Python - 列表/矩阵连接/组合

Question

我有 2 个列表/矩阵

a = [[1,   2,  3],
     [4,   5,  6],
     [7,   8,  9]]
b = [[10,  11,  12],
     [13,  14,  15],
     [16,  17,  18]]

我想得到这样的
结果

result = 
[[(1, 10), (1, 11), (1, 12), (2, 10), (2, 11), (2, 12), (3, 10), (3, 11), (3, 12)], 
 [(4, 13), (4, 14), (4, 15), (5, 13), (5, 14), (5, 15), (6, 13), (6, 14), (6, 15)], 
 [(7, 16), (7, 17), (7, 18), (8, 16), (8, 17), (8, 18), (9, 16), (9, 17), (9, 18)]]

在 Python 中该怎么做？？ 请帮忙

Answer 1

这不是用python编写列表的正确方法。我还冒昧地假设每一行都是一个列表。因此，a和b是列表的列表。

a = [
[1,   2,  3],
[4,   5,  6],
[7,   8,  9]
]
b= [
[10,  11,  12],
[13,  14,  15],
[16,  17,  18]
]
result = []
for ix, i in enumerate(a):
    temp = []
    for j in i:
        for k in b[ix]:
            temp.append((j,k))
    result.append(temp)
print(result)

[[(1， 10(， (1， 11(， (1， 12(， (2， 10(， (2， 11(， (2， 12(， (3

， 10(， (3， 11(， (3， 12(]， [(4， 13(， (4， 14(， (4， 15(， (5， 13(， (5，
14(， (5， 15(， (6， 13(， (6， 14(， (6， 15(]， [(7， 16(， (7， 17(， (7， 18(， (8，
16(， (8， 17(， (8， 18(， (9， 16(， (9， 17(， (9， 18(]]

Answer 2

您可以使用itertools.product来实现此目的：

from itertools import product
a = [[1,   2,  3],
     [4,   5,  6],
     [7,   8,  9]]
b = [[10,  11,  12],
     [13,  14,  15],
     [16,  17,  18]]
res = [list(product(a[i], b[i])) for i in range(len(a))]

结果：

[[(1, 10), (1, 11), (1, 12), (2, 10), (2, 11), (2, 12), (3, 10), (3, 11), (3, 12)], 
 [(4, 13), (4, 14), (4, 15), (5, 13), (5, 14), (5, 15), (6, 13), (6, 14), (6, 15)], 
 [(7, 16), (7, 17), (7, 18), (8, 16), (8, 17), (8, 18), (9, 16), (9, 17), (9, 18)]]