遍历python中每个嵌套列表的第二个值

Question

Probably something really simple, but I'm having some real trouble with it. I have a list that's like so:

[[1, 2500],[3,4319],[8,3292],[3,34590]]

where the first value in each nested list is a score of some kind and the second value is the user id that the score corresponds to.

I'm trying to do some simple arithmetic on the list to remove all user IDs in a second list. however, I'm finding that I can't address only the second element of each nested list.

newlist = list(set(oldlist[][1]) - set(to_be_removed))

am I trying to get too creative with the brackets, or am I just missing something very simple?

Answer 1

在这里使用列表理解将是最简单的解决方案：

new_list = [x for x in old_list if x[1] not in to_be_removed]

Answer 2

Something like this?

old_list = [[1, 2500],[3,4319],[8,3292],[3,34590]]
removals = set([2500, 34590])
new_list = [[x, y] for x, y in old_list if y not in removals]