Weighted average where one weight is infinite

Question

Using NumPy's weighted average, I expected an element with infinite weighting to dominate the result, but instead it returns NaN,

>>> np.average([1,2], weights=[np.inf, 1])
nan

Was this an intentional design? It seems counter-intuitive.

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EDIT: here's a more simple example:

>>> np.average([1], weights=[np.inf])
nan

Answer 1

Though not intentional, it is mathematically correct.

You end up with a formula like infinity/infinity. The result depends on which infinitiy is larger. And that is nonsense.

You need concrete numbers as weights, so you could use very large ones.