生成2-d陣列使用Python

Question

例如我已經通過使用列表解析以下2-d陣列

t = [[1,2,3], 
    [4,5], 
    [6,7]] 

我得到

>>> [[x, y, z] for x in t[2] for y in t[1] for z in t[0]] 
[[6, 4, 1], 
[6, 4, 2], 
[6, 4, 3], 
[6, 5, 1], 
[6, 5, 2], 
[6, 5, 3], 
[7, 4, 1], 
[7, 4, 2], 
[7, 4, 3], 
[7, 5, 1], 
[7, 5, 2], 
[7, 5, 3]] 

但如何，如果輸入具有更多的迭代比3個列表？我的意思是，我不想硬編碼t [2]等等。我想以包含任意數量的列表作爲輸入的t。無論如何，使用列表解析來做到這一點？

提前致謝！

Answer 1

使用itertools.product：

>>> import itertools 
>>> t = [[1,2,3], [4,5], [6,7]] 
>>> [x for x in itertools.product(*t[::-1])] 
[(6, 4, 1), 
(6, 4, 2), 
(6, 4, 3), 
(6, 5, 1), 
(6, 5, 2), 
(6, 5, 3), 
(7, 4, 1), 
(7, 4, 2), 
(7, 4, 3), 
(7, 5, 1), 
(7, 5, 2), 
(7, 5, 3)] 
>>> [list(x) for x in itertools.product(*t[::-1])] 
[[6, 4, 1], 
[6, 4, 2], 
[6, 4, 3], 
[6, 5, 1], 
[6, 5, 2], 
[6, 5, 3], 
[7, 4, 1], 
[7, 4, 2], 
[7, 4, 3], 
[7, 5, 1], 
[7, 5, 2], 
[7, 5, 3]] 

Answer 2

使用itertools.product：

In [1]: import itertools 

In [2]: t = [[1,2,3], [4,5], [6,7]] 

In [3]: list(itertools.product(*t[::-1])) 
Out[3]: 
[(6, 4, 1), 
(6, 4, 2), 
(6, 4, 3), 
(6, 5, 1), 
(6, 5, 2), 
(6, 5, 3), 
(7, 4, 1), 
(7, 4, 2), 
(7, 4, 3), 
(7, 5, 1), 
(7, 5, 2), 
(7, 5, 3)] 

Answer 3

看一看itertools模塊裏。 itertools.product函數做你想要的， ，除了你可能想要扭轉輸入順序。