List Overlap

[Josh_Python890]

I'm doing online problems to get me back into python after not touching it for a couple of years.

Problem comes from http://www.practicepython.org/exercise/2...erlap.html

Take two lists, say for example these two:

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
and write a program that returns a list that contains only the elements that are common between the lists (without duplicates). Make sure your program works on two lists of different sizes.

Extras:

Randomly generate two lists to test this
Write this in one line of Python (don’t worry if you can’t figure this out at this point - we’ll get to it soon)

This is my attempt but it doesn't work:

def exc5():
    a = random.sample(range(20), 10)
    b = random.sample(range(20), 10)
    c = [i for i in a if a in b]
    print(a)
    print(b)
    print(c)


exc5()

Sample output:

Output:
[9, 17, 6, 4, 12, 16, 18, 5, 1, 7]
[4, 5, 19, 12, 9, 18, 14, 2, 8, 11]
[]

As you can see the numbers which are in both lists should be in the third list but it's an empty list.

I am aware you can do this with sets but I'm interested as to why this doesn't work.

[micseydel]

You want "if i in b" not "if a in b". Also the in operator in this case will not be efficient for large inputs. Happy to talk about that if you want but waiting so as to not hijack your thread.

[perfringo]

“without duplicates” requirement directs my thoughts toward sets. Sonething like: convert lists to sets, do set operation, convert result to list.

Edit: need to read more carefully, you already considered sets. My mistake.

[micseydel]

Yes, sets are exactly what you'd use to improve the efficiency.

[kotter]

I believe your fourth line is wrong.
You have:

c = [i for i in a if a in b]

But it should be:

c = [i for i in a if i in b]

Edit: This was already said, never mind