CODEWITHSUNDEEP
pythonarraysnumpylogical-operators
dada
dada

Reputation: 1493

Logical OR without using numpy.logical_or

To know the elements of a numpy array that verifies two conditions, one can use the operator *: 

>>> a = np.array([[1,10,2],[2,-6,8]])
>>> a
array([[ 1, 10,  7],
       [ 2, -6,  8]])
>>> (a <= 6) * (a%2 == 0) # elements that are even AND inferior or equal to 6
array([[False, False, False],
       [ True,  True, False]], dtype=bool)

But how about OR? I tried to do this:

>>> (a%2 == 0) + (a <= 6) - (a%2 == 0) * (a <= 6)
array([[ True,  True, False],
       [False, False,  True]], dtype=bool)

but the result is false for the elements that verifies both conditions. I don't understand why.

Upvotes: 2

Views: 2277

Answers (2)

Alex Riley
Alex Riley

Reputation: 177098

@plonser's answer is the correct one: use +.

If you wanted to use multiplication again, you could remember that one of De Morgan's laws tells you that

A or B

is logically equivalent to

not ( not A and not B )

So in NumPy you could write:

>>> ~(~(a%2 == 0) * ~(a <= 6))
array([[ True,  True,  True],
       [ True,  True,  True]], dtype=bool)

But this isn't particularly readable.

Upvotes: 2

plonser
plonser

Reputation: 3363

You don't need the subtraction. The point is that + already behaves like the or operator

>>(a%2==0)+(a<=6)
array([[ True,  True,  True],
       [ True,  True,  True]], dtype=bool)

because "True+True=True".

When you subtract (a<=6)*(a%2==0) you turn all elements which satisfy both conditions into false.

It is easiest when you just do

>>(a<=6)|(a%2==0)
array([[ True,  True,  True],
       [ True,  True,  True]], dtype=bool)

Upvotes: 3

Related Questions