[atraub]

Here's a challenge that's short and sweet.

Write a function that takes two lists of ints as input and outputs a single list that contains only the values that appear in both lists. Shoot for efficiency!

[codeprada]

Does it have to be in Python?

[Brewer]

def doppleganger(listone, listtwo):
    return set(listone).intersection(set(listtwo))

Thats my answer

[atraub]

Haha don't be lame brewer! Create an actual algorithm for solving the problem

Quote

Does it have to be in Python?

Any high level language will do fine... but no functional languages :)

Put it in any language you damn well please.

This post has been edited by atraub: 28 July 2011 - 09:42 PM
Reason for edit:: Clarification

[baavgai]

Spoiler

def mergeDup(list1, list2):
	return list(set(list1 + list2))

Edit: Hmmm, got it backwards. brb.

This post has been edited by baavgai: 27 July 2011 - 12:31 PM

[Brewer]

This sounds like fun, I'll give it a shot when I get off work.

[baavgai]

Right, sorry about that.

Spoiler

def bothIn(list1, list2):
	return [ n for n in list1 if n in list2 ]

Actually an easier question.

[atraub]

haha these are some creative answers for sure. Not exactly what I was hoping for though

[Motoma]

How should it behave when there are multiples of an integer in a list? What's the output for: ([1,2,3,3,3,4,4], [3,1,1,6])?

Here's a less-traditional way of doing this; not sure if dict's hashing of the key will give a performance gain or loss:

Spoiler

def intersect(list1, list2):
    bucket = {}
    ret = []
    for i in list1:
        bucket[i] = 1
    for i in list2:
        if bucket.has_key(i):
            ret.append(i)
    return ret

[gregwhitworth]

PHP

function dg($listone, $listtwo)
{
    return array_merge($listone, $listtwo);
}

[Raynes]

atraub, on 27 July 2011 - 08:26 PM, said:

Haha don't be lame brewer! Create an actual algorithm for solving the problem

Quote

Does it have to be in Python?

Any high level language will do fine... but no functional languages

Why no functional languages? If you don't use core functions, you still have to write an algorithm for this. It isn't cheating by any means.

[baavgai]

Hmm... what if the lists look like [ 2, 5, 5, 5, 8, 3, 3, 12 ], [ 12, 5, 5, 1, 3, 3, 3, 4 ].

The answer I'd want ( order doesn't matter): [3, 3, 12, 5, 5]

That's far more fun:

Spoiler

def bothIn(list1, list2):
	d = {}
	for n in list1:
		if n in list2:
			if n in d:
				d[n][0] += 1
			else:
				d[n] = [1,0]
	for n in list2:
		if n in d:
			d[n][1] += 1
	result = []
	for k,v in d.items():
		result.extend([k] * min(v))
	return result

I'm sure someone can do better than that.

[atraub]

Fine, functional languages too

[Nallo]

Another O(len(seq1) + len(seq2)) solution:

Spoiler

def doppelgaenger(seq1, seq2):
    set1 = set(seq1)
    return list(filter(lambda x: x in set1, seq2))

Without using a hash table, i.e. a dictionary or set, it would be troublesome.

[atraub]

am I the only one who did not use a set or a dictionary?