创建不同列表的元组列表[重复]

Question

这个问题在这里已经有了答案：

得到一系列列表的笛卡尔积？ (16 个答案)

5 个月前关闭。

假设我有以下三个列表（l1、l2、l3）。 如何创建一个新列表，其中每个元素都是列表元素的元组（l_desired）？ 它实际上是 python 中 zip 方法的扩展版本。

简而言之，给定l1, l2, l3 ，我该如何创建l_desired ？

l1 = [1,2,3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']

l_desiered = [(1, 'a', 'Jess'), (1, 'b', 'Jess'), (1, 'c', 'Jess'), 
              (1, 'a', 'Muss'), (1, 'b', 'Muss'), (1, 'c', 'Muss'), 
              (2, 'a', 'Jess'), (2, 'b', 'Jess'), (2, 'c', 'Jess'), ...]

`

Answer 1

正如其他人已经说过的，使用itertools.product() ：

>>> l1 = [1,2,3]
>>> l2 = ['a', 'b', 'c']
>>> l3 = ['Jess', 'Muss']
>>> list(itertools.product(l1, l2, l3))
[(1, 'a', 'Jess'), (1, 'a', 'Muss'), (1, 'b', 'Jess'), (1, 'b', 'Muss'), (1, 'c', 'Jess'), (1, 'c', 'Muss'), (2, 'a', 'Jess'), (2, 'a', 'Muss'), (2, 'b', 'Jess'), (2, 'b', 'Muss'), (2, 'c', 'Jess'), (2, 'c', 'Muss'), (3, 'a', 'Jess'), (3, 'a', 'Muss'), (3, 'b', 'Jess'), (3, 'b', 'Muss'), (3, 'c', 'Jess'), (3, 'c', 'Muss')]

要实现问题中指定的排序顺序，您可以对结果进行如下排序：

>>> from operator import itemgetter
>>> sorted(itertools.product(l1, l2, l3), key=itemgetter(0,2,1))
[(1, 'a', 'Jess'), (1, 'b', 'Jess'), (1, 'c', 'Jess'), (1, 'a', 'Muss'), (1, 'b', 'Muss'), (1, 'c', 'Muss'), (2, 'a', 'Jess'), (2, 'b', 'Jess'), (2, 'c', 'Jess'), (2, 'a', 'Muss'), (2, 'b', 'Muss'), (2, 'c', 'Muss'), (3, 'a', 'Jess'), (3, 'b', 'Jess'), (3, 'c', 'Jess'), (3, 'a', 'Muss'), (3, 'b', 'Muss'), (3, 'c', 'Muss')]

Answer 2

使用itertools.product从一组列表中获取笛卡尔积（每个可能的值组合）

import itertools
l1 = [1, 2, 3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']
for element in itertools.product(l1, l2, l3):
    print(element)

或者您可以使用列表理解

[element for element in itertools.product(l1, l2, l3)]

或者试试这个，

list(itertools.product(l1, l2, l3))

Answer 3

如果这对您很重要，这将产生与 l_desired 相同的l_desired ，否则itertools.product解决方案会更整洁。

例子：

from pprint import pprint

l1 = [1,2,3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']

l_desired = [
    (l1_elem, l2_elem, l3_elem)
    for l1_elem in l1
    for l3_elem in l3
    for l2_elem in l2
]

pprint(l_desired)

Output：

[(1, 'a', 'Jess'),
 (1, 'b', 'Jess'),
 (1, 'c', 'Jess'),
 (1, 'a', 'Muss'),
 (1, 'b', 'Muss'),
 (1, 'c', 'Muss'),
 (2, 'a', 'Jess'),
 (2, 'b', 'Jess'),
 (2, 'c', 'Jess'),
 (2, 'a', 'Muss'),
 (2, 'b', 'Muss'),
 (2, 'c', 'Muss'),
 (3, 'a', 'Jess'),
 (3, 'b', 'Jess'),
 (3, 'c', 'Jess'),
 (3, 'a', 'Muss'),
 (3, 'b', 'Muss'),
 (3, 'c', 'Muss')]

Answer 4

如果顺序无关紧要，您可以使用 itertools：

from itertools import product


l1 = [1,2,3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']
l_desired = list(product(l1,l2,l3))

output：

[(1, 'a', 'Jess'), (1, 'a', 'Muss'), (1, 'b', 'Jess'), (1, 'b', 'Muss'), (1, 'c', 'Jess'), (1, 'c', 'Muss'), (2, 'a', 'Jess'), (2, 'a', 'Muss'), (2, 'b', 'Jess'), (2, 'b', 'Muss'), (2, 'c', 'Jess'), (2, 'c', 'Muss'), (3, 'a', 'Jess'), (3, 'a', 'Muss'), (3, 'b', 'Jess'), (3, 'b', 'Muss'), (3, 'c', 'Jess'), (3, 'c', 'Muss')]

Answer 5

您也可以简单地执行此操作，而无需仅通过list comprehension导入itertools ：

l1 = [1,2,3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']

最后：

l_desiered=list(((x,y,z) for x in l1 for y in l2 for z in l3))

现在如果你打印l_desired你会得到你想要的 output

Answer 6

嵌套 for 循环：

l1 = [1,2,3]
l2 = ['a', 'b', 'c']
l3 = ['Jess', 'Muss']
out = []
for l in l1:
    for l_2 in l2:
        for l_3 in l3:
            out.append((l, l_2, l_3))
print(out)
'''
[
(1, 'a', 'Jess'), (1, 'a', 'Muss'), (1, 'b', 'Jess'),
(1, 'b', 'Muss'), (1, 'c', 'Jess'), (1, 'c', 'Muss'), 
(2, 'a', 'Jess'), (2, 'a', 'Muss'), (2, 'b', 'Jess'), 
(2, 'b', 'Muss'), (2, 'c', 'Jess'), (2, 'c', 'Muss'), 
(3, 'a', 'Jess'), (3, 'a', 'Muss'), (3, 'b', 'Jess'), 
(3, 'b', 'Muss'), (3, 'c', 'Jess'), (3, 'c', 'Muss')
]'''