Logical OR without using numpy.logical_or

Question

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.

Arithmetic operations on numpy booleans are cheesy, it is not the best of ideas to rely on them. Plus they make your code obscure and hard to interpret. If dealing with boolean arrays, use the bitwise operators: & for and, | for or, ^ for xor and ~ for not. — Jaime
– Jaime, Commented Mar 6, 2015 at 16:54
In case an answer solved your problem, please accept it as the solution like here — plonser
– plonser, Commented Mar 9, 2015 at 17:06

plonser · Accepted Answer · 2015-03-06 14:42:20Z

3

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)

edited Mar 6, 2015 at 14:42

answered Mar 6, 2015 at 14:35

plonser

3,3632 gold badges21 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

AnnanFay Over a year ago

My opinion only, but | should always be preferred over +. When I see + I start wondering why we are adding stuff together, or if in fact we are concatenating stuff. As soon as I see | I think of bitwise_or which I closely associate with logical_or.

Alex Riley · Accepted Answer · 2015-10-13 15:07:49Z

2

@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.

edited Oct 13, 2015 at 15:07

answered Mar 6, 2015 at 14:47

Alex Riley

178k46 gold badges274 silver badges247 bronze badges

Collectives™ on Stack Overflow

Logical OR without using numpy.logical_or

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related