Python list: exchange every n-th value with the (n+1)th

Question

What is the best way to do this:

>>> replace2([1, 2, 3, 4, 5, 6])
[2, 1, 4, 3, 6, 5]

Do you need to handle cases where there are an odd number of elements? — Kenan Banks, Apr 09 '13 at 15:50
There's a really good discussion of going over multiple list elements here: http://stackoverflow.com/questions/5389507/iterating-over-every-two-elements-in-a-list — rongenre, Apr 09 '13 at 15:54

Martijn Pieters · Accepted Answer · 2013-04-09T16:10:46.870

def replace2inplace(lst):
    lst[1::2], lst[::2] = lst[::2], lst[1::2]

This uses slice assignment and slice step sizes to swap every pair in the list around, in-place:

>>> somelst = [1, 2, 3, 4, 5, 6]
>>> replace2inplace(somelst)
>>> somelst
[2, 1, 4, 3, 6, 5]

Otherwise you could use some itertools tricks:

from itertools import izip, chain

def replace2copy(lst):
    lst1, lst2 = tee(iter(lst), 2)
    return list(chain.from_iterable(izip(lst[1::2], lst[::2])))

which gives:

>>> replace2([1, 2, 3, 4, 5, 6])
[2, 1, 4, 3, 6, 5]

with the list() call optional; if you only need to loop over the result the generator is enough:

from itertools import izip, chain, islice, tee

def replace2gen(lst):
    lst1, lst2 = tee(iter(lst))
    return chain.from_iterable(izip(islice(lst1, 1, None, 2), islice(lst2, None, None, 2)))

for i in replace2gen([1, 2, 3, 4, 5, 6]):
    print i

where replace2gen() can take arbitrary iterators too.

score 0 · Answer 2 · answered Apr 09 '13 at 15:50

0

Out-of-place version:

def replace2(lst):
    return [x for pair in zip(lst[1::2], lst[::2]) for x in pair]

answered Apr 09 '13 at 15:50

Pavel Anossov

60,842
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score 0 · Answer 3 · answered Apr 09 '13 at 15:54

0

My choice:

x = range(1,7)
res = [e for e in itertools.chain(*zip(x[1::2],x[0::2]))]

answered Apr 09 '13 at 15:54

volcano

3,578
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Lee · Answer 4 · 2013-04-10T08:50:19.403

>>> a = [1, 2, 3, 4, 5, 6]
>>> sum([[x+1,x] for x in a if x&1 == True],[])
[2, 1, 4, 3, 6, 5]

EDIT: Some further explanation was requested:

The code steps through each element in the list a and, if the element is odd (x&1 == True) it puts that element and the next element into a list in reverse order ([x+1,x]).

With out the sum(...,[]) function we would have

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

The sum(...,[]) function removes the internal square brackets giving

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

This can be done more generally by using the index of the list rather than its value:

>>> a = [1, 2, 3, 4, 5, 6]
>>> sum([[a[x],a[x-1]] for x in range(len(a)) if x&1 == True],[])
[2, 1, 4, 3, 6, 5]

However, this will remove the last element of the list if its length is not even.

Python list: exchange every n-th value with the (n+1)th

4 Answers4