How should I grab pairs from a list in python?

Question

Say I have a list that looks like this:

['item1', 'item2', 'item3', 'item4', 'item5', 'item6', 'item7', 'item8', 'item9', 'item10']

Using Python, how would I grab pairs from it, where each item is included in a pair with both the item before and after it?

['item1', 'item2']
['item2', 'item3']
['item3', 'item4']
['item4', 'item5']
['item5', 'item6']
['item6', 'item7']
['item7', 'item8']
['item8', 'item9']
['item9', 'item10']

Seems like something i could hack together, but I'm wondering if someone has an elegant solution they've used before?

Answer 1

This should do the trick:

[(foo[i], foo[i+1]) for i in xrange(len(foo) - 1)]

Answer 2

A quick and simple way of doing it would be something like:

a = ['item1', 'item2', 'item3', 'item4', 'item5', 'item6', 'item7', 'item8', 'item9', 'item10']

print zip(a, a[1:])

Which will produce the following:

[('item1', 'item2'), ('item2', 'item3'), ('item3', 'item4'), ('item4', 'item5'), ('item5', 'item6'), ('item6', 'item7'), ('item7', 'item8'), ('item8', 'item9'), ('item9', 'item10')]

Answer 3

>>>a = ['item1', 'item2', 'item3', 'item4', 'item5', 'item6', 'item7', 'item8', 'item9', 'item10']
>>>[[a[i],a[i+1]] for i in range(len(a)-1)]

[['item1', 'item2'], ['item2', 'item3'], ['item3', 'item4'], ['item4', 'item5'], ['item5', 'item6'], ['item6', 'item7'], ['item7', 'item8'], ['item8', 'item9'], ['item9', 'item10']]

Answer 4

Check out the pairwise recipe in the itertools documentation.

Answer 5

this is also do the trick:

list_of_tupples = map(None,the_list[::2], the_list[1::2])