列表列表的所有组合

I'm basically looking for a python version of Combination of List<List<int>>

我基本上是在寻找一个python版本的List >的组合

Given a list of lists, I need a new list that gives all the possible combinations of items between the lists.

给定一个列表列表，我需要一个新的列表，它给出列表之间所有可能的项目组合。

[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]

The number of lists is unknown, so I need something that works for all cases. Bonus points for elegance!

列表的数量未知，所以我需要一些适用于所有情况的东西。为优雅加分!

5 个解决方案

#1

282

you need itertools.product:

你需要itertools.product:

>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

#2

The most elegant solution is to use itertools.product in python 2.6.

最优雅的解决方案是使用迭代工具。在python 2.6的产品。

If you aren't using Python 2.6, the docs for itertools.product actually show an equivalent function to do the product the "manual" way:

如果您没有使用Python 2.6，则使用itertools的文档。产品实际上显示了一个等价的功能来做产品的“手工”方式:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

#3

listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for list in itertools.product(*listOLists):
  print list;

I hope you find that as elegant as I did when I first encountered it.

我希望你能像我第一次见到它时那样优雅。

#4

Numpy can do it:

Numpy能做到:

 >>> import numpy
 >>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
 >>> [list(x) for x in numpy.array(numpy.meshgrid(*a)).T.reshape(-1,len(a))]
[[ 1, 4, 7], [1, 5, 7], [1, 6, 7], ....]

#5

Nothing wrong with straight up recursion for this task, and if you need a version that works with strings, this might fit your needs:

这个任务直接递归没有问题，如果您需要一个使用字符串的版本，这可能适合您的需要:

combinations = []

def combine(terms, accum):
    last = (len(terms) == 1)
    n = len(terms[0])
    for i in range(n):
        item = accum + terms[0][i]
        if last:
            combinations.append(item)
        else:
            combine(terms[1:], item)


>>> a = [['ab','cd','ef'],['12','34','56']]
>>> combine(a, '')
>>> print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']

#1

282