On Mon, 2008-07-28 at 16:00 -0700, iu2 wrote: > On Jul 29, 12:10 am, John Krukoff <[EMAIL PROTECTED]> wrote: > > On Mon, 2008-07-28 at 16:24 -0500, Ervan Ensis wrote: > > > My programming skills are pretty rusty and I'm just learning Python so > > > this problem is giving me trouble. > > > > > I have a list like [108, 58, 68]. I want to return the sorted indices > > > of these items in the same order as the original list. So I should > > > return [2, 0, 1] > > > > > For a list that's already in order, I'll just return the indices, i.e. > > > [56, 66, 76] should return [0, 1, 2] > > > > > Any help would be appreciated. > > > > > -- > > >http://mail.python.org/mailman/listinfo/python-list > > > > If your lists aren't so large that memory is an issue, this might be a > > good place for a variation of decorate, sort, undecorate. > > > > >>> listToSort = [ 108, 58, 68 ] > > >>> decorated = [ ( data, index ) for index, data in > > > > enumerate( listToSort ) ]>>> decorated > > > > [(108, 0), (58, 1), (68, 2)]>>> result = [ None, ] * len( listToSort ) > > >>> for sortedIndex, ( ignoredValue, originalIndex ) in > > > > enumerate( sorted( decorated ) ): > > ... result[ originalIndex ] = sortedIndex > > ...>>> result > > > > [2, 0, 1] > > > > -- > > John Krukoff <[EMAIL PROTECTED]> > > Land Title Guarantee Company > > Inspired by your idea and the above one, here is another try: > > >>> a0 = [108, 58, 68, 108, 58] > >>> a1 = [(x, y) for x, y in enumerate(a0)]
You know this line is a no-op, right? > >>> a1 > [(0, 108), (1, 58), (2, 68), (3, 108), (4, 58)] > >>> a2 = sorted(a1, lambda x, y: cmp(x[1], y[1])) If you're going to do the unpacking here for the sort, just use enumerate directly. Since this is a simple case, should use the key parameter instead of the cmp parameter for speed. Can also use the operator module to avoid a bit of lambda overhead: >>> import operator >>> a2 = sorted( enumerate( a0 ), key = operator.itemgetter( 1 ) ) > >>> a2 > [(1, 58), (4, 58), (2, 68), (0, 108), (3, 108)] > >>> a3 = [a2.index(x) for x in a1] > >>> a3 > [3, 0, 2, 4, 1] > > The idea is to make each item unique (by making it a tuple), and then > apply the naive solution. > > -- > http://mail.python.org/mailman/listinfo/python-list Using index is still kinda evil, due to the exponential time required since you're rescanning half the list (on average) to find each index value. -- John Krukoff <[EMAIL PROTECTED]> Land Title Guarantee Company -- http://mail.python.org/mailman/listinfo/python-list
