Converting a set into list

[Peter Otten]

Daniel Kluev wrote:

>> Both solutions seem to be equivalent in that concerns the number of
>> needed loop runs, but this two-step operation might require one less loop
>> over list1. The set&set solution, in contrary, might require one loop
>> while transforming to a set and another one for the & operation.
> 
> python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)"
> "l3 = list(set(l1) & set(l2))"
> 100 loops, best of 3: 2.19 msec per loop
> 
> python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)"
> "s=set(l2); l3 = [i for i in l1 if i in s]"
> 100 loops, best of 3: 2.45 msec per loop
> 
> python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)"
> "l3 = list(set(l1) & set(l2))"
> 10 loops, best of 3: 28 msec per loop
> 
> python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)"
> "s=set(l2); l3 = [i for i in l1 if i in s]"
> 10 loops, best of 3: 28.1 msec per loop
> 
> So even with conversion back into list set&set is still marginally faster.

If you are looking for speed, consider

s = set(l1)
s.intersection_update(l2)
l3 = list(s) 

$ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" 
"list(set(l1) & set(l2))"
100 loops, best of 3: 4 msec per loop

$ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "s = 
set(l1); s.intersection_update(l2); list(s)"
100 loops, best of 3: 1.99 msec per loop