Iterate over all pairs of consecutive items in a list [duplicate]

Answer #1:

Just use zip

>>> l = [1, 7, 3, 5]
>>> for first, second in zip(l, l[1:]):
...     print first, second
...
1 7
7 3
3 5

As suggested you might consider using the izip function in itertools for very long lists where you don’t want to create a new list.

import itertools

for first, second in itertools.izip(l, l[1:]):
    ...

Answer #2:

Look at pairwise at itertools recipes: http://docs.python.org/2/library/itertools.html#recipes

Quoting from there:

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

A General Version

A general version, that yields tuples of any given positive natural size, may look like that:

def nwise(iterable, n=2):                                                      
    iters = tee(iterable, n)                                                     
    for i, it in enumerate(iters):                                               
        next(islice(it, i, i), None)                                               
    return izip(*iters)   

Answer #3:

I would create a generic grouper generator, like this

def grouper(input_list, n = 2):
    for i in xrange(len(input_list) - (n - 1)):
        yield input_list[i:i+n]

Sample run 1

for first, second in grouper([1, 7, 3, 5, 6, 8], 2):
    print first, second

Output

1 7
7 3
3 5
5 6
6 8

Sample run 1

for first, second, third in grouper([1, 7, 3, 5, 6, 8], 3):
    print first, second, third

Output

1 7 3
7 3 5
3 5 6
5 6 8

Answer #4:

Generalizing sberry’s approach to nwise with comprehension:

def nwise(lst, k=2):
    return list(zip(*[lst[i:] for i in range(k)])) 

Eg

nwise(list(range(10)),3)

[(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6,
7), (6, 7, 8), (7, 8, 9)]

Answer #5:

A simple means to do this without unnecessary copying is a generator that stores the previous element.

def pairs(iterable):
    """Yield elements pairwise from iterable as (i0, i1), (i1, i2), ..."""
    it = iter(iterable)
    try:
        prev = next(it)
    except StopIteration:
        return
    for item in it:
        yield prev, item
        prev = item

Unlike index-based solutions, this works on any iterable, including those for which indexing is not supported (e.g. generator) or slow (e.g. collections.deque).

Answer #6:

You could use a zip.

>>> list(zip(range(5), range(2, 6)))
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

Just like a zipper, it creates pairs. So, to to mix your two lists, you get:

>>> l = [1,7,3,5]
>>> list(zip(l[:-1], l[1:]))
[(1, 7), (7, 3), (3, 5)]

Then iterating goes like

for x, y in zip(l[:-1], l[1:]):
    pass

Answer #7:

If you wanted something inline but not terribly readable here’s another solution that makes use of generators. I expect it’s also not the best performance wise :-/

Convert list into generator with a tweak to end before the last item:

gen = (x for x in l[:-1])

Convert it into pairs:

[(gen.next(), x) for x in l[1:]]

That’s all you need.