On Mar 31, 1:09 pm, Raymond Hettinger <pyt...@rcn.com> wrote: > On Mar 30, 4:25 pm, s...@sig.for.address (Victor Eijkhout) wrote: > > > I have two arrays, made with numpy. The first one has values that I want > > to use as sorting keys; the second one needs to be sorted by those keys. > > Obviously I could turn them into a dictionary of pairs and sort by the > > first member, but I think that's not very efficient, at least in space, > > and this needs to be done as efficiently as possible. > > Alf's recommendation is clean and correct. Just make a list of > tuples. > > FWIW, here's a little hack that does the work for you: > > >>> values = ['A', 'B', 'C', 'D', 'E'] > >>> keys = [50, 20, 40, 10, 30] > >>> keyiter = iter(keys) > >>> sorted(values, key=lambda k: next(keyiter)) > > ['D', 'B', 'E', 'C', 'A'] >
Another option: [values[i] for i in sorted(range(len(keys)), key=lambda i: keys[i])] Sort the indexes according to keys values, then use indexes to get the values. It might read more clearly when broken out into two lines: >>> sorted_indexes = sorted(range(len(keys)), key = lambda i: keys[i]) >>> sorted_indexes [3, 1, 4, 2, 0] >>> [values[i] for i in sorted_indexes] ['D', 'B', 'E', 'C', 'A'] The advantage of Raymond's solution is that he only creates one new Python list, whereas my solutions create an intermediate Python list of integers. I don't think my solution really is that space-wasteful, though, since by the time the second list gets created, any internal intermediate lists from CPython's sorted() implementation will probably have been cleaned up. -- http://mail.python.org/mailman/listinfo/python-list
