I got an array of arrays:
temp = np.empty(5, dtype=np.ndarray)
temp[0] = np.array([0,1])
I want to check if np.array([0,1]) in temp, which in the above example clearly is but the code returns false. I also tried temp.__contains__(np.array([0,1])) but also returns false. Why is this? Shouldn't it return true?
EDIT:
So __contain__ wont work. Is there any other way of checking?
One thing you need to understand, in python in general, is that, semantically, __contains__ is based on __eq__, i.e. it looks for an element which satisfies the == predicate. (Of course one can override the __contains__ operator to do other things, but that's a different story).
Now, with numpy arrays, the __eq__ does not return bool at all. Every person using numpy encountered this error at some point:
if temp == temp2:
print 'ok'
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
Which means, given the conflicting semantics of the __contains__ and ndarray.__eq__, it's not surprising this operation does not do what you want.
With the code you posted, there is no clear advantage to setting temp to be a np.array over a list. In either case, you can "simulate" the behavior of __contains__ with something like:
temp2 = np.array([0,1])
any( (a == temp2).all() for a in temp if a is not None )
If you explain why you choose to use an hetrogeneous np.array in the first place, I might come up with a more detailed solution.
And, of course, this answer would not be complete without @user2357112's link to this question.
__contains__in NumPy. It does crazy, useless shit.dtype=objectare not very well developed. Basic things like indexing work. But binary operations are hit and miss.