从数据集中删除不匹配的项目

Question

I have two datasets consisting of lists of nested lists such that each item in the list looks like list1[i]= [a, x, yb] and list2[j] = [c, x, y, d] and where the length of the two lists does not necessarily match. I'd like to be able to go through the lists, preserve their order, and eliminate any of the sub-lists that do not contain matching x values. In the end, I want to get two lists of identical length and where for each index, the x value is the same in corresponding sub lists.

Right now I have a somewhat messy code that assumes that the set of x values in list2 is a subset of those in list1 (true at the moment) and then proceeds to remove items where the x values don't match.

    len_diff = len(list1) - len(list2)
    if len_diff > 0:
        removed = []
        for (counter, row) in enumerate(list2):
            while list1[counter][1] != list2[counter][1]:
                removed.append(list1.pop(counter))
        new_len_diff = len(list1) - len(list2)
        if new_len_diff < 0:
            raise IndexError('Data sets do not completely overlap')
        else:
            for i in range(new_len_diff):
                removed.append(temp_data.pop())

So basically I'm removing any items that don't x values match until they start matching again and then removing the end of list1 beyond the x values in list2 (raising an exception if I've cut too much out of list1 ).

Is there a better way to do this?

I don't necessarily need to relax the assumption that all x values in list2 are in list1 at the moment but it would make this code more useful to me in the future for other data manipulations. The biggest hole in my code now is that if there is a gap in my list1 data, I'll remove my entire list.

Answer 1

You should try this:

list1 = list2 = [x for x in list1 if x[1] in zip(*list2)[1]]

EDIT

Based on the comments below, the OP adapted this answer to do what was wanted by doing

list1 = [x for x in list1 if x[1] in zip(*list2)[1]]
list2 = [x for x in list2 if x[1] in zip(*list1)[1]]