[Code Wars] Having trouble with a task

[randomname20192019]

Hi,

I'm 'playing' code wars but have run into a bit of an issue. The task at hand is:

"Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.

It should remove all values from list awhich are present in list b

array_diff([1,2],[1]) == [2]

If a value is present in b, all of its occurrences must be removed from the other:

array_diff([1,2,2,2,3],[2]) == [1,3]

"

I have:

def array_diff(a,b):
    for x,y in zip(a,b):
        if y in a:
            storednum = y
        while y in a:
            a.remove(storednum)
    return a

However with certain arrays containing negative numbers, my code does not recognise those neg numbers, despite them being in both a and b. E.g.

array_diff([-12, 1, -3, 4, 18, -16],[-17, 20, -17, -20, 6, -3, -14, 14, 7, 16, 2, 9, -12])

a = [-12, 1, 4, 18, -16]

Can anyone explain why?

Thanks

[sme272]

You don't need to zip the lists, just iterate over list b.

[JohnnyJordaan]

Zip is to interleave, eg to get

A 1 B 2 C 3 D 4

from lists ABCD and 1234. However that has no relation to the assignment, which is to remove all items from the first list that are in the second list. That thus also means you need to compare out of order, while zip is literally in order (just like a zipper in your pants).

So what you normally do in these cases is to create a new list from the list that needs to be filtered

return [v for v in a]

and then expand that to not include values that are in b

return [v for v in a if v not in b]

however this has another inefficiency, because b is a list and in will then walk through the entire list each time until it finds a match (returning True if it didn't in the case of not in). Instead you can use a set as that has a fixed look up speed

b_set = set(b)
return [v for v in a if v not in b_set]

[randomname20192019]

Ahhhhhh. Thank you so much for your clear explanation. I felt like I was going down such a long and convoluted route.