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]

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).

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):
    ...

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]]

Nearly verbatim from https://stackoverflow.com/questions/1257413/iterate-over-pairs-in-a-list-circular-fashion-in-python/1257446#1257446:

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

>>> 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