import numpy as np
a = np.eye(2)
b = np.array([1,1],[0,1])
my_list = [a, b]
a in my_list returns true, but b in my_list returns "ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()". I can get around this by converting the arrays to strings or lists first, but is there a nicer (more Pythonic) way of doing it?
The problem is that in numpy the == operator returns an array:
>>> a == b
array([[ True, False],
[ True, True]], dtype=bool)
You use .array_equal() to compare arrays to a pure boolean value.
>>> any(np.array_equal(a, x) for x in my_list)
True
>>> any(np.array_equal(b, x) for x in my_list)
True
>>> any(np.array_equal(np.array([a, a]), x) for x in my_list)
False
>>> any(np.array_equal(np.array([[0,0],[0,0]]), x) for x in my_list)
False
a in my_list above) due to short-circuiting: if every element in the array is true on the first check it doesn't check the rest of the list?
any() function is short-circuiting too, btw). Also, if all elements are True, it is unambiguous that the boolean value is True. But if it's a mix of True and False then NumPy can't decide the implicit conversion, thus raising an Error.More info about the problem. If you form my_list with
my_list = [b,a]
the one that fails is a... Interesting problem.
