Iterate consecutive elements in a list in Python such that the last element combines with first

Question

I have a list:

L = [1,2,3,4,5,6,7,8]

I want to iterate consecutive elements in the list such that, when it comes to last element i'e 8 it pairs with the first element 1 .

The final output I want is:

[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8],[8,1]

I tried using this way:

for first,second in zip(L, L[1:]):
    print([first,second])

But I am getting only this result:

[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8]

How do I make a pair of last element with first? I have heard about the negative indexing property of a list.

Answer 1

You can simply extend the second list in zip() with a list with only the first item, something like:

for first, second in zip(L, L[1:] + L[0:1]):  # or simply zip(L, L[1:] + L[:1])
    print([first, second])

Answer 2

You can use cycle to cycle the lists (in combination with islice to skip the first element):

from itertools import cycle, islice

L = [1,2,3,4,5,6,7,8]
rest_L_cycled = islice(cycle(L), 1, None)
items = zip(L, rest_L_cycled)
print(list(items))

This is easily extensible. Note that it relies on the fact that zip halts on the shorter list (the second argument is an infinite cycle). It also does everything lazily and does not create any intermediate list (well, except for the print ed list) :-)

Answer 3

您还可以遍历L的索引，并且对于输出元组的第二项的索引，只需使用L的长度的剩余部分：

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

Answer 4

You can just append the front element(s) to the back.

for first,second in zip(L, L[1:] + L[:1]):
    print([first,second])

Answer 5

You can simply concatenate the list resulting from zip(L, L[1:]) with the pair formed by the last element L[-1] and first one L[0] and iterate over the result

for first,second in zip(L, L[1:]) + [(L[-1],L[0])]:
    print ([first,second])

It gives the desired outcome.

Answer 6

It's not a one-liner, but:

>>> L = [1,2,3,4,5,6,7,8]
>>> z = list(zip(L[:-1], L[1:]))
>>> z.append((L[-1], L[0]))
>>> z
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 1)]

Answer 7

This is a version (over-)using itertools :

from itertools import islice, cycle
L = [1,2,3,4,5,6,7,8]

for a, b in zip(L, islice(cycle(L), 1, None)):
    print(a, b)

The idea is to cycle over the second argument - that way zip runs until L itself is exhausted. This does not create any new lists.

Answer 8

for i in range(len(L)):
   first = L[i]
   second = L[(i+1) % len(L)]

Answer 9

You can simply use itertools.zip_longest with a fillvalue of the first item.

from itertools import zip_longest
list(map(tuple, zip_longest(L, L[1:], fillvalue=L[0]))