Ron Adam <[EMAIL PROTECTED]> wrote: > [EMAIL PROTECTED] wrote: > > Your solution Steven Bethard looks very intelligent, here is a small > > speed test, because sorting a list according another one is a quite > > common operation. > > (Not all solutions are really the same, as Alex has shown). > > Try this one. > > def psort10(s1, s2): > d = dict(zip(s2,s1)) > s1[:] = (d[n] for n in sorted(d.keys())) > > It's faster on my system because d.keys() is already sorted. But that > may not be the case on other versions of python.
If there are duplicates in s2, this solution will silently lose some items from s1. I would at least include an assert len(s2)==len(d) as the second statement to get some insurance that this doesn't occur. Alex -- http://mail.python.org/mailman/listinfo/python-list
