将列表分成两个列表的所有可能性 [英] All possibilities to split a list into two lists

问题描述

我有一个包含一些元素的列表，并且想要遍历所有可能的方式将该列表分为两个列表.我的意思是所有组合，因此顺序并不重要(即元素1和3可以在一个列表中，元素2可以在另一个列表中).目前，我这样做是这样的，其中facs是我的初始列表:

I have a list with some elements and want to iterate over all possible ways to divide this list into two lists. By that I mean all combinations, so the order doesn't matter (i.e. Element 1 and 3 could be in the one list and Element 2 in the other). Currently I do it like this, where facs is my initial list:

patterns = []
for i in range(2**(len(facs)-1)):
    pattern = []
    for j in range((len(facs)-1)):
        pattern.append(i//(2**j)%2)
    patterns.append(pattern)

for pattern in patterns:
    l1 = [facs[-1]]
    l2 = []
    for i in range(len(pattern)):
        if pattern[i] == 1:
            l1.append(facs[i])
        else:
            l2.append(facs[i])

因此，我基本上创建了一个长度为2^(len(facs)-1)的列表，并用一和零的每种可能组合填充它.然后我用facs'覆盖'每个模式，除了facs的最后一个元素总是在l1中，否则我将获得两次结果，因为我处理两个列表相同，无论如何列表是l1或l2.

So I basically create a list of length 2^(len(facs)-1) and fill it with every possible combination of ones and zeros. I then 'overlay' every pattern with facs, except for the last element of facs which is always in l1, as I'd otherwise get every result twice, as I handle two lists the same, no matter what lists is l1 or l2.

是否有更快，更优雅(更短/更pythonic)的方法来实现?

Is there a faster and more elegant (shorter/more pythonic) way to do this?

推荐答案

itertools 具有product()可用于生成掩码，而izip()可将列表结合起来以便于过滤.另外，由于它们返回迭代器，因此不会占用太多内存.

itertools has product() which could be used to generate the masks and izip() which could combine the lists for easy filtering. As a bonus, since they return iterators, they don't use much memory.

from itertools import *

facs = ['one','two','three']

l1 = []
l2 = []
for pattern in product([True,False],repeat=len(facs)):
    l1.append([x[1] for x in izip(pattern,facs) if x[0]])
    l2.append([x[1] for x in izip(pattern,facs) if not x[0]])

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