I have a list similar to:

[1 2 1 4 5 2 3 2 4 5 3 1 4 2] 

I want to create a list of x random elements from this list where none of the chosen elements are the same. The difficult part is that I would like to do this by using list comprehension... So possible results if x = 3 would be:

[1 2 3]
[2 4 5]
[3 1 4]
[4 5 1]

etc...

Thanks!

I should have specified that I cannot convert the list to a set. Sorry! I need the randomly selected numbers to be weighted. So if 1 appears 4 times in the list and 3 appears 2 times in the list, then 1 is twice as likely to be selected...

Disclaimer: the "use a list comprehension" requirement is absurd.

Moreover, if you want to use the weights, there are many excellent approaches listed at Eli Bendersky's page on weighted random sampling.

The following is inefficient, doesn't scale, etc., etc.

That said, it has not one but two (TWO!) list comprehensions, returns a list, never duplicates elements, and respects the weights in a sense:

>>> s = [1, 2, 1, 4, 5, 2, 3, 2, 4, 5, 3, 1, 4, 2]
>>> [x for x in random.choice([p for c in itertools.combinations(s, 3) for p in itertools.permutations(c) if len(set(c)) == 3])]
[3, 1, 2]
>>> [x for x in random.choice([p for c in itertools.combinations(s, 3) for p in itertools.permutations(c) if len(set(c)) == 3])]
[5, 3, 4]
>>> [x for x in random.choice([p for c in itertools.combinations(s, 3) for p in itertools.permutations(c) if len(set(c)) == 3])]
[1, 5, 2]

.. or, as simplified by FMc:

>>> [x for x in random.choice([p for p in itertools.permutations(s, 3) if len(set(p)) == 3])]
[3, 5, 2]

(I'll leave the x for x in there, even though it hurts not to simply write list(random.choice(..)) or just leave it as a tuple..)