How to loop over a circular list, while peeking ahead and behind current element?

Question

With the following example list: L = ['a','b','c','d']

I'd like to achieve the following output:

>>> a d b
>>> b a c
>>> c b d
>>> d c a

Pseudo-code would be:

for e in L:
    print(e, letter_before_e, letter_after_e

Answer 1

You could just loop over L and take the index i minus and plus 1 modulo len(L) to get the previous and next element.

Answer 2

You're pretty much there

for i, e in enumerate(L):
    print(e, L[i-1], L[(i+1) % len(L)])

EDITED to add mod

Answer 3

it would probably be overkill in this case, but this is the general use-case for a circular doubly-linked list http://ada.rg16.asn-wien.ac.at/~python/how2think/english/chap17.htm

Answer 4

In my code I would use a moving window of 3 elements over the list prepended by the last element and appended by the first element:

from itertools import tee, izip, chain

def window(iterable,n):
    '''Moving window
    window([1,2,3,4,5],3) -> (1,2,3), (2,3,4), (3,4,5)
    '''
    els = tee(iterable,n)
    for i,el in enumerate(els):
        for _ in range(i):
            next(el, None)
    return izip(*els)


def chunked(L):
    it = chain(L[-1:], L, L[:1]) # (1,2,3,4,5) -> (5,1,2,3,4,5,1)
    for a1,a2,a3 in window(it,3): # (3,1,2,3,1) -> (3,1,2), (1,2,3), (2,3,1)
        yield (a2,a1,a3)


## Usage example ##
L = ['a','b','c','d']

for t in chunked(L):
    print(' '.join(t))

Answer 5

It's often simpler conceptually to keep track items you've already seen than to look ahead. The deque class is ideal for keeping track of n previous items because it lets you set a maximum length; appending new items automatically pushes old items off.

from collections import deque

l = ['a','b','c','d']
d = deque(l[-2:], maxlen=3)

for e in l:
    d.append(e)
    print d[1], d[0], d[2]

The only difference in this solution is that d c a will come first instead of last. If that matters, you can start out as though you've already seen one iteration:

from collections import deque

l = ['a','b','c','d']
d = deque(l[-1:] + l[:1], maxlen=3)

for e in l[1:] + l[:1]:
    d.append(e)
    print d[1], d[0], d[2]