从Python的子列表中获取所有非项目的索引？

Question

As per the title, I have a nested lists like so (the nested list is a fixed length):

        # ID,  Name, Value
list1 = [[ 1, "foo",    10],
         [ 2, "bar",  None],
         [ 3, "fizz",   57],
         [ 4, "buzz", None]]

I'd like to return a list (the number of items equal to the length of a sub-list from list1 ), where the sub-lists are the indices of rows without None as their Xth item, ie:

[[non-None ID indices], [non-None Name indices], [non-None Value indices]]

Using list1 as an example, the result should be:

[[0, 1, 2, 3], [0, 1, 2, 3], [0, 2]]

My current implementation is:

indices = [[] for _ in range(len(list1[0]))]
for i, row in enumerate(list1):
    for j in range(len(row)):
        if not isinstance(row[j], types.NoneType):
            indices[j].append(i)

...which works, but can be slow (the lengths of the lists are in the hundreds of thousands).

Is there a better/more efficient way to do this?

EDIT:

I've refactored the above for loops into nested list comprehensions (similar to SilentGhost's answer). The following line gives the same result as the my original implementation, but runs approximately 10x faster.

[[i for i in range(len(list1)) if list1[i][j] is not None] for j in range(len(log[0]))]

Answer 1

>>> [[i for i, j in enumerate(c) if j is not None] for c in zip(*list1)]
[[0, 1, 2, 3], [0, 1, 2, 3], [0, 2]]

in python-2.x you could use itertools.izip instead of zip to avoid generating intermediate list.

Answer 2

[[i for i in range(len(list1)) if list1[i] is not None] for _ in range(len(log[0]))]

上面的内容似乎比我的原始帖子快10倍。

Answer 3

import numpy as np

map(lambda a: np.not_equal(a, None).nonzero()[0], np.transpose(list1))
# -> [array([0, 1, 2, 3]), array([0, 1, 2, 3]), array([0, 2])]