Using list comprehension to get a list of entries not in another list, preserving order and quantity

Question

So I have a list, let's say

a = [1, 2, 3, 4, 2, 1, 2, 4]

1 appears twice, 2 appears three times, 4 appears twice. Now I define

b = [4, 2, 2]

Now I want a new list, c, that has the entries of a that are not in b. I tried using list comprehension:

c = [x for x in a if x not in b]

However, this omits the entry if it is in b, rather than seeing how many of each entry are in b and removing that many from a.

c == [1, 3, 1]

I would want it to be

c == [1, 3, 1, 2, 4]

Can anyone provide some help?

This definetely can help, see difflib http://stackoverflow.com/a/15646006/7086023 — lizaveta, Dec 04 '16 at 00:54

score 4 · Answer 1 · answered Dec 04 '16 at 00:48

4

You can loop through list b and remove each element from list a:

for i in b:
    a.remove(i)

a
# [1, 3, 1, 2, 4]

answered Dec 04 '16 at 00:48

Psidom

1

Thank you! Asking the question made me think about this route and I came to the same conclusion. – J. Doe Dec 04 '16 at 00:54
Nice & simple answer! One thing of note is the O(n*m) complexity, which can be reduced to O(n + m) but not with such a clear, concise implementation. To do that you'd probably want to use a histogram via `collections.Counter`. – orip Dec 04 '16 at 00:58

1 Answers1