从两个列表中对分布的夫妇进行排序

from itertools import islice, izip, cycle

list_1 = ['a','b','c']
list_2 = [1,2]

list(islice(izip(cycle(list_1), cycle(list_2)), len(list_1)*len(list_2)))

退换商品 [('a', 1), ('b', 2), ('c', 1), ('a', 2), ('b', 1), ('c', 2)]

你应该退房 itertools.product()

编辑:我是说 product() permutations()

import itertools

list_1 = ['a','b','c']
list_2 = [1,2]

# To pair list_1 with list_2 
paired = list(itertools.product(list_1, list_2))
# => [('a', 1), ('a', 2), ('b', 1), ('b', 2), ('c', 1), ('c', 2)]

# To get the sorting you desired:
wanted = sorted(paired, key=lambda x: x[1])
# [('a', 1), ('b', 1), ('c', 1), ('a', 2), ('b', 2), ('c', 2)]

自 产品() 返回一个迭代器,您不需要将其强制转换为 list() (对于较大的列表,您可能不应该这样做。)我刚刚添加了一个示例,为了便于演示,您应该决定自己测试它并打印值。

以您想要的方式执行,您可以翻转值:

>>> z = [[k, l] for l in x for k in y]
>>> z
[[1, 'a'], [2, 'a'], [3, 'a'], [1, 'b'], [2, 'b'], [3, 'b'], [1, 'c'], [2, 'c'], [3, 'c']]
>>> z.sort()
>>> z
[[1, 'a'], [1, 'b'], [1, 'c'], [2, 'a'], [2, 'b'], [2, 'c'], [3, 'a'], [3, 'b'], [3, 'c']]
>>> T = [[x[1], x[0]] for x in z]
>>> T
[['a', 1], ['b', 1], ['c', 1], ['a', 2], ['b', 2], ['c', 2], ['a', 3], ['b', 3], ['c', 3]]

在伪代码中(不太清楚python的语法):

给定大小为n的列表\u 1和大小为m的列表\u 2:

couples = []
for i=0 to n-1
   for j=0 to m-1
      couples.append( [ list_1[i], list_2[ (i+j) % m] ] )