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

Question

I just discovered a logical bug in my code which was causing all sorts of problems. I was inadvertently doing a bitwise AND instead of a logical AND

Answer 1

r is a numpy (rec)array. So r["dt"] >= startdate is also a (boolean) array. For numpy arrays the & operation returns the elementwise-and of the two boolean arrays.

The NumPy developers felt there was no one commonly understood way to evaluate an array in boolean context: it could mean True if any element is True, or it could mean True if all elements are True, or True if the array has non-zero length, just to name three possibilities.

Since different users might have different needs and different assumptions, the NumPy developers refused to guess and instead decided to raise a ValueError whenever one tries to evaluate an array in boolean context. Applying and to two numpy arrays causes the two arrays to be evaluated in boolean context (by calling __bool__ in Python3 or __nonzero__ in Python2).

Your original code

mask = ((r["dt"] >= startdate) & (r["dt"] <= enddate))
selected = r[mask]

looks correct. However, if you do want and, then instead of a and b use (a-b).any() or (a-b).all().