Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
Groups > comp.lang.python > #103532
| From | Peter Otten <__peter__@web.de> |
|---|---|
| Newsgroups | comp.lang.python |
| Subject | Re: Cycling through iterables diagonally |
| Date | 2016-02-26 10:59 +0100 |
| Organization | None |
| Message-ID | <mailman.143.1456480794.20994.python-list@python.org> (permalink) |
| References | <CAB_tDZwN8Onys+GLQme2cwKO2Fdo68b-erBpzky7Lv+r0ANNtA@mail.gmail.com> |
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 ;)
Back to comp.lang.python | Previous | Next | Find similar | Unroll thread
Re: Cycling through iterables diagonally Peter Otten <__peter__@web.de> - 2016-02-26 10:59 +0100
csiph-web