How to remove element from list without using for loop?-CodePudding

I have two list of tuples,A and B, that stores the pairs of data ids, and I want to remove pairs from A if the pair (x,y) or (y,x) in B are also in A.

I tried to do it by using for loop like below,

A = [(321, 33), (56, 991), (645, 2), (8876, 556)]
B = [(33, 321), (645, 2)]

for pair in B:
    if pair in A: 
        A.remove(pair)
    elif (pair[1], pair[0]) in A:
        A.remove((pair[1], pair[0]))

print(A)  # [(56, 991), (8876, 556)]

but when the elements in list is large, this code runs very slowly.

So I want to make this code faster, possibly avoid using for loops or completely different way. Can someone help me to solve this problem?

CodePudding user response：

If A has unique items, you can convert both lists to sets and use set difference, with the caveat that you add the reversed tuples into the B_set:

set_B = set(B).union([(j,i) for (i,j) in B])
out = list(set(A) - set_B)

Output:

[(321, 33), (645, 2)]

If A doesn't have unique items and you want to keep duplicate items, you can use list comprehension for the second line instead:

set_B = set(B).union([(j,i) for (i,j) in B])
out = [tpl for tpl in A if tpl in set_B]

CodePudding user response：

Provided that you use sets instead of lists and that the second set contains both variants of elements ((x, y) and (x, y))…

A = {(321, 33), (56, 991), (645, 2), (8876, 556)}
B = {(33, 321), (645, 2), (321, 33), (2, 645)}

You only need a set subtraction:

print(A - B)

CodePudding user response：

Instead of removing, create a new list by excluding the common elements:

[(x,y) for x,y in A if (x,y) not in B and (y,x) not in B]

Output:

[(56, 991), (8876, 556)]