How to make my code more simple by using a loop or someting else?

March 24, 2022

I have to return a symbol from a dictionary of lists when I am given a [x, y] coordinate. The dictionary of lists is this :

GOBBLET_REPRESENTATION = {
    1: ["▫", "◇", "◯", "□"],
    2: ["▪", "◆", "●", "■"],
}

Here is what I tried:

def formate_gobblet(gobblet):
    if gobblet == []:
        return "   "
    else:
        if gobblet == [1,0]:
            return f' {GOBBLET_REPRESENTATION.get(1)[0]} '
        elif gobblet == [1,1]:
            return f' {GOBBLET_REPRESENTATION.get(1)[1]} '
        elif gobblet == [1,2]:
            return f' {GOBBLET_REPRESENTATION.get(1)[2]} '
        elif gobblet == [1,3]:
            return f' {GOBBLET_REPRESENTATION.get(1)[3]} '
        elif gobblet == [2,0]:
            return f' {GOBBLET_REPRESENTATION.get(2)[0]} '
        elif gobblet == [2,1]:
            return f' {GOBBLET_REPRESENTATION.get(2)[1]} '
        elif gobblet == [2,2]:
            return f' {GOBBLET_REPRESENTATION.get(2)[2]} '
        elif gobblet == [2,3]:
            return f' {GOBBLET_REPRESENTATION.get(2)[3]} '

Having formate_gobblet(gobblet) with gobblet = [1,2] should return ◯ . This code does that but, I was wandering if there was a way to make my code simpler for the else statement since it looks like this could be resolved using a loop.

>Solution :

You do not need a loop for this. In fact, this can be done in one line using indexing itself.

def formate_gobblet(gobblet):
    if gobblet == []:
        return "   "
    else:
        return return f' {GOBBLET_REPRESENTATION.get(gobblet[0])[gobblet[1]]}  '

What the code above does is it checks for an empty gobbletotherwise returns the representation which is at the position pointed by the first element gobblet[0] and the second element gobblet[1]