allclose returns True when comparing an empty array to a scalar

In Numpy v 1.9.1

import numpy as np
np.allclose(np.array([]), 1)

returns True. The value of the scalar doesn’t matter. It also returns True when comparing an empty array to an array with only one value, e.g.

np.allclose(np.array([]), np.array([123])

also returns True. It should return False unless comparing two empty arrays if you ask me.

[eric-wieser]

Adding a note to the docstring may be helpful.

I agree - lets leave this open as a reminder to do that

[spencerahill]

As stated in the original post, this isn’t a bug

I was agreeing with the sentiment by @gmcastil that “np.allclose should return an error of some sort if the lengths of the arrays are not the same.”

You’d expect np.allclose(a, 1) to have the same meaning as np.all(np.isclose(a, 1)). When a == []. then isclose also returns [], and all([]) is defined as `True. So that’s why you get the result you do.

Thanks, that is a nice explanation of the logic.

This is useful to check something like if every point in a list of points is close to another point, ie np.allclose(points_2d, point_1d)

That’s a reasonable use case that I hadn’t considered.

Can you give an argument for why broadcasting isn’t desirable here?

I’ll walk back my statement. I see now that the existing behavior is also consistent with that of exact equality:

>>>np.all(np.equal(np.ones([5, 1]), np.ones([2])))
True

There is certainly nothing in the docstring to suggest the kind of behavior you are seeing.

Adding a note to the docstring may be helpful. The crux of my confusion was that I had incorrectly assumed that these required matching array shapes. Closing this Issue also seems appropriate – the combination of nothing in the docstring about this and an open Issue on it led me to believe that it was indeed a bug.

Thanks very much for your help.