how fast is Python code - another detail
skip at pobox.com
Thu Mar 4 20:46:58 CET 2004
>>> if sig not in d.keys():
>> if sig not in d.iterkeys():
if sig not in d:
The first must build the list of keys then search it sequentially. The
second doesn't have to build the entire list, but still searches
sequentially. The third effectively tries d[sig] and catches the KeyError
exception of sig is not a key in d. The third is O(1) in the general case.
The others are O(len(d)).
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