Why is numpy.ndarray([0]).all() returning True while numpy.array([0]).all() returning False?

Question

This seems surprising to me:

import numpy as np
assert np.ndarray([0]).all()
assert not np.array([0]).all()

What is going on here?

np.ndarray([0]).all() returns empty array where as np.array([0]) returns array with 0 as element. np.all() checks if there is any False i.e (0) — user10289025
– user10289025, Commented Jul 24, 2022 at 19:08
ndarray is a lower level constructor that we don't usually use. It's best reserved for cases where you have a preexisting data buffer. Here's it's more like np.empty. — hpaulj
– hpaulj, Commented Jul 24, 2022 at 20:33

sj95126 · Accepted Answer · 2022-07-24 19:08:32Z

1

Consider what those two calls produce:

>>> np.ndarray([0])
array([], dtype=float64)

This is an empty 1-D array, because you specified a single dimension of length 0.

>>> np.array([0])
array([0])

This is a 1-D array containing a single element, 0.

The definition of all() isn't just "all elements are True", it's also "no elements are False", as seen here:

>>> np.array([]).all()
True

So:

np.ndarray([0]).all()

is True because it's an empty array, while:

np.array([0]).all()

is False because it's an array of one non-True element.

answered Jul 24, 2022 at 19:08

sj95126

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1 Comment

Ah, I assumed the signatures of ndarray and array were compatible... didn't realize the first arg to ndarray was actually a shape. Thank you!