Mistake in list comprehension-CodePudding

Problem Solved!

I have two python lists of integers that are randomly generated. I want to find the numbers that are common between the lists.

Using list comprehension, I've come up with this:

new_list = [a for a in list_one for b in list_two if a == b and a not in new_list]

However, this returns an empty list. I have a working program using loops that works perfectly:

for a in list_one:
    for b in list_two:
        if a == b and a not in new_list:
            new_list.append(a)

What mistakes have I made converting it to a list comprehension?

CodePudding user response：

I do not think list comprehension is required for intersection of 2 lists. Anyways you can modify your code to-

list_one = [1,3,4,5]
list_two = [1,5]
new_list = []
new_list = [a for a in list_one for b in list_two if a == b and a not in new_list]
print(new_list)

you can also do-

list_one = [1,3,4,5]
list_two = [1,5]
new_list = []
new_list = [a for a in list_one if a in list_two and a not in new_list]
print(new_list)

You can convert it into set and use set1.intersection(set2).

list_one = [1,3,4,5]
list_two = [1,5]
new_list = []
new_list = list(set(list_one).intersection(set(list_two)))
print(new_list)

CodePudding user response：

You can't reference the same list inside of a list comprehension.

Thanks @Keith from the comments of the OP for pointing it out.

CodePudding user response：

This is a simple approach with sets,

list_1 = [1,2,3,4,5]
list_2 = [5,4,7,7,5,1,2,8,9]

new_list = list(set(list_1) & set(list_2))

print(new_list)

It uses the set intersection has a number of benefits compared to a list comprehension:

The input lists do not need to be sorted (though you can sort them by using sorted() function.
The input lists can have different lengths.
Presence of duplicates is taken care of automatically (sets don't admit those).