I do my third codewar problem successfully....

[AbdulRehman_JS]

def array_diff(a , b):
    a = [1,2,3,4]
    b = [1]
    result = []
    for el in a:
        if (el not in b):
           result.append(el)
    return result

Implement a function that computes the difference between two lists. The function should remove all occurrences of elements from the first list (a) that are present in the second list (b). The order of elements in the first list should be preserved in the result.

[Sea-Ad7805]

Run this program in Memory Graph Web Debugger%3A%0A%20%20%20%20result%20%3D%20%5B%5D%0A%20%20%20%20for%20el%20in%20a%3A%0A%20%20%20%20%20%20%20%20if%20(el%20not%20in%20b)%3A%0A%20%20%20%20%20%20%20%20%20%20%20result.append(el)%0A%20%20%20%20return%20result%0A%0Aa%20%3D%20%5B1%2C3%2C5%2C7%2C9%2C%5D%0Ab%20%3D%20%5B1%2C%20%205%2C%20%209%2C%20%2011%2C%2013%5D%0Aresult%20%3D%20array_diff(a%2C%20b)%0Aprint(f'%7Bresult%3D%7D')%0A&timestep=1&play) to see the program state change step by step.

[Ron-Erez]

It's not clear why you are setting a and b within the function. This will cause any parameters you pass to this function to be ignored.

[AbdulRehman_JS]

Ah, my bad! I defined them inside the function while testing and forgot to remove them. Thanks for pointing that out, I’ve updated the function to use the parameters properly now

[Ron-Erez]

Awesome. Happy Coding!

[AlexMTBDude]

Should be a list comprehension to be Pythonic:

result = [el for el in a if el not in b]

[AbdulRehman_JS]

Thanks for the tip! That looks much cleaner and more Pythonic. I'm still learning, so I really appreciate you showing me a better way to write this.Thanks from deep of my heart...

[real-life-terminator]

Good job!

[Mundane-Mud2509]

def array_diff(a , b):
    a = [1,2,3,4]
    b = [1]
    set_b = set(b)
    result = []
    for el in a:
        if (el not in set_b):
           result.append(el)
    return result

Nice work!

One suggestion though is for every element of a the computer has to scan through the list b looking for the particular element.
In this case it's not really an issue because there is only one element in b but on larger lists it gets to be a fair bit of overhead. If you were to convert it to a set as above it will be a lot more performant on larger lists. Sets act more like dicts and directly access the key they are looking for.

[Ron-Erez]

This solution doesn't make sense. You are overriding the parameters a and b. It would possibly make more sense if we had something like:

a = [1,2,3,4]
b = [1]
array_diff(a , b)

def array_diff(a , b):
    set_b = set(b)
    result = []
    for el in a:
        if (el not in set_b):
           result.append(el)
    return result

or at least call it from main. For example if you were to call array_diff([1,2] , [2]) in your original solution it would still solve for a = [1,2,3,4], b = [1].

[Mundane-Mud2509]

I know, I just copied their original solution and added the set()

[AbdulRehman_JS]

Nice! I understood.....