Iterate through pairs of items in a Python list

Is it possible to iterate a list in the following way in Python (treat this code as pseudocode)?

a = [5, 7, 11, 4, 5]
for v, w in a:
    print [v, w]

And it should produce

[5, 7]
[7, 11]
[11, 4]
[4, 5]

Answers:

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Method 1

You can zip the list with itself sans the first element:

a = [5, 7, 11, 4, 5]

for previous, current in zip(a, a[1:]):
    print(previous, current)

This works even if your list has no elements or only 1 element (in which case zip returns an empty iterable and the code in the for loop never executes). It doesn’t work on generators, only sequences (tuple, list, str, etc).

Method 2

From the itertools recipes:

from itertools import tee

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

for v, w in pairwise(a):
    ...

Method 3

To do that you should do:

a =  [5, 7, 11, 4, 5]
for i in range(len(a)-1):
    print [a[i], a[i+1]]

Method 4

Nearly verbatim from Iterate over pairs in a list (circular fashion) in Python:

def pairs(seq):
    i = iter(seq)
    prev = next(i)
    for item in i:
        yield prev, item
        prev = item

Method 5

>>> a = [5, 7, 11, 4, 5]
>>> for n,k in enumerate(a[:-1]):
...     print a[n],a[n+1]
...
5 7
7 11
11 4
4 5

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0