Unexpected Unresolved attribute reference 'all' for class 'bool'

When I do a matrix multiplication with numpy (Python 3.12.1, numpy 1.26.4, PyCharm 2024.3.1 (Professional Edition)), I get this warning which I believe is wrong:

Unresolved attribute reference 'all' for class 'bool'

Proof:

Minimial Reproducible Example

import numpy as np

a_matrix = np.array([[1, 2], [3, 4]])
b_matrix = np.array([[10], [20]])
a = [5.0]
b = [2.0]

if ((a_matrix @ np.array([[round(a[0], 0)], [round(b[0], 0)]])) == b_matrix).all():
    print("Success")

The issue is with the == operator. It appears the numpy class definition in the numpy library specifies that it returns a typing.Any with a type hint, so pycharm assumes it was the usual return type of builtins.bool.

Here is the line from the numpy library that pycharm is using, copied for context and posterity and abbreviated to the key components for brevity:

class _ArrayOrScalarCommon:
    ...
    def __eq__(self, other: Any, /) -> Any: ...
    ...
class ndarray(_ArrayOrScalarCommon, Generic[_ShapeT_co, _DType_co]):
    ...

I confirmed that the typing.Any hint causes this issue by creating a custom class and seeing what the return type hint suggested and it matched exactly. This code runs without errors for the same reason yours does, it returns a numpy.ndarray which supports the .all method.

from typing import Any
import numpy


class A:
    def __eq__(self, other) -> Any:
        return numpy.array([])


a = A()
(a == a).all()

Rearranging doesn't remove the issue, but makes the issue more obvious for the next person to come along and check your code with a static type checker by very clearly indicating what you expect the return type of that operation to be.

from typing import Any
import numpy


class A:
    def __eq__(self, other) -> Any:
        return numpy.array([])


a = A()
equal_result: numpy.ndarray = a == a
equal_result.all()

Whether you consider this to be a pycharm error by assuming that a return type of typing.Any is a builtins.bool or if it is a numpy error by type hinting typing.Any on a == between two numpy.ndarray error is up to you.

This has been discussed in the numpy community before, here is what appears to be the main discussion on the topic.