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Groups > comp.lang.python > #103533
| From | Mark Lawrence <breamoreboy@yahoo.co.uk> |
|---|---|
| Newsgroups | comp.lang.python |
| Subject | Re: Cycling through iterables diagonally |
| Date | 2016-02-26 11:00 +0000 |
| Message-ID | <mailman.144.1456484470.20994.python-list@python.org> (permalink) |
| References | <CAB_tDZwN8Onys+GLQme2cwKO2Fdo68b-erBpzky7Lv+r0ANNtA@mail.gmail.com> <nap7md$ktd$1@ger.gmane.org> |
On 26/02/2016 09:59, Peter Otten wrote: > Pablo Lucena wrote: > >> Say I have a group of 4 lists as follows: >> >> l1 = ['a1', 'a2', 'a3', 'a4'] >> l2 = ['b1', 'b2', 'b3', 'b4'] >> l3 = ['c1', 'c2', 'c3', 'c4'] >> l4 = ['d1', 'd2', 'd3', 'd4'] >> >> I would like to cycle through these lists "diagonally" in groups of >> len(list) (in this example, each list has 4 items). > >> Prior to this I was mucking around with index counting while looping, and >> popping lists out of a deque, popping an item out of the list, and >> appending the list back into the deque during each iteration. >> >> Is there a better/cleaner way to do this? I was hoping for some cool >> itertools logic =) > > I have a weak spot for the itertools myself, but I think in terms of clarity > it is hard to beat the conventional > > def diagonals(a): > N = len(a) > for i in range(N): > for k in range(N): > yield a[k][(k+i)%N] > > print(list(diagonals([l1, l2, l3, l4]))) > > Of course that's as uncool as it can get ;) > It might be uncool, but at least I can read it, unlike the earlier Jussi Piitulainen answer that made my head spin :) -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence
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Re: Cycling through iterables diagonally Mark Lawrence <breamoreboy@yahoo.co.uk> - 2016-02-26 11:00 +0000
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