Question

I want to use boolean algebra instead of several conditionals, for example

def boo(x,y):
    if x==3 and y==1: return 4
    if x==3 and y==4: return 1
    if x==4 and y==1: return 2
    if x==4 and y==2: return 1
    if x==5 and y==1: return 3

what I want to do is

def simple(x,y):
    return x#y

and there are total 12 equations, I just want to directly return (x#y) where # is a boolean operator. I did this to a smaller problem where I luckily found out a relation. I want to do the same in this case also, how do I proceed with it?

Does this have any performance gains, because its not going through several if conditionals?
Is this normal practice?

sample:

 x    y   output
 1    2     3
 1    3     2
 1    4     5
 1    5     4

here a simple bitwise xor gate will do

def(x,y): return x^y

Answer 1

I don’t know whether trying to find a terser expression of the above logic would lead to more readable code; probably not. But you can rework the logic as-is to a more mathematical formulation:

def boo(x, y):
    p = (x, y)
    return (1 if p in ((3, 4), (4, 2)) else
            2 if p == (4, 1) else
            3 if p == (5, 1) else
            4 if p == (3, 1) else
            None)

Another option is to use a dictionary:

def boo(x, y):
    return {(3,4):1, (4,2):1, (4,1):2, (5,1):3, (3,1):4}.get((x, y), None)

If you know that all values will match the specified cases, you can write [(x, y)] instead of .get((x, y), None).