NumPy ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Solution 1:

The error message explains it pretty well:

ValueError: The truth value of an array with more than one element is ambiguous. 
Use a.any() or a.all()

What should bool(np.array([False, False, True])) return? You can make several plausible arguments:

(1) True, because bool(np.array(x)) should return the same as bool(list(x)), and non-empty lists are truelike;

(2) True, because at least one element is True;

(3) False, because not all elements are True;

and that's not even considering the complexity of the N-d case.

So, since "the truth value of an array with more than one element is ambiguous", you should use .any() or .all(), for example:

>>> v = np.array([1,2,3]) == np.array([1,2,4])
>>> v
array([ True,  True, False], dtype=bool)
>>> v.any()
True
>>> v.all()
False

and you might want to consider np.allclose if you're comparing arrays of floats:

>>> np.allclose(np.array([1,2,3+1e-8]), np.array([1,2,3]))
True

Solution 2:

As it says, it is ambiguous. Your array comparison returns a boolean array. Methods any() and all() reduce values over the array (either logical_or or logical_and). Moreover, you probably don't want to check for equality. You should replace your condition with:

np.allclose(A.dot(eig_vec[:,col]), eig_val[col] * eig_vec[:,col])