Iterate over pairs in a list (circular fashion) in

Question

The problem is easy, I want to iterate over each element of the list and the next one in pairs (wrapping the last one with the first).

I've thought about two unpythonic ways of doing it:

def pairs(lst):
    n = len(lst)
    for i in range(n):
        yield lst[i],lst[(i+1)%n]

and:

def pairs(lst):
    return zip(lst,lst[1:]+[lst[:1]])

expected output:

>>> for i in pairs(range(10)):
    print i

(0, 1)
(1, 2)
(2, 3)
(3, 4)
(4, 5)
(5, 6)
(6, 7)
(7, 8)
(8, 9)
(9, 0)
>>> 

any suggestions about a more pythonic way of doing this? maybe there is a predefined function out there I haven't heard about?

also a more general n-fold (with triplets, quartets, etc. instead of pairs) version could be interesting.

Answer 1

This infinitely cycles, for good or ill, but is algorithmically very clear.

from itertools import tee, cycle

def nextn(iterable,n=2):
    ''' generator that yields a tuple of the next n items in iterable.
    This generator cycles infinitely '''
    cycled = cycle(iterable)
    gens = tee(cycled,n)

    # advance the iterators, this is O(n^2)
    for (ii,g) in zip(xrange(n),gens):
        for jj in xrange(ii):
            gens[ii].next()

    while True:
        yield tuple([x.next() for x in gens])


def test():
    data = ((range(10),2),
        (range(5),3),
        (list("abcdef"),4),)
    for (iterable, n) in data:
        gen = nextn(iterable,n)
        for j in range(len(iterable)+n):
            print gen.next()            


test()

gives:

(0, 1)
(1, 2)
(2, 3)
(3, 4)
(4, 5)
(5, 6)
(6, 7)
(7, 8)
(8, 9)
(9, 0)
(0, 1)
(1, 2)
(0, 1, 2)
(1, 2, 3)
(2, 3, 4)
(3, 4, 0)
(4, 0, 1)
(0, 1, 2)
(1, 2, 3)
(2, 3, 4)
('a', 'b', 'c', 'd')
('b', 'c', 'd', 'e')
('c', 'd', 'e', 'f')
('d', 'e', 'f', 'a')
('e', 'f', 'a', 'b')
('f', 'a', 'b', 'c')
('a', 'b', 'c', 'd')
('b', 'c', 'd', 'e')
('c', 'd', 'e', 'f')
('d', 'e', 'f', 'a')

Answer 2

i=(range(10))

for x in len(i):
    print i[:2]
    i=i[1:]+[i[1]]

more pythonic than this is impossible

Answer 3

[(i,(i+1)%len(range(10))) for i in range(10)]

replace range(10) with the list you want.

In general "circular indexing" is quite easy in python; just use:

a[i%len(a)] 

Answer 4

I, as always, like tee:

from itertools import tee, izip, chain

def pairs(iterable):
    a, b = tee(iterable)
    return izip(a, chain(b, [next(b)]))

Answer 5

This might be satisfactory:

def pairs(lst):
    for i in range(1, len(lst)):
        yield lst[i-1], lst[i]
    yield lst[-1], lst[0]

>>> a = list(range(5))
>>> for a1, a2 in pairs(a):
...     print a1, a2
...
0 1
1 2
2 3
3 4
4 0

If you like this kind of stuff, look at python articles on wordaligned.org. The author has a special love of generators in python.

Answer 6

def pairs(lst):
    i = iter(lst)
    first = prev = item = i.next()
    for item in i:
        yield prev, item
        prev = item
    yield item, first

Works on any non-empty sequence, no indexing required.