在python中转置嵌​​套列表

Question

我喜欢将此列表中的每个项目移动到另一个嵌套列表，有人可以帮助我吗？

a = [['AAA', '1', '1', '10', '92'], ['BBB', '262', '56', '238', '142'], ['CCC', '86', '84', '149', '30'], ['DDD', '48', '362', '205', '237'], ['EEE', '8', '33', '96', '336'], ['FFF', '39', '82', '89', '140'], ['GGG', '170', '296', '223', '210'], ['HHH', '16', '40', '65', '50'], ['III', '4', '3', '5', '2']]

最后，我会像这样列出：

[['AAA', 'BBB', 'CCC', 'DDD', 'EEE', 'FFF'.....],
['1', '262', '86', '48', '8', '39', ...],
['1', '56', '84', '362', '33', '82', ...],
['10', '238', '149', '205', '96', '89', ...],
...
...]

Answer 1

将zip与*和map：

一起使用

>>> map(list, zip(*a))
[['AAA', 'BBB', 'CCC', 'DDD', 'EEE', 'FFF', 'GGG', 'HHH', 'III'],
 ['1', '262', '86', '48', '8', '39', '170', '16', '4'],
 ['1', '56', '84', '362', '33', '82', '296', '40', '3'],
 ['10', '238', '149', '205', '96', '89', '223', '65', '5'],
 ['92', '142', '30', '237', '336', '140', '210', '50', '2']]

请注意map在Python 3中返回一个地图对象，因此您需要list(map(list, zip(*a)))

将list comprehension与zip(*...)一起使用，这在Python 2和3中都可以正常工作。

[list(x) for x in zip(*a)]

---

NumPy方式：

>>> import numpy as np
>>> np.array(a).T.tolist()
[['AAA', 'BBB', 'CCC', 'DDD', 'EEE', 'FFF', 'GGG', 'HHH', 'III'],
 ['1', '262', '86', '48', '8', '39', '170', '16', '4'],
 ['1', '56', '84', '362', '33', '82', '296', '40', '3'],
 ['10', '238', '149', '205', '96', '89', '223', '65', '5'],
 ['92', '142', '30', '237', '336', '140', '210', '50', '2']]

Answer 2

通过列表理解：

[[x[i] for x in mylist] for i in range(len(mylist[0]))]

Answer 3

您也可以使用

a= np.array(a).transpose().tolist()

Answer 4

你也可以这样做：

row1 = [1,2,3]
row2 = [4,5,6]
row3 = [7,8,9]

matrix = [row1, row2, row3]
trmatrix = [[row[0] for row in matrix],[row[1] for row in matrix],  [row[2] for row in matrix]]