在Python中迭代列表（循环方式）中的对

问题描述

问题很简单，我想迭代遍历列表中的每个元素和成对的下一个元素（用最后一个元素包装第一个元素）。

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已经考虑过两种非语言方式：

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]

和：

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

预期输出：

>>> 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)
>>>

有关更多pythonic方式的任何建议吗？也许有一个我没有听说过的预定义函数？

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

也是一个更通用的n-fold（有三连音，四重奏等而不是对）版本可能很有趣。

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

推荐答案

我自己编写了元组通用版本，我喜欢第一个版本，因为它简洁，简洁，我看得更多，它对我来说感觉越Pythonic ......毕竟，Pythonic比带有zip的单线程，星号参数扩展，列表理解，列表切片，列表连接和范围更多？

I've coded myself the tuple general versions, I like the first one for it's ellegant simplicity, the more I look at it, the more Pythonic it feels to me... after all, what is more Pythonic than a one liner with zip, asterisk argument expansion, list comprehensions, list slicing, list concatenation and "range"?

def ntuples(lst, n):
    return zip(*[lst[i:]+lst[:i] for i in range(n)])

即使对于大型列表，itertools版本也应该足够高效。来自itertools import的

The itertools version should be efficient enough even for large lists...

from itertools import *
def ntuples(lst, n):
    return izip(*[chain(islice(lst,i,None), islice(lst,None,i)) for i in range(n)])

非i的版本ndexable序列：

And a version for non-indexable sequences:

from itertools import *
def ntuples(seq, n):
    iseq = iter(seq)
    curr = head = tuple(islice(iseq, n))
    for x in chain(iseq, head):
        yield curr
        curr = curr[1:] + (x,)

无论如何，感谢大家的建议！ ： - ）

Anyway, thanks everybody for your suggestions! :-)

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