python：选择一个具有列表长度限制的子列表

Question

players = [ a long list of (id, float) tuples with each id unique, order by highest float]

on_teams = [ a list of unique ids, every on_teams id is also in the players list]

picked = set(on_teams)
best_remaining = []
for best_player in players:
    if best_player[0] not in picked:
        best_remaining.append(best_player)
    if len(best_remaining) == 5: break

When I use six lines of code to do a simple thing, such as "get the five best remaining players", I wonder if there isn't a more elegant, pythonic solution. It's a simple problem, no doubt, but is there a better way to code it?

UPDATE: My code run 100,000 times, runs in 0.24 secs. When I run:

best_remaining = [(id, float) for id, float in players if id not in picked][:5]

The code runs in 4.61 secs (100,000x). So the code looks and scans nicer, but it create the whole list then slices it. So now my question is a little different. With speed as a constraint, is there a better way to code up the search for the '5 best remaining players`?

UPDATE:

best_remaining = list(islice((p for p in players if p[0] not in picked), 5))

This code runs trivially longer than my code. To me at least, it has the value of a list comprehension. And best of all, it shows me a good place to work on my code habits. Thanks

Answer 1

Use a generator, and then islice it to a max of 5 results...

from itertools import islice
picked = set(on_teams)
players = list(islice((p for p in players if p[0] not in picked), 5))

Answer 2

numpy is great for something like this

players = numpy.array([[1,1.1],[2,1.2],[3,3.2],[4,4.4]])
on_teams = [1,3]
print players[~numpy.in1d(players[:,0],on_teams)]  #all players not on teams (in example 2,4)
# now you can just select the top n players