How can I iterate over overlapping (current, next) pairs of values from a list?

Question

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?.}

Answer 1

The documentation for 3.8 provides this recipe:

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, use itertools.izip instead of zip to get the same kind of lazy iterator (zip will instead create a list):

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, a and 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 a and 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.

Since tee() can take an n parameter (the number of iterators to produce), the same technique can be adapted to produce a larger "window". For example:

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)

Caveat: If one of the iterators produced by tee advances further than the others, then the implementation needs to keep the consumed elements in memory until every iterator has consumed them (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.

Answer 2

Roll your own!

def pairwise(iterable):
    it = iter(iterable)
    a = next(it, None)

    for b in it:
        yield (a, b)
        a = b

Answer 3

Starting in Python 3.10, this is the exact role of the pairwise function:

from itertools import pairwise

list(pairwise([1, 2, 3, 4, 5]))
# [(1, 2), (2, 3), (3, 4), (4, 5)]

or simply pairwise([1, 2, 3, 4, 5]) if you don't need the result as a list.

Answer 4

I’m just putting this out, I’m very surprised no one has thought of enumerate().

for (index, thing) in enumerate(the_list):
    if index < len(the_list):
        current, next_ = thing, the_list[index + 1]
        #do something

Answer 5

Since the_list[1:] actually creates a copy of the whole list (excluding its first element), and zip() creates a list of tuples immediately when called, in total three copies of your list are created. If your list is very large, you might prefer

from itertools import izip, islice
for current_item, next_item in izip(the_list, islice(the_list, 1, None)):
    print(current_item, next_item)

which does not copy the list at all.

Answer 6

Iterating by index can do the same thing:

#!/usr/bin/python
the_list = [1, 2, 3, 4]
for i in xrange(len(the_list) - 1):
    current_item, next_item = the_list[i], the_list[i + 1]
    print(current_item, next_item)

Output:

(1, 2)
(2, 3)
(3, 4)

Answer 7

I am really surprised nobody has mentioned the shorter, simpler and most importantly general solution:

Python 3:

from itertools import islice

def n_wise(iterable, n):
    return zip(*(islice(iterable, i, None) for i in range(n)))

Python 2:

from itertools import izip, islice

def n_wise(iterable, n):
    return izip(*(islice(iterable, i, None) for i in xrange(n)))

It works for pairwise iteration by passing n=2, but can handle any higher number:

>>> for a, b in n_wise('Hello!', 2):
>>>     print(a, b)
H e
e l
l l
l o
o !

>>> for a, b, c, d in n_wise('Hello World!', 4):
>>>     print(a, b, c, d)
H e l l
e l l o
l l o
l o   W
o   W o
  W o r
W o r l
o r l d
r l d !

Answer 8

This is now a simple Import As of 16th May 2020

from more_itertools import pairwise
for current, next in pairwise(your_iterable):
  print(f'Current = {current}, next = {nxt}')

Docs for more-itertools Under the hood this code is the same as that in the other answers, but I much prefer imports when available.

If you don't already have it installed then: pip install more-itertools

Example

For instance if you had the fibbonnacci sequence, you could calculate the ratios of subsequent pairs as:

from more_itertools import pairwise
fib= [1,1,2,3,5,8,13]
for current, nxt in pairwise(fib):
    ratio=current/nxt
    print(f'Curent = {current}, next = {nxt}, ratio = {ratio} ')

Answer 9

As others have pointed out, itertools.pairwise() is the way to go on recent versions of Python. However, for 3.8+, a fun and somewhat more concise (compared to the other solutions that have been posted) option that does not require an extra import comes via the walrus operator:

def pairwise(iterable):
  a = next(iterable)
  yield from ((a, a := b) for b in iterable)

Answer 10

Pairs from a list using a list comprehension

the_list = [1, 2, 3, 4]
pairs = [[the_list[i], the_list[i + 1]] for i in range(len(the_list) - 1)]
for [current_item, next_item] in pairs:
    print(current_item, next_item)

Output:

(1, 2)
(2, 3)
(3, 4)

Answer 11

A basic solution:

def neighbors( list ):
  i = 0
  while i + 1 < len( list ):
    yield ( list[ i ], list[ i + 1 ] )
    i += 1

for ( x, y ) in neighbors( list ):
  print( x, y )

Answer 12

temp=[-39.5, -27.5, -15.5, -3.5, 8.5, 20.5, 32.5, 44.5, 56.5, 68.5, 80.5, 92.5,104.5]

li_tup = []
count=0
i=0
new_tup=()
while i<len(temp):
    if count<1:
        new_tup = new_tup + (temp[i],)
        count=count+1
        i=i+1
    else:
        new_tup = new_tup + (temp[i],)
        count=0
        li_tup.append(new_tup)
        new_tup=()
print(li_tup)

One way to do it would be as above

Answer 13

code = '0016364ee0942aa7cc04a8189ef3'
# Getting the current and next item
print  [code[idx]+code[idx+1] for idx in range(len(code)-1)]
# Getting the pair
print  [code[idx*2]+code[idx*2+1] for idx in range(len(code)/2)]