difference array.any() vs. np.any(array())

Question

Simple question, confusing output: np.array([-1, 0, 1]).any() < 0 out: False (why??) but: np.any(np.array([-1, 0, 1])) out: True (as expected)

Acoording to documentation both commands should be equivalent, but they aren't.

np.any and np.ndarray.any do the same thing, your example just has `< 0` for one of them, but not the other — Chrysophylaxs, Jan 02 '23 at 13:39
what is `True<0`? Or did you mean to apply the `<` to the array first `(np.array([-1, 0, 1])< 0` — hpaulj, Jan 02 '23 at 15:52

Vincent Bénet · Answer 1 · 2023-01-02T14:24:53.783

Numpy is used to provide a method and a function that do the exact same thing.

assert my_array.any() == numpy.any(my_array)

Here my_array.any() is the method and numpy.any(my_array) is the function.

It both return a Boolean. Here you ask why np.array([-1, 0, 1]).any() < 0 returns False because np.array([-1, 0, 1]).any() is True which is equal to the value 1 and you ask if it is < 0 which is False.

import numpy as np
my_array = np.array([-1, 0, 1])
assert my_array.any() == True
assert my_array.any() == 1
assert my_array.any() > 0
assert np.any(my_array) == True
assert my_array.any() == np.any(my_array)

difference array.any() vs. np.any(array())

1 Answers1