[Python] Strange numpy divide behaviour for scalars

I have been trying to upgrade a library which has a bunch of geometric operations for scalars so they will work with numpy arrays as well. While doing this I noticed some strange behaviour with numpy divide.

In original code checks a normalised difference between to variables if neither variable is zero, swapping across to numpy this ended up looking something like:

import numpy as np

a = np.array([0, 1, 2, 3, 4])
b = np.array([1, 2, 3, 0, 4])

o = np.zeros(len(a))
o = np.divide(np.subtract(a, b), b, out=o, where=np.logical_and(a != 0, b != 0))
print(f'First implementation: {o}')


where I passed in a output buffer initialised to zero for instances which could not be calculated; this returns:

First implementation: [ 0. -0.5 -0.33333333 0. 0. ]


I had to slightly modify this for scalars as out required an array, but it seemed fine.

a = 0
b = 4
o = None if np.isscalar(a) else np.zeros(len(a))
o = np.divide(np.subtract(a, b), b, out=o, where=np.logical_and(b != 0, a != 0))
print(f'Modified out for scalar: {o}')


returns

Modified out for scalar: 0.0.


Then ran this through some test functions and found a lot of them failed. Digging into this, I found that the first time I call the divide with a scalar with where set to False the function returns zero, but if I called it again, the second time it returns something unpredictable.

a = 0
b = 4
print(f'First divide: {np.divide(b, a, where=False)}')
print(f'Second divide: {np.divide(b, a, where=False)}')


returns

First divide: 0.0
Second divide: 4.0


Looking at the documentation, it says "locations within it where the condition is False will remain uninitialized", so I guess numpy as some internal buffer which is initially set to zero then subsequently it ends up carrying over an earlier intermediate value.

I am struggling to see how I can use divide with or without a where clause; if I use where I get an unpredictable output and if I don't I can't protect against divide by zero. Am I missing something or do I just need to have a different code path in these cases? I realise I'm half way to a different code path already with the out variable.

I would be grateful for any advice.

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