Simplifying my code

[ilondire05]

Alright, I had a problem where I needed to make two random lists, then make a third list which would contain the elements that the first 2 lists shared. I did that:

import random

a = []
b = []
c = []
i = 0
while i < 13:
    y = random.randint(0, 20)
    a.append(y)
    i += 1
i = 0
while i < 14:
    y = random.randint(0, 20)
    b.append(y)
    i += 1
print("This is list a: " + str(a))
print("This is list b: " + str(b))
for x in a:
    if x in b:
        if x not in c:
            c.append(x)
print(str(c) + " are element in both a and b lists")

Now lines 7 through 15 where I make the random lists feel slightly clunky and make me feel that there is a better way to do what I did in them. Is that correct, if so how?
Secondly, there is apparently a way to solve this problem with 1 line of code. How? What is the trick for that?

[metulburr]

This is much shorter if that is what you are looking for.

import random

a = [random.randint(0,20) for num in range(13)]
b = [random.randint(0,20) for num in range(13)]

result = set(a).intersection(b)
print(result)

You could mesh that into one line, but then it would be hard to read.

A couple of things about your code:
You should never increment a counter in python while loops. That is what for loops with enumerate() is for. That alone would of shortened it.
https://python-forum.io/misc.php?action=help&hid=53

And you should use the format method (or f-strings) if you are using python3.6+
https://python-forum.io/Thread-Basic-str...xpressions

[ichabod801]

Basic simplification would be to use while loops instead of for loops, use sets instead of lists, add the random numbers to the sets directly instead of assigning them to a variable and then adding/appending, and union the two sets to find the elements of both.

If you create the two sets using list comprehensions, you could do it all in one line.

[metulburr]

Basic simplification would be to use while loops instead of for loops

You mean this in reverse?

[ichabod801]

You mean this in reverse?

Not only did I meant that in reverse, but it's time to go to bed.

[scidam]

Another simplification could be proposed on the basis of theory of probability.
It could be shown, that probability of the event that an arbitrary random number x belonging to A = (a1, ..., a13) and B = (b1, ..., b13) simultaneously is equal 169/400 (13/20 * 13/20).

So, we can write:

result = set([random.randint(0,20) for x in range(13) if random.choice([1]*169 + [0]*(400-169))])

We can roughly verify this solution, e.g. by comparing average lengths of sets for direct solution (that uses set intersection) and the above, e.g.

res1 = list()
for k in range(100):
    res1.append(len(set([random.randint(0,20) for x in range(13) if random.choice([1]*169 + [0]*(400-169))])))

avg_len1 = sum(res1) / 100
res2 = list()
for k in range(100):
    res2.append(len(set([random.randint(0,20) for num in range(13)])& set([random.randint(0,20) for num in range(13)])))

avg_len2 = sum(res2) / 100

I got the following values:

Output:
>>> avg_len2
4.83
>>> avg_len1
4.53

Not sure, if this is simplification. Direct method based on set's intersection is clear and simpler.