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Fast way to remove a few items from a list/queue

This is a follow up to a similar question which asked the best way to write

for item in somelist:
    if determine(item):
         code_to_remove_item

and it seems the consensus was on something like

somelist[:] = [x for x in somelist if not determine(x)]

However, I think if you are only removing a few items, most of the items are being copied into the same object, and perhaps that is slow. In an answer to another related question, someone suggests:

for item in reversed(somelist):
    if determine(item):
        somelist.remove(item)

However, here the list.remove will search for the item, which is O(N) in the length of the list. May be we are limited in that the list is represented as an array, rather than a linked list, so removing items will need to move everything after it. However, it is suggested here that collections.dequeue is represented as a doubly linked list. It should then be possible to remove in O(1) while iterating. How would we actually accomplish this?

Update: I did some time testing as well, with the following code:

import timeit
setup = """
import random
random.seed(1)
b = [(random.random(),random.random()) for i in xrange(1000)]
c = []
def tokeep(x):
        return (x[1]>.45) and (x[1]<.5)
"""
listcomp = """
c[:] = [x for x in b if tokeep(x)]
"""
filt = """
c = filter(tokeep, b)
"""
print "list comp = ", timeit.timeit(listcomp,setup, number = 10000)
print "filtering = ", timeit.timeit(filt,setup, number = 10000)

and got:

list comp =  4.01255393028
filtering =  3.59962391853

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Answer

The list comprehension is the asymptotically optimal solution:

somelist = [x for x in somelist if not determine(x)]

It only makes one pass over the list, so runs in O(n) time. Since you need to call determine() on each object, any algorithm will require at least O(n) operations. The list comprehension does have to do some copying, but it’s only copying references to the objects not copying the objects themselves.

Removing items from a list in Python is O(n), so anything with a remove, pop, or del inside the loop will be O(n**2).

Also, in CPython list comprehensions are faster than for loops.

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