Tim Chase <[EMAIL PROTECTED]> writes: > Is there an obvious/pythonic way to remove duplicates from a > list (resulting order doesn't matter, or can be sorted > postfacto)? My first-pass hack was something of the form > > >>> myList = [3,1,4,1,5,9,2,6,5,3,5] > >>> uniq = dict([k,None for k in myList).keys() > > or alternatively > > >>> uniq = list(set(myList)) > > However, it seems like there's a fair bit of overhead > here...
It seems you haven't timed these to find out which one actually takes longer; the first one gives syntax errors. Rather than guessing about overhead, and things we can't put our finger on, let's actually finger it: $ python2.4 -m timeit -c 'raw = [3,1,4,1,5,9,2,6,5,3,5]; uniq = dict([(k,None) for k in raw]).keys()' 10000 loops, best of 3: 45 usec per loop $ python2.4 -m timeit -c 'raw = [3,1,4,1,5,9,2,6,5,3,5]; uniq = list(set(raw))' 10000 loops, best of 3: 21 usec per loop The set method is faster, in this example. It also seems to do what you want. Why do you believe you even need to create a list from the set? $ python2.4 -m timeit -c 'raw = [3,1,4,1,5,9,2,6,5,3,5]; uniq = set(raw)' 100000 loops, best of 3: 13.1 usec per loop Why not just create a set and use that? -- \ "It is forbidden to steal hotel towels. Please if you are not | `\ person to do such is please not to read notice." -- Hotel | _o__) sign, Kowloon, Hong Kong | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list
