How to remove all items in a list that are in another list in Python 3?

Question

I’m trying to write a function that takes as input two lists: def array_diff(a, b):. In this function, I would like to return a filtered a list that will have been removed all items that are also present in list b. It will work as such:

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

Or:

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

I have written the following code:

def array_diff(a, b):
    for i in a:
        if i in b:
            a.remove(i)      
    return a

But even though I get no errors, when I try to run this function and list a has two items with the same value, and it is present in list b, it doesn’t filter it properly. I’m not sure why this is happening, but I already tried using a.pop(i) as well, but also didn’t work. I also tried iterating over the list with for i in range(len(a)), but I get errors saying I went over the list’s index.

Hopefully, you can help me, thank you!

Asked By: John Rayburn

||

Source

Answer 1

Use a list comprehension to create a new list with all the items of a that are not in b like this:

def array_diff(a, b):
    return [item for item in a if item not in b]

ls1 = [1, 2, 3, 4, 5]
ls2 = [2, 5]

print(array_diff(ls1, ls2))  # should be [1, 3, 4]
print(array_diff(ls2, ls1))  # should be []

Alternatively, you can use the filter function to filter out any item in a that is also in b.

def array_diff(a, b):    
    return list(filter(lambda item: item not in b, a))

ls1 = [1, 2, 3, 4, 5]
ls2 = [2, 5]

print(array_diff(ls1, ls2))  # should be [1, 3, 4]
print(array_diff(ls2, ls1))  # should be []

Answered By: Alex Mandelias

Answer 2

there is a simpler approach
eg.

a = [1,2,2,2,2,3]
b = [2]
c = list(set(a)-set(b)) # for output in list
print(c)

Answered By: Aryman Deshwal

Answer 3

If you need only unique values then:

def array_diff(a, b):     
    return set(a) - set(b)

Answered By: busfighter

How to remove all items in a list that are in another list in Python 3?

Question:

Answers: