intersection() method to get common elements of the sets by using & operator

All set methods

Returns a new set with common elements only. Original set remains unchanged.

intersection using & operator

All Common elements in sets
Using ampersand ( & ) operator

A={1,2,3}
B={3,4,5}
print(A & B)

Output ( Note 3 is the only common element )

{3}

We can check the output by using type()

A={1,2,3}
B={3,4,5}
x=A & B 
print(type(x))

Output

<class 'set'>

A={'a','b','c'}
B={'a','y','z'}
print(A.intersection(B))

Output

{'a'}

We can use any number of sets with intersection()

A={'a','b','c'}
B={'a','y','z'}
C={'a','k','l'}
print(A & B & C)

Output

{'a'}

Using string ( iterable object ) with intersection() method

A={'a','b','c','x','y'}
B='Alex'
print(A.intersection(B))

Output

{'x'}

Note : we can't use iterable object by using ampersand ( & ), the intersection operator

Using list with intersection() method.

A={'a','b','c'}
B=['a','x','y']
print(A.intersection(B))

Output

{'a'}

Below code will generate error.

print(A & B)

 TypeError: unsupported operand type(s) for &: 'set' and 'list'

In all above code by using intersection() we created a new set. By using intersection_update() method we can change the original set with common elements.
All set methods Questions with solutions on set
union() difference() symmetric_difference()