Circular pairs from array?

Question

I can't believe this is nowhere to be found, but: I want all consecutive pairs from an array, including the last element with the first one. I tried:

[(a, b) for a, b in zip(list, list[1:])]

What is the most pythonic and efficient way to do it?

Eric Duminil · Accepted Answer · 2017-02-13T22:28:59.553

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Another approach would be to use modulo to jump back from the last element to the first one :

l = [1,2,3,4,5,6]
n = len(l)
[(l[i], l[(i+1) % n]) for i in range(n)]

It returns :

[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 1)]

edited Feb 13 '17 at 22:28

answered Feb 13 '17 at 22:20

Eric Duminil

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upvotes for everyone! This is likely the most efficient way, although, maybe not the easiest to understand immediately. – juanpa.arrivillaga Feb 13 '17 at 22:22
not the sexiest, but maybe the fastest solution since it doesn't create temporary lists. In C you would do like this since creating a deque or a list is just annoying. – Jean-François Fabre Feb 13 '17 at 22:22
@Jean-FrançoisFabre what are you talking about, Jean? Modular arithmetic is sexy! – juanpa.arrivillaga Feb 13 '17 at 22:23
from now on: no `lambda` in answer: I downvote :) – Jean-François Fabre Feb 13 '17 at 22:23
@EricDuminil one remark: quit python 2 and `xrange`. that's becoming old. – Jean-François Fabre Feb 13 '17 at 22:25
@Jean-FrançoisFabre: I'm just a Python noobie. But for every Python expert telling me Python 3 is the way to go, there's another one telling me that Python 2 will always be more often used than 3. – Eric Duminil Feb 13 '17 at 22:28
Yay, this will be the question for which I get my first python badge! :) – Eric Duminil Feb 13 '17 at 22:30
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I'd like to see noobies like you more often. `range` is less efficient in python 2 but works at least (you could mention: "use `xrange` for better efficiency" for python 2 lovers) – Jean-François Fabre Feb 13 '17 at 22:31
I have posted an answer that I believe achieves the same objective and faster. @Jean-FrançoisFabre it doesn't have a `lambda` so I'll get a downvote I guess :P This is a nice answer, but not sure it pays off. Did I represent your code fairly, Eric? – roganjosh Feb 14 '17 at 00:22

Jean-François Fabre · Answer 2 · 2017-02-13T22:29:18.493

10

you just have to add the first element to the second list:

l = [1,2,3,4,5,6]

r = [(a, b) for a, b in zip(l, l[1:]+l[:1])]

result:

[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 1)]

Aside: "pythonic" also means not using list as a variable name.

Variant: use itertools.ziplongest instead of zip and fill with first element (as a bonus, also works with numpy arrays since no addition):

import itertools
r = [(a, b) for a, b in itertools.zip_longest(l, l[1:], fillvalue=l[0])]

edited Feb 13 '17 at 22:29

answered Feb 13 '17 at 22:10

Jean-François Fabre

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Thanks :) Do you have any idea why this breaks down when I have an array instead of a list? I know they are mostly interchangeable, but not always. – John Feb 13 '17 at 22:15
@John What do you mean by an "array"? A `numpy.array`? – juanpa.arrivillaga Feb 13 '17 at 22:15
Yes, if l is a numpy.array this does not work and gives me the first two pairs only. – John Feb 13 '17 at 22:17
@John Because the `+` operator does not concatenate arrays, instead, it results in vectorized addition, broadcasting where necessary – juanpa.arrivillaga Feb 13 '17 at 22:19
`+` doesn't work the same for numpy arrays. Let me dig. Ok, you could do `r = [(a, b) for a, b in zip(l, l[1:].tolist()+l[:1].tolist())]` (convert to list to add) – Jean-François Fabre Feb 13 '17 at 22:20
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Use `np.concatenate([arr[1:],arr[:1]])` – juanpa.arrivillaga Feb 13 '17 at 22:20
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@juanpa.arrivillaga added a variant which is array-agnostic (and probably a tad faster, but still probably not as fast as Eric answer) – Jean-François Fabre Feb 13 '17 at 22:32

score 7 · Answer 3 · edited Feb 13 '17 at 22:17

7

Here's another approach using a collections.deque:

>>> from collections import deque
>>> x = list(range(7))
>>> d = deque(x)
>>> d.rotate(-1)
>>> list(zip(x,d))
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 0)]
>>>

edited Feb 13 '17 at 22:17

Moinuddin Quadri

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answered Feb 13 '17 at 22:16

juanpa.arrivillaga

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using `deque` and `rotate` for an actual useful purpose. Rare! – Jean-François Fabre Feb 13 '17 at 22:17

roganjosh · Answer 4 · 2017-02-14T09:17:47.180

5

I'm not disputing the coolness of some of these answers but nobody seems to have timed them and, in terms of being "Pythonic", they can get a little obscure compared to just appending or extending the circular part to the end.

l = xrange(1, 10000)

def simple_way(list_input):
    a = [(list_input[i], list_input[i+1]) for i in xrange(len(list_input)-1)]
    a.append((list_input[-1], list_input[0]))
    return a


timeit simple_way(l)
1000 loops, best of 3: 1.14 ms per loop

def top_rated_answer(list_input):
    n = len(list_input)
    a = [(list_input[i], list_input[(i+1) % n]) for i in xrange(n)]
    return a


timeit top_rated_answer(l)
1000 loops, best of 3: 1.37 ms per loop

edited Feb 14 '17 at 09:17

answered Feb 14 '17 at 00:08

roganjosh

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try `a.append((l[-1], l[0]))` for a potentially even faster answer. – Jean-François Fabre Feb 14 '17 at 06:54
@Jean-FrançoisFabre it's giving me the same speed either way – roganjosh Feb 14 '17 at 08:26
ok, but it avoids the creation of a temp list. Less characters & same speed => better :) – Jean-François Fabre Feb 14 '17 at 08:27
@Jean-FrançoisFabre that's true, I will update. I just wanted to demo that no fancy tricks were needed to make this fast (although it does take up a second line though :) ). – roganjosh Feb 14 '17 at 08:30
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one-liners are sexy but sometimes a good old C-like coding is faster, yes. Even, modulo is costly. – Jean-François Fabre Feb 14 '17 at 08:35
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Deserves more upvotes due to analysis if nothing else. Want the best way to do something? test! – Baldrickk Feb 14 '17 at 09:08
@Baldrickk indeed. I realised I had a couple of global references in the tests as I packaged them into functions for testing. I've updated the answer and the approach by Eric has improved more than my simple approach, but it's still appreciably faster to go with the approach that's easier to read. – roganjosh Feb 14 '17 at 09:20

Circular pairs from array?

4 Answers4