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

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### Question :

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

Given a list

``````l = [1, 7, 3, 5]
``````

I want to iterate over all pairs of consecutive list items `(1,7), (7,3), (3,5)`, i.e.

``````for i in xrange(len(l) - 1):
x = l[i]
y = l[i + 1]
# do something
``````

I would like to do this in a more compact way, like

``````for x, y in someiterator(l): ...
``````

Is there a way to do do this using builtin Python iterators? I’m sure the `itertools` module should have a solution, but I just can’t figure it out.

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:]):
...
``````

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)
``````

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
``````

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)]

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`).

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
``````

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.