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Syntax –
set_name.issubset(iterable)
Parameter –
Example 1-
set1 = {1, 2, 3}
set2 = {1, 2, 3, 4, 5}
print("Set 1:",set1)
print("Set 2:",set2)
print("set2 is superset of set1:",set1.issuperset(set2))
# Output -
# Set 1: {1, 2, 3}
# Set 2: {1, 2, 3, 4, 5}
# set2 is superset of set1: True
.issuperset() method and print the output.
Example 2-
s={'a','b','c','d','e','f','g','h'}
s1={'a','c','e','h'}
print("First Set:",s)
print("Second set:",s1)
print("s is a superset of s1:",s.issuperset(s1))
# Output -
# First Set: {'f', 'g', 'd', 'e', 'c', 'h', 'b', 'a'}
# Second set: {'h', 'e', 'a', 'c'}
# s is a superset of s1: True
The .issuperset() method with another type of collection –
Example 3-
set1 = [10, 20, 30]
set2 = {10, 20, 30, 40, 50}
print("Set 1:",set1)
print("Set 2:",set2)
print("set2 is superset of set1:",set1.issuperset(set2))
# Output -
# Set 1: [10, 20, 30]
# Set 2: {50, 20, 40, 10, 30}
# set2 is superset of set1: True
how to check whether a set is a superset of another set without using built-in method –
Example 4-
def is_superset(set1, set2):
# Check if every element in set2 is also in set1
for element in set2:
if element not in set1:
return False
return True
# Example usage:
set1 = {1, 2, 3, 4, 5}
set2 = {1, 2, 3}
result = is_superset(set1, set2)
print("set1 is superset of set2:",result)
# Output -
# set1 is superset of set2: True
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