python嵌套列表索引_python - 在嵌套列表中，重复列表的索引_performance_酷徒编程知识库...

考虑numpy来解决这个问题：import numpy as np

y = [

[1, 2, 3],

[1, 2, 3],

[3, 4, 5],

[6, 5, 4],

[4, 2, 5],

[4, 2, 5],

[1, 2, 8],

[1, 2, 3]

]

# Returns unique values of array, indices of that

# array, and the indices that would rebuild the original array

unique, indices, inverse = np.unique(y, axis=0, return_index=True, return_inverse=True)

下面是每个变量的打印输出：unique = [

[1 2 3]

[1 2 8]

[3 4 5]

[4 2 5]

[6 5 4]]

indices = [0 6 2 4 3]

inverse = [0 0 2 4 3 3 1 0]

如果我们看看逆变量，可以看到我们确实得到了[0.1.7]作为第一个唯一元素[1 ,2 ,3]的索引位置，我们现在需要做的就是对它们进行适当的分组。new_list = []

for i in np.argsort(indices):

new_list.append(np.where(inverse == i)[0].tolist())

输出：new_list = [[0, 1, 7], [2], [3], [4, 5], [6]]