复制迭代器并生成无序自笛产品 [英] copying iterators and producing unordered self-cartesian product

问题描述

假设我有一个列表，我想生成所有唯一元素对的列表，而不考虑订单。一种方法是：

Say I have a list, and I want to produce a list of all unique pairs of elements without considering the order. One way to do this is:

mylist = ['W','X','Y','Z']
for i in xrange(len(mylist)):
    for j in xrange(i+1,len(mylist)):
        print mylist[i],mylist[j]
W X
W Y
W Z
X Y
X Z
Y Z

我想用迭代器做这个，我想到了以下内容，即使它没有简洁：

I want to do this with iterators, I thought of the following, even though it doesn't have brevity:

import copy
it1 = iter(mylist)
for a in it1:
    it2 = copy.copy(it1)
    for b in it2:
        print a,b

但这甚至不起作用。什么是更加pythonic和有效的方式，使用迭代器或zip等等？

But this doesn't even work. What is a more pythonic and efficient way of doing this, with iterators or zip, etc.?

推荐答案

这已经完成并且从Python 2.6开始包含在标准库中：

This has already been done and is included in the standard library as of Python 2.6:

import itertools

mylist = ['W', 'X', 'Y', 'Z']
for pair in itertools.combinations(mylist, 2):
    print pair        # pair is a tuple of 2 elements

对我来说好像是Pythonic; - ）

Seems pretty Pythonic to me ;-)

请注意，即使是如果你计算了很多组合，那么组合（）函数会返回一个迭代器，以便你可以立即开始打印它们。请参阅文档。

Note that even if you're calculating a lot of combinations, the combinations() function returns an iterator so that you can start printing them right away. See the docs.

此外，您将结果称为列表与其自身之间的笛卡尔积，但这并非严格正确：笛卡尔积将具有16个元素（4x4）。您的输出是其中的一部分，即只有2个元素的组合（不允许重复）列表的值。

Also, you are referring to the result as a Cartesian product between the list and itself, but this is not strictly correct: The Cartesian product would have 16 elements (4x4). Your output is a subset of that, namely only the 2-element combinations (with repetition not allowed) of the values of the list.

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