Python "Every Other Element" Idiom

Question

I feel like I spend a lot of time writing code in Python, but not enough time creating Pythonic code. Recently I ran into a funny little problem that I thought might have an easy, idiomatic solution. Paraphrasing the original, I needed to collect every sequential pair in a list. For example, given the list [1,2,3,4,5,6], I wanted to compute [(1,2),(3,4),(5,6)].

I came up with a quick solution at the time that looked like translated Java. Revisiting the question, the best I could do was

l = [1,2,3,4,5,6]
[(l[2*x],l[2*x+1]) for x in range(len(l)/2)]

which has the side effect of tossing out the last number in the case that the length isn't even.

Is there a more idiomatic approach that I'm missing, or is this the best I'm going to get?

related "What is the most “pythonic” way to iterate over a list in chunks?" http://stackoverflow.com/questions/434287/what-is-the-most-pythonic-way-to-iterate-over-a-list-in-chunks — jfs, May 21 '10 at 17:51

score 87 · Accepted Answer · answered Apr 13 '10 at 16:15

87

This will do it a bit more neatly:

>>> data = [1,2,3,4,5,6]
>>> zip(data[0::2], data[1::2])
[(1, 2), (3, 4), (5, 6)]

(but it's arguably less readable if you're not familiar with the "stride" feature of ranges).

Like your code, it discards the last value where you have an odd number of values.

answered Apr 13 '10 at 16:15

RichieHindle

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You can do a `data += [None]` before doing the above to handle the odd number of items scenario. – badp Apr 13 '10 at 16:22
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`some_list += [foo]` is written `some_list.append(foo)`. – Mike Graham Apr 13 '10 at 17:14
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How many times does this iterate over the data? Take a look at [this answer](http://stackoverflow.com/questions/5389507/iterating-over-every-two-elements-in-a-list/5389547#5389547) – johnsyweb Mar 22 '11 at 10:15

score 49 · Answer 2 · edited Jan 01 '19 at 12:45

49

The one often-quoted is:

zip(*[iter(l)] * 2)

I prefer this more readable version of the iter solution:

it = iter(l)
list(zip(it, it))
# [(1, 2), (3, 4), (5, 6)]

edited Jan 01 '19 at 12:45

cs95

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answered Apr 13 '10 at 16:20

Ignacio Vazquez-Abrams

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awsome, but reads like perl ;-) – Kamil Szot Apr 13 '10 at 16:28
True, but it's also the most efficient solution presented so far, I think. I'll test, brb. – jemfinch Apr 13 '10 at 16:30
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Yes, it's faster than RichieHindle's solution by about 10%, since it doesn't require any allocation of memory and only one iteration over the input list. – jemfinch Apr 13 '10 at 16:32
+1 because I never knew about * to unpack an argument list before! – Dana Apr 13 '10 at 16:45
This is a slick approach, but obviously less readable, especially to less advanced Python developers. This is why I like using `grouper` to abstract away what it does. – Mike Graham Apr 13 '10 at 17:13
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Know that, while RichieHindle won the accepted answer (really hard to ignore readability), this answer won my heart. – Matt Luongo Apr 15 '10 at 22:18
1

It's also worth noting that this is straight out of [the Python documentation](https://docs.python.org/3/library/functions.html#zip). – Chris Martin Oct 31 '14 at 19:26

score 9 · Answer 3 · answered Apr 13 '10 at 16:16

9

How about using the step feature of range():

[(l[n],l[n+1]) for n in range(0,len(l),2)]

answered Apr 13 '10 at 16:16

Isaac

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score 9 · Answer 4 · answered Apr 13 '10 at 16:23

I usually copy the grouper recipe from the itertools documentation into my code for this.

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

prasanna · Answer 5 · 2010-04-13T16:30:37.337

4

try this

def pairs(l, n):
    return zip(*[l[i::n] for i in range(n)])

So,

pairs([1, 2, 3, 4], 2) gives

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

edited Apr 13 '10 at 16:30

answered Apr 13 '10 at 16:22

prasanna

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score 4 · Answer 6 · answered Apr 13 '10 at 23:08

The right thing is probably not to compute lists, but to write an iterator->iterator function. This is more generic -- it works on every iterable, and if you want to "freeze" it into a list, you can use the "list()" function.

def groupElements(iterable, n):
    # For your case, you can hardcode n=2, but I wanted the general case here.
    # Also, you do not specify what to do if the 
    # length of the list is not divisible by 2
    # I chose here to drop such elements
    source = iter(iterable)
    while True:
        l = []
        for i in range(n):
            l.append(source.next())
        yield tuple(l)

I'm surprised the itertools module does not already have a function for that -- perhaps a future revision. Until then, feel free to use the version above :)

What's this like performance-wise? – LondonRob Aug 14 '15 at 06:52 — LondonRob, Aug 14 '15 at 06:52

score 4 · Answer 7 · answered Aug 11 '15 at 17:37

4

toolz is a well-built library with many functional programming niceties overlooked in itertools. partition solves this (with an option to pad the last entry for lists of odd length)

>>> list(toolz.partition(2, [1,2,3,4,5,6]))
[(1, 2), (3, 4), (5, 6)]

answered Aug 11 '15 at 17:37

beardc

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score 3 · Answer 8 · answered Apr 13 '10 at 21:08

3

If you don't want to lose elements if their number in list is not even try this:

>>> l = [1, 2, 3, 4, 5]
>>> [(l[i],  l[i+1] if i+1 < len(l) else None)  for i in range(0, len(l), 2)]
[(1, 2), (3, 4), (5, None)]

answered Apr 13 '10 at 21:08

Kamil Szot

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Python "Every Other Element" Idiom

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