Python Random List Comprehension

Question

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...

Answer 1

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..)

Answer 2

Generally, you don't want to do this sort of thing in a list comprehension -- It'll lead to much harder to read code. However, if you really must, we can write a completely horrible 1 liner:

>>> values = [random.randint(0,10) for _ in xrange(12)]
>>> values
[1, 10, 6, 6, 3, 9, 0, 1, 8, 9, 1, 2]
>>> # This is the 1 liner -- The other line was just getting us a list to work with.
>>> [(lambda x=random.sample(values,3):any(values.remove(z) for z in x) or x)() for _ in xrange(4)]
[[6, 1, 8], [1, 6, 10], [1, 0, 2], [9, 3, 9]]

Please never use this code -- I only post it for fun/academic reasons.

Here's how it works:

I create a function inside the list comprehension with a default argument of 3 randomly selected elements from the input list. Inside the function, I remove the elements from values so that they aren't available to be picked again. since list.remove returns None, I can use any(lst.remove(x) for x in ...) to remove the values and return False. Since any returns False, we hit the or clause which just returns x (the default value with 3 randomly selected items) when the function is called. All that is left then is to call the function and let the magic happen.

The one catch here is that you need to make sure that the number of groups you request (here I chose 4) multiplied by the number of items per group (here I chose 3) is less than or equal to the number of values in your input list. It may seem obvious, but it's probably worth mentioning anyway...

Here's another version where I pull shuffle into the list comprehension:

>>> lst = [random.randint(0,10) for _ in xrange(12)]
>>> lst
[3, 5, 10, 9, 10, 1, 6, 10, 4, 3, 6, 5]
>>> [lst[i*3:i*3+3] for i in xrange(shuffle(lst) or 4)]
[[6, 10, 6], [3, 4, 10], [1, 3, 5], [9, 10, 5]]

This is significantly better than my first attempt, however, most people would still need to stop, scratch their head a bit before they figured out what this code was doing. I still assert that it would be much better to do this in multiple lines.