Possible to return two lists from a list comprehension?

>>> rr,tt = zip(*[(i*10, i*12) for i in xrange(4)])
>>> rr
(0, 10, 20, 30)
>>> tt
(0, 12, 24, 36)

Creating two comprehensions list is better (at least for long lists). Be aware that, the best voted answer is slower can be even slower than traditional for loops. List comprehensions are faster and clearer.

python -m timeit -n 100 -s 'rr=[];tt = [];' 'for i in range(500000): rr.append(i*10);tt.append(i*12)' 
10 loops, best of 3: 123 msec per loop

> python -m timeit -n 100 'rr,tt = zip(*[(i*10, i*12) for i in range(500000)])' 
10 loops, best of 3: 170 msec per loop

> python -m timeit -n 100 'rr = [i*10 for i in range(500000)]; tt = [i*10 for i in range(500000)]' 
10 loops, best of 3: 68.5 msec per loop

It would be nice to see list comprehensionss supporting the creation of multiple lists at a time.

However,

if you can take an advantage of using a traditional loop (to be precise, intermediate calculations), then it is possible that you will be better of with a loop (or an iterator/generator using yield). Here is an example:

$ python3 -m timeit -n 100 -s 'rr=[];tt=[];' "for i in (range(1000) for x in range(10000)): tmp = list(i); rr.append(min(tmp));tt.append(max(tmp))" 
100 loops, best of 3: 314 msec per loop

$ python3 -m timeit -n 100 "rr=[min(list(i)) for i in (range(1000) for x in range(10000))];tt=[max(list(i)) for i in (range(1000) for x in range(10000))]"
100 loops, best of 3: 413 msec per loop

Of course, the comparison in these cases are unfair; in the example, the code and calculations are not equivalent because in the traditional loop a temporary result is stored (see tmp variable). So, the list comprehension is doing much more internal operations (it calculates the tmp variable twice!, yet it is only 25% slower).