Trying to index or remove a numpy array item from a python list, does not fail as expected on first item.
import numpy as np
# works:
lst = [np.array([1,2]), np.array([3,4])]
lst.index(lst[0])
# fails with: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
lst = [np.array([1,2]), np.array([3,4])]
lst.index(lst[1])
I understand why the second fails, I would like to understand why the first one works.
Because it's not leveraging the array's eq method, but rather just comparing the ids of the two objects:
lst = [np.array([1,2]), np.array([3,4])]
lst[0] => array([1, 2])
lst[1] => array([3, 4])
lst[0] == lst[1] => array([False, False], dtype=bool)
id(lst[0]) => 140232956554776
id(lst[1]) => 140232956554960
lst.index(lst[0]) => 0
lst.index(lst[1]) => 1
lst.index(lst[0]) == lst.index(lst[1]) => False
To get the behavior you want, you need to make a comparison that leverages the array's eq, such as:
lst.index(lst[0].tolist()) => 0
lst.index(lst[1].tolist()) => 1
Or:
lst.index(lst[0].tostring()) => 0
lst.index(lst[1].tostring()) => 1
But this is not going to be an efficient way to work with the data, since you'll be converting the arrays to lists or strings.
id(lst[0]) == id([lst[0])), if that fails an elementwise check is performed. That fails forlst[0] == lst[1]if those arenumpyarrays.Guarantees that identity implies equalityis what a buggy logic...l = [inf, nan]; i = float("inf"); n = float("nan"); print(l.index(i), 'Works'); print(l.index(nan), 'Works'); print(l.index(n), 'Fails')from numpy import nan, inf