Suppose I have 3 x 2 matrix
A = np.arange(3*2).reshape(3,2)
and wish to select elements by index array
I = [0, 1, 0]
to get
[[0],[3],[4]]
How would I do this?
Writing this way
A[:,[0,1,0]]
gives something completely different (what?)
Add a comment
|
1 Answer
What you can do is pass an iterable of the first dimesion value, and an iterable (e.g. a list) of the second dimension. Something like:
I = [0, 1, 0]
A[range(len(I)),I]
This produces:
>>> A[range(len(I)),I]
array([0, 3, 4])
In case you want it as a 2d array, you can use an additional reshape:
>>> A[range(len(I)),I].reshape(-1,1)
array([[0],
[3],
[4]])
A[:,[0,1,0]]gives something completely different (what?)
It creates a matrix where the first column is the first (0) column of A, the second column is the second (1) column of A, and the third column is again the first (0) column of A.
answered Sep 26, 2017 at 11:58
willeM_ Van Onsem
482k33 gold badges483 silver badges624 bronze badges
4 Comments
Dims
I there any function in
numpy with aggregation sematics like min, max and etc?willeM_ Van Onsem
np.min, np.max, etc.Dims
yes, but doing select, like
np.select(A, I, axis=1, keepdims=True)?willeM_ Van Onsem
AFAIK not,
np.choose is probably the closest, but it works on the first dimension, so you can transpose first, and then do the processing.