A.intersection(B) returns a new set containing elements common to both inputs. The & operator is the set-to-set equivalent. The original sets are not changed.
The method form can accept iterable arguments; operator forms require set-like operands. Use intersection_update() when the original set should be reduced in place. See the Set operations overview.
All distinct elements from both sets A={1,2,3}
B={3,4,5}
print(A & B)
Output ( Note 3 is the only common element )
{3}
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'}
A={'a','b','c'}
B={'a','y','z'}
C={'a','k','l'}
print(A & B & C)
Output
{'a'}
A={'a','b','c','x','y'}
B='Alex'
print(A.intersection(B))
Output
{'x'}
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'
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