I'm trying to dynamically construct a 2-D matrix with numpy based on the values of an array, like this:
In [113]: A = np.zeros((5,5),dtype=bool)
In [114]: A
Out[114]: array([[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False]], dtype=bool)
In [116]: B = np.array([0,1,3,0,2])
In [117]: B
Out[117]: array([0, 1, 3, 0, 2])
Now, I'd like to use the values of B to assign the first n values of each row to A to True. For this A and B, the correct output would be:
In [118]: A
Out[118]: array([[False, False, False, False, False],
[ True, False, False, False, False],
[ True, True, True, False, False],
[False, False, False, False, False],
[ True, True, False, False, False]], dtype=bool)
The length of B will always equal the number of rows of A, and the the values of B will always be less than or equal to the number of columns of A. The size of A and the values of B are constantly changing, so I need to build these on the fly.
I'm certain that this has a simple(-ish) solution in numpy, but I've spent the last hour banging my head against variations of repeat, tile, and anything else I can think of. Can anyone help me out before I give myself a concussion? :)
EDIT: I'm going to need to do this a lot, so speed will be an issue. The only version that I can come up with for now is something like:
np.vstack([ [True]*x + [False]*(500-x) for x in B ])
but I expect that this will be slow due to the for loop (I would time it if I had anything to compare it to).
How about:
>>> A = np.zeros((5, 7),dtype=bool)
>>> B = np.array([0,1,3,0,2])
>>> (np.arange(len(A[0])) < B[:,None])
array([[False, False, False, False, False, False, False],
[ True, False, False, False, False, False, False],
[ True, True, True, False, False, False, False],
[False, False, False, False, False, False, False],
[ True, True, False, False, False, False, False]], dtype=bool)
(I changed the shape from (5,5) because I was getting confused about which axis was which, and I wanted to make sure I was using the right one.)
[Simplified from (np.arange(len(A[0]))[:,None] < B).T
-- if we expand B
and not A
, there's no need for the transpose.]
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