How to access a 'window' of elements from a list in every iteration?

Question

Suppose I have a list [a,b,c,d,e], I want to create a list of tuples [(a,b,c),(b,c,d),(c,d,e)]. Is it possible without using array indices as my loop variable? What is the best way of doing this?

score 1 · Answer 1 · answered Sep 10 '11 at 14:48

1

zip(li, li[1:],li[2:])

This zips together the elements. Each iteration corresponds to li[i], li[i+1], li[i+2], hence solving the original problem.

answered Sep 10 '11 at 14:48

Karthick

4,456
7
28
34

This copies the list over and over, and doesn't work for an iterator / generator rather than list. It also creates all the windows immediately on Python 2, consuming width of window * length of list memory. – agf Sep 10 '11 at 14:52

score 0 · Answer 2 · answered Sep 10 '11 at 14:49

0

Not sure this satisfies your constraints, but it should do it.

the_list = [1,2,3,4,5,6,7]
tuple_size = 3
tuple_list = [tuple(the_list[i-tuple_size:i]) for i in range(tuple_size, len(the_list))]

answered Sep 10 '11 at 14:49

Alec Munro

223
1
6

_without using array indices as my loop variable_ was in the question. – agf Sep 10 '11 at 14:54

score 0 · Answer 3 · answered Sep 10 '11 at 16:39

I keep this handy function around for when I want to do this.

import itertools

def nwise(it, n, wrapping=False):
    # nwise('ABCD', 2) --> AB BC CD
    # nwise('ABCDEF', 3) --> ABC BCD CDE DEF
    # nwise('ABCD', 2, wrapping=True) --> AB BC CD DA
    # nwise('ABCD', 3, wrapping=True) --> ABC BCD CDA DAB
    copies = itertools.tee(it, n)
    columns = []
    for index, copy in enumerate(copies):
        front = list(itertools.islice(copy, 0, index))
        if wrapping:
            columns.append(itertools.chain(copy, front))
        else:
            columns.append(copy)
    return itertools.izip(*columns)

How to access a 'window' of elements from a list in every iteration?

3 Answers3