Question

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Iterate a list as pair (current, next) in Python

I have a list like this:

list = [A, B, C, D, E, F, G]

How can I group this to get the following Python output

[(A,B), (B,C), (C, D), (D,E), (E,F), (F,G)]

So the values are grouped by the secound value but the order is preserved…

Answer 1

Try using zip():

zip(lst, lst[1:])

Also, you shouldn’t use the name list, as you will override the built-in list type.

Example:

>>> lst = ['A', 'B', 'C', 'D', 'E', 'F', 'G']
>>> zip(lst, lst[1:])
[('A', 'B'), ('B', 'C'), ('C', 'D'), ('D', 'E'), ('E', 'F'), ('F', 'G')]

For a version that will work with generators or other one-pass iterables, you can use the pairwise recipe from the itertools docs:

from itertools import tee, izip

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