numpy logical_and：意外行为

Question

Let us say I give you the following boolean arrays:

b1 = np.array([ True,  True, False,  True ])
b2 = np.array([ True, False, False,  True ])
b3 = np.array([ True,  True,  True, False ])

If you AND them together, you would expect the following result:

b4 = np.array([ True, False, False, False ])

Right? If not, please explain. If we agree, then, why does the following happen?

>>> np.logical_and(b1, b2, b3)
array([ True, False, False,  True ])

np.logical_and(np.logical_and(b1, b2), b3) does give the expected result.

Answer 1

Look at the documentation of np.logical_and . Like most of the NumPy operator functions, the third parameter is an out parameter, specifying a destination array. It is not an operand! Putting b3 there will simply overwrite the contents of b3 .

Using & is clearer and simpler in most cases:

b4 = b1 & b2 & b3

Answer 2

The third argument to np.logical_and is the optional out parameter, which stores the result of the operation.

That is, calling np.logical_and(b1, b2, b3) overwrites b3 with the result of np.logical_and(b1, b2) .

Output arguments are useful for determining output type and general efficiency.

Answer 3

logical_and() is a binary operator, for your question, you can use:

np.all([b1, b2, b3], axis=0)
np.logical_and.reduce([b1, b2, b3], axis=0)