I searched prior questions and didn't find one that quite fits this.
I am setting the value of all cells in A's jth column to 0 whose row indeces are less than m[j]:
import numpy as np
n = 8
A = np.ones([n,n])
m = np.array([1,1,1,2,2,2,3,4])
for j in range(n):
for i in range(m[j]):
A[i,j] = 0
How do I rewrite that script without 'for' loops?
You can use broadcasted NumPy comparison—
>>> A[np.arange(n)[:, None] < m] = 0
>>>
array([[0., 0., 0., 0., 0., 0., 0., 0.],
[1., 1., 1., 0., 0., 0., 0., 0.],
[1., 1., 1., 1., 1., 1., 0., 0.],
[1., 1., 1., 1., 1., 1., 1., 0.],
[1., 1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1., 1.]])
Here, calling [:, None] augments the shape of np.arange(n) such that the comparison < is broadcasted across every element of m, for each item in the range. This generates a boolean mask of the same shape as A which is then used to set values to 0.
Note—If A is guaranteed to be an array of ones, I would recommend Divakar's solution, which is very similar to this.
Use outer-broadcasted comparison against the ranged array of length n to get directly get A -
A = (m <= np.arange(n)[:,None]).astype(float)
To boost up the performance, we could use the type conversion to lower precision ones, like .astype(np.int) or even .astype(np.uint8). If you are okay with a boolean array, skip the type conversion altogether.
For int dtypes, we can also simply view into it, that saves us some memory -
(m <= np.arange(n)[:,None]).view(np.int8)
