On Wed, Jun 17, 2009 at 12:01 AM, Jeff Rush <j...@taupro.com> wrote: > Abhishek Tiwari wrote: > > > > *Ans. 1* > > > > values, items = list(zip(*sorted(zip(values,items), reverse=True))) > > > > *Ans. 2* > > new_values = sorted(values, reverse=True) > > new_items = [items[x] for x in map(values.index,new_values)] > > > > I would like to know which method is better and why? > > The first one is better because the second one has a bug. ;-) > > Because of the way the second one uses values.index, it assumes there is > a unique mapping between items and values. When a duplicate integer > appears in values, the second solution returns the wrong answer.
Here is a 4th solution. [item[1] for item in sorted(zip(values, items))][-1::-1] If you are too lazy to write a timeit function, you can use my timeit wrapper script at ASPN cookbook. http://code.activestate.com/recipes/389767/ I timed all the solutions provided so far. def f1(): list(zip(*sorted(zip(values,items), reverse=True))) def f2(): new_values = sorted(values, reverse=True); [items[x] for x map(values.index,new_values)] def f3(): [item[1] for item in sorted(zip(values, items))][-1::-1] def f4(): d = dict(zip(items, values)) new_items = sorted(items, key=d.__getitem__, reverse=True) Here are the times printed in order of f1...f4 for 10,000 passes. [an...@localhost python]$ python listsort.py 6.97 usec/pass 6.95 usec/pass 4.65 usec/pass 9.27 usec/pass Don't use sorted(..., reverse=True), it is a killer in terms of time. This is the reason why f4(...) does not work anywhere close to what is expected out of a dict(...) solution and why f1(...) performs much below par. For example, let us change f4(...) to not to use sorted(..., reverse=True) def f4(): d = dict(zip(items, values)) new_items = sorted(items, key=d.__getitem__)[-1::-1] The performance improves a bit, [an...@localhost python]$ python listsort.py 7.06 usec/pass 7.04 usec/pass 4.61 usec/pass 9.04 usec/pass To see the real effect of sorted(..., reverse=True), let us modify f3(...) to use it instead of reversing the list at the end. def f3(): [item[1] for item in sorted(zip(values, items), reverse=True)] [an...@localhost python]$ python listsort.py 7.19 usec/pass 7.13 usec/pass 6.39 usec/pass* 9.72 usec/pass The function f3() slows down almost 1.3 times. Don't use sorted(..., reverse=True). Instead reverse the list in place by using reverse slicing of l[-1::-1], which is about 1.5 times faster. > > -Jeff > _______________________________________________ > BangPypers mailing list > BangPypers@python.org > http://mail.python.org/mailman/listinfo/bangpypers > -- -Anand
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