Asked By – dcrosta
I sometimes need to iterate a list in Python looking at the “current” element and the “next” element. I have, till now, done so with code like:
for current, next in zip(the_list, the_list[1:]): # Do something
This works and does what I expect, but is there’s a more idiomatic or efficient way to do the same thing?
Some answers to this problem can simplify by addressing the specific case of taking only two elements at a time. For the general case of N elements at a time, see Rolling or sliding window iterator?.
Now we will see solution for issue: How can I iterate over overlapping (current, next) pairs of values from a list?
Here’s a relevant example from the itertools module docs:
import itertools def pairwise(iterable): "s -> (s0, s1), (s1, s2), (s2, s3), ..." a, b = itertools.tee(iterable) next(b, None) return zip(a, b)
For Python 2, you need
itertools.izip instead of
import itertools def pairwise(iterable): "s -> (s0, s1), (s1, s2), (s2, s3), ..." a, b = itertools.tee(iterable) next(b, None) return itertools.izip(a, b)
How this works:
First, two parallel iterators,
b are created (the
tee() call), both pointing to the first element of the original iterable. The second iterator,
b is moved 1 step forward (the
next(b, None)) call). At this point
a points to s0 and
b points to s1. Both
b can traverse the original iterator independently – the izip function takes the two iterators and makes pairs of the returned elements, advancing both iterators at the same pace.
One caveat: the
tee() function produces two iterators that can advance independently of each other, but it comes at a cost. If one of the iterators advances further than the other, then
tee() needs to keep the consumed elements in memory until the second iterator comsumes them too (it cannot ‘rewind’ the original iterator). Here it doesn’t matter because one iterator is only 1 step ahead of the other, but in general it’s easy to use a lot of memory this way.
tee() can take an
n parameter, this can also be used for more than two parallel iterators:
def threes(iterator): "s -> (s0, s1, s2), (s1, s2, s3), (s2, s3, 4), ..." a, b, c = itertools.tee(iterator, 3) next(b, None) next(c, None) next(c, None) return zip(a, b, c)
This question is answered By – Rafał Dowgird
This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0