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!

`it doesn't filter it properly`. please explain, give a minimal reproducible example — AAAlex123, Oct 06 '20 at 13:46
Does this answer your question? [Remove all the elements that occur in one list from another](https://stackoverflow.com/questions/4211209/remove-all-the-elements-that-occur-in-one-list-from-another) — code11, Oct 06 '20 at 13:47
[`set`](https://docs.python.org/3/library/stdtypes.html#set-types-set-frozenset)s may make this significantly easier. — 0x5453, Oct 06 '20 at 13:49
@AAAlex123 here: `a was [1,2,2], b was [2], expected [1]: [1, 2] should equal [1]` — John Rayburn, Oct 06 '20 at 13:54

score 2 · Accepted Answer · answered Oct 06 '20 at 13:49

2

Use the filter function to filter out any items present in both lists like this:

def array_diff(a, b):    
    return list(filter(lambda item: item not in ls2, ls1))

ls1 = [1, 2, 3, 4, 5]
ls2 = [2, 5]
print(array_diff(ls1, ls2))

answered Oct 06 '20 at 13:49

AAAlex123

347
1
8

score 0 · Answer 2 · answered Oct 06 '20 at 13:50

0

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 Oct 06 '20 at 13:50

Aryman Deshwal

99
5

1

This would remove duplicates from list a that are not in list b as well – rbf Oct 06 '20 at 13:53
Oh yeah, i didnt consider that, I'll edit it soon. – Aryman Deshwal Oct 06 '20 at 15:28

score 0 · Answer 3 · answered Oct 06 '20 at 13:52

0

If you need only unique values then:

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

answered Oct 06 '20 at 13:52

busfighter

256
1
6

Hey! This didn't work... – John Rayburn Oct 06 '20 at 13:54
@JohnRayburn "didn't work" is suuuuuch a useful description of how it didn't work. – superb rain Oct 06 '20 at 13:55
@superbrain, this didn't give out any errors but all of the outputs were wrong. – John Rayburn Oct 06 '20 at 13:56
@JohnRayburn what do you mean "didn't work"? – busfighter Oct 06 '20 at 13:56
@busfighter I answered this in the previous comment. Thanks! – John Rayburn Oct 06 '20 at 13:57
@JohnRayburn "were wrong" is suuuuuch a useful description of how they were wrong. – superb rain Oct 06 '20 at 13:57
@JohnRayburn can you provide an example of wrong behaviour? – busfighter Oct 06 '20 at 13:58
Hey @busfighter, AAAlex123's answered worked. Thank you for your support! – John Rayburn Oct 06 '20 at 14:00
It also works but faster because set operations are faster in python) I think you can't provide an example because you pass the result into some checking system and actually dont' have those examples. Checking system doesn't take the result because it's of type `set` not `list` but the result made by my solution is right. – busfighter Oct 06 '20 at 14:07

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

3 Answers3