I am trying to apply a method that takes in two vectors and returns a resulting vector.

Below is an example input and expected result. Valid values are 0, 1, and -1:

a = [ 1,  0, -1,  0,  0,  1]
b = [ 0,  0, -1,  0,  1, -1]

result = my_func(a, b)
# [ 1,  0, -1,  0,  1, 0]

I am trying to apply the following logic to get a resulting vector. For each index of both vector:

1. If either a[i] or b[i] is zero and the other is either -1 or 1, set result[i] as the non-zero value.
2. If both are the same value, set result[i] as the shared value.
3. If a[i] and b[i] are (1, -1) or (-1, 1) respectively, set result[i] as zero.

My intial intuition is that a bitwise logic would be able to meet the above specifications, but I haven't been able to solve it so far:

np.bitwise_or(a,b)
# array([ 1,  0, -1,  0,  1, -1])

np.bitwise_and(a,b)
# array([ 0,  0, -1,  0,  0,  1])

np.bitwise_xor(a,b)
# array([ 1,  0,  0,  0,  1, -2])

If a logical pairing is not possible, would the most efficient method of meeting the specification be applying a lambda to each index?

CodePudding user response：

You can sum and

Code used to run the benchmark

perfplot.show(
    setup = lambda n: (np.random.choice([1, 0, -1], n).astype(np.int8),
                       np.random.choice([1, 0, -1], n).astype(np.int8)),
    kernels = [lambda a, b: (a | b) * (-a ^ b).view('bool'),
               lambda a, b: np.clip(a b, -1, 1)],
    labels=['or and multiply xor', 'add and clip'],
    n_range=[2**k for k in range(10,25,2)],
    equality_check=np.array_equal
)