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python - 在python中转置嵌​​套列表

转载 作者:IT老高 更新时间:2023-10-28 20:51:20 25 4
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我喜欢将此列表中的每个项目移动到另一个嵌套列表中,有人可以帮我吗?

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', ...],
...
...]

最佳答案

zip*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']]

注意 map 在 Python 3 中返回一个 map 对象,所以你需要 list(map(list, zip(*a)))

使用 list comprehensionzip(*...),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']]

关于python - 在python中转置嵌​​套列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21444338/

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