How do I check if one list contains 2 elements of another list?

Question

I want to be able to check if one list contains 2 elements of another list (that in total has 3 elements)

eg:

list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]

if #list2 contains any 2 elements of list1:
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")

Answer 1

I would use a similar description of the one described in the following question, only this time check the length of the sets intersection: How to find list intersection?

list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]

if len(list(set(a) & set(b))) == 2:
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")

Answer 2

You could write a function to check the elements in common between the two lists, and after verify if quantity of this elements is equals to 2.

For example:

def intersection(list1, list2): 
     lst3 = [value for value in list1 if value in list2] 
     return lst3 


list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]

if len(intersection(list1, list2) == 2):
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")

Answer 3

def my_func(list1, list2):
    count = 0
    for i in list1:
        for j in list2:
            if(i == j) : count += 1
    print(count)
    if(count >= 2) : return True
    else: return False


list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]

if my_func(list1, list2):#list2 contains any 2 elements of list1:
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")

Answer 4

You can use itertools.combinations() to get all sets of two elements from list2 and see if they're subsets of list1 :

import itertools

if any(set(comb).issubset(set(list1)) for comb in itertools.combinations(list2, 2)):
    print("yes, ...")
...

Answer 5

count = 0
for ele in list2:
    if ele in list1:
        count += 1
if count == 2:
    print ("yes, list 2 contains 2 elements of list 1")
else:
    print ("no, list 2 does not contain 2 elements of list 1")

Answer 6

list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]

if len(set(list1).intersection(set(list2))) == 2:
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")

The.intersection method takes two sets and finds what is common between them. If what is common between them is 2 elements, then you have what you want.

Answer 7

This works:

list1 = ["a", "b", "c"]
list2 = ["a", "f", "g", "b"]
val = 0
for idx in list1:
    for idx2 in list2:
        if idx == idx2:
            val += 1
if val == 2:
    print("yes, list 2 contains 2 elements of list 1")
else:
    print("no, list 2 does not contain 2 elements of list 1")