Numpy : The Truth Value Of An Array With More Than One Element Is Ambiguous

I am really confused on why this error is showing up. Here is my code: import numpy as np x = np.array([0, 0]) y = np.array([10, 10]) a = np.array([1, 6]) b = np.array([3, 7]) poi

Solution 1:

Numpy arrays define a custom equality operator, i.e. they are objects that implement the __eq__ magic function. Accordingly, the == operator and all other functions/operators that rely on such an equality call this custom equality function.

Numpy's equality is based on element-wise comparison of arrays. Thus, in return you get another numpy array with boolean values. For instance:

x = np.array([1,2,3])
y = np.array([1,4,5])
x == y

returns

array([ True, False, False], dtype=bool)

However, the in operator in combination with lists requires equality comparisons that only return a single boolean value. This is the reason why the error asks for all or any. For instance:

any(x==y)

returns True because at least one value of the resulting array is True. In contrast

all(x==y) 

returns False because not all values of the resulting array are True.

So in your case, a way around the problem would be the following:

other_pairs = [p for p in points if all(any(p!=q) for q in max_pair)]

and print other_pairs prints the expected result

[array([1, 6]), array([3, 7])]

Why so? Well, we look for an item p from points where any of its entries are unequal to the entries of all items q from max_pair.

Solution 2:

Reason behind is that they're two different objects completely. np.array has it's own operators that work on it.

They are named the same as the global operators any and all but don't work exactly the same, and that distinction is reflected in the fact they are methods of np.array.

>>> x = np.array([0,9])
>>> x.any(axis=0)
True
>>> y = np.array([10, 10])
>>> y.all()
True
>>> y.all(axis=0)
True

meanwhile:

>>> bool([])
False
>>> bool([[]])
True
>>> bool([[]][0])
False

Notice how first result is false (in python2) an empty list is deemed False. However a list with another list in it, even if that one is empty, is not False but True. Evaluating the inside list returns False again because it's empty. Since any and all are defined over the conversion to bool the results you see are different.

>>> help(all)
all(...)
    all(iterable) -> bool
    Return True if bool(x) is True for all values x in the iterable.

>>> help(any)
any(...)
    any(iterable) -> bool
    Return True if bool(x) is True for any x in the iterable.

See a better explanations for logical numpy operators here