python set in clude in set code example

Example 1: python set &

>>> A = {0, 2, 4, 6, 8};
>>> B = {1, 2, 3, 4, 5};

>>> print("Union :", A | B)  
Union : {0, 1, 2, 3, 4, 5, 6, 8}

>>> print("Intersection :", A & B)
Intersection : {2, 4}

>>> print("Difference :", A - B)
Difference : {0, 8, 6}

# elements not present both sets
>>> print("Symmetric difference :", A ^ B)   
Symmetric difference : {0, 1, 3, 5, 6, 8}

Example 2: unique element intersection python

#get name1 and name2 as a list input
name1 =list("".join(str(x)for x in input("Enter name1").replace(" ","")))
name2 =list("".join(str(x)for x in input("Enter name2").replace(" ","")))
#check using list comprehension if x in name1 is in name2
#this will return multiple instances of the same character from name1 that matches with the name2
common = [x for x in name1 if x in name2]
#create a set out of the output so as to have only unique values of the repeated characters
unique = set(common)
#thus the above set will have common unrepeated characters from both names
#create a variable and initialize it to zero
d=0
#run a loop that checks the minimum occurrence of 
#the character from the set in name1 & name2
#Minimum because for ex: a might exist thrice in name1, but only twice in name2
#we will need to take only 2 common occurrences from the name2
#thus finding the minimum occurrence of the character from both names
for x in unique:
    d = d + min(name1.count(x),name2.count(x))
#multiplying by two, because if one character from name 1 matches with one,
#character from name2, then it makes two in total
difference = (len(name1) + len(name2)) - d*2
print(difference)