如何比较Python的两个列表列表并通过列表中的单个元素删除重复项

Question

If you have two list of lists, and you would like to remove lists from one list of lists, that have a matching element for lists in the other list of lists, how do you achieve this?

Say you have

a = [[a,b,c],[d,e,f]]
b = [[g,h,i],[k,l,m],[n,o,c]]

How do you check the 3rd element in each list in b against the 3rd element in each list in a , and if there is a match then remove the list containing that match from b .

The desired end result here is

b = [[g,h,i],[k,l,m]]

because "c" is the 3rd element in one of the lists from a , and also is the 3rd element in one of the lists from b , so that list in b is removed.

I've tried

c = [x for x in b for y in a if not x[2] == y[2]]
c = [y for x,y in zip(a,b) if not x[2] == y[2]]
c = [[x for x in b if not x[2] == y[2]] for y in a]

so far, with all failing.

I've researched online and on SOF, and although there might be an answer for this, I've not been able to find one which deals with this specifically.

Answer 1

This might help.

a = [["a","b","c"],["d","e","f"]] 
b = [["g","h","i"],["k","l","m"],["n","o","c"]]

check = [i[2] for i in a]
print([i for i in b if i[2] not in check])

Output :

[['g', 'h', 'i'], ['k', 'l', 'm']]