I have a list I filter using another list and I would like this to be
as fast as possible
right now I do like this:
[x for x in list1 if x not in list2]
i tried using the method filter:
filter(lambda x: x not in list2, list1)
but it didn't make much difference, because of lambda I guess
is there any way I can speed this up 9 1622
lars_woetmann wrote: I have a list I filter using another list and I would like this to be as fast as possible right now I do like this:
[x for x in list1 if x not in list2]
i tried using the method filter:
filter(lambda x: x not in list2, list1)
but it didn't make much difference, because of lambda I guess is there any way I can speed this up
Both of these techniques are O(n^2). You can reduce it to O(n log n)
by using sets: set2 = set(list2) [x for x in list1 if x not in set2]
Checking to see if an item is in a set is much more efficient than a
list.
--Ben
lars_woetmann wrote: I have a list I filter using another list and I would like this to be as fast as possible right now I do like this:
[x for x in list1 if x not in list2]
i tried using the method filter:
filter(lambda x: x not in list2, list1)
but it didn't make much difference, because of lambda I guess is there any way I can speed this up
if list2 is a list object, "not in list2" is an O(N) operation.
maybe you should use sets instead ? does the following work
better ?
set2 = set(list2)
result = [x for x in list1 if x not in set2]
?
</F>
Lars Woetmann wrote: I have a list I filter using another list and I would like this to be as fast as possible right now I do like this:
[x for x in list1 if x not in list2]
i tried using the method filter:
filter(lambda x: x not in list2, list1)
but it didn't make much difference, because of lambda I guess is there any way I can speed this up
If you use a reasonably new python version, you could use sets:
#v+ a = set(range(10)) a
set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b = set(range(5, 15)) b
set([5, 6, 7, 8, 9, 10, 11, 12, 13, 14]) a.difference(b)
set([0, 1, 2, 3, 4]) a-b
set([0, 1, 2, 3, 4]) list(a-b)
[0, 1, 2, 3, 4]
#v-
Cheers,
--
Klaus Alexander Seistrup
SubZeroNet, Copenhagen, Denmark http://magnetic-ink.dk/
> Both of these techniques are O(n^2). You can reduce it to O(n log n) by using sets:
set2 = set(list2) [x for x in list1 if x not in set2]
Checking to see if an item is in a set is much more efficient than a list.
Is the set-lookup reliably O(log n)? I was under the impression that it is
hash-based, and this should be O(1) usually, but couldbve O(n) worst-case
(hash the same for _all_ entries).
Regards,
Diez
Diez B. Roggisch wrote: Both of these techniques are O(n^2). You can reduce it to O(n log n) by using sets:
>set2 = set(list2) >[x for x in list1 if x not in set2]
Checking to see if an item is in a set is much more efficient than a list.
Is the set-lookup reliably O(log n)? I was under the impression that it is hash-based, and this should be O(1) usually, but couldbve O(n) worst-case (hash the same for _all_ entries).
That's largely a theoretical concern. Google for something like
'''dict worst-case performance "tim peters"'''
to learn more. (The third article there (no doubt obsolete in some
ways, given that it was in 2000) says that Python "keeps at least 1/3 of
the internal hash table entries unused, making collisions very rarely a
problem... It's possible to contrive keys that will cause collisions
systematically ... but unlikely to happen by accident in 2.0")
-Peter
Thanks all, sets work like i charm
comparing
[x for x in list1 if x not in list2]
with
set1, set2 = set(list1), set(list2)
list(set1-set2)
gives something like
len(list2) speedup
------------------------------
100 10
1000 100
10000 1000
the speedup is constant for different len(list1) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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