) [a generator expression already inside () needs no additional parens -- as long as there is no additional arg -- see PEP 289 http://www.python.org/dev/peps/pep-0289/] ------ la vita e estrany _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
