Getting the indices of all non-None items from a sub-list in Python?
As per the title, I have a nested lists like so (the nested list is a fixed length):
# ID, Name, Value
list1 = [[ 1, "foo", 开发者_如何学Go 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, i.e.:
[[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]))]
>>> [[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.
[[i for i in range(len(list1)) if list1[i] is not None] for _ in range(len(log[0]))]
The above seems to be about 10x faster than my original post.
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])]
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