Populate numpy matrix dynamically from array values?

Question

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).

DSM · Accepted Answer · 2013-12-20 05:58:49Z

3

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.]

edited Dec 20, 2013 at 5:58

answered Dec 20, 2013 at 5:52

DSM

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1 Comment

DSM Over a year ago

Note that you don't need to actually build A-- I only did it to check the shape. All you need is the number len(A[0]) and B.

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