Remove sublist from list

Question

I want to do the following in Python:

A = [1, 2, 3, 4, 5, 6, 7, 7, 7]
C = A - [3, 4]  # Should be [1, 2, 5, 6, 7, 7, 7]
C = A - [4, 3]  # Should not be removing anything, because sequence 4, 3 is not found

So, I simply want to remove the first appearance of a sublist (as a sequence) from another list. How can I do that?

Edit: I am talking about lists, not sets. Which implies that ordering (sequence) of items matter (both in A and B), as well as duplicates.

Answer 1

Use sets:

C = list(set(A) - set(B))

In case you want to mantain duplicates and/or oder:

filter_set = set(B)
C = [x for x in A if x not in filter_set]

Answer 2

If you want to remove exact sequences, here is one way:

Find the bad indices by checking to see if the sublist matches the desired sequence:

bad_ind = [range(i,i+len(B)) for i,x in enumerate(A) if A[i:i+len(B)] == B]
print(bad_ind)
#[[2, 3]]

Since this returns a list of lists, flatten it and turn it into a set:

bad_ind_set = set([item for sublist in bad_ind for item in sublist])
print(bad_ind_set)
#set([2, 3])

Now use this set to filter your original list, by index:

C = [x for i,x in enumerate(A) if i not in bad_ind_set]
print(C)
#[1, 2, 5, 6, 7, 7, 7]

The above bad_ind_set will remove all matches of the sequence. If you only want to remove the first match, it's even simpler. You just need the first element of bad_ind (no need to flatten the list):

bad_ind_set = set(bad_ind[0])

---

Update: Here is a way to find and remove the first matching sub-sequence using a short circuiting for loop. This will be faster because it will break out once the first match is found.

start_ind = None
for i in range(len(A)):
    if A[i:i+len(B)] == B:
        start_ind = i
        break

C = [x for i, x in enumerate(A) 
     if start_ind is None or not(start_ind <= i < (start_ind + len(B)))]
print(C)
#[1, 2, 5, 6, 7, 7, 7]