Set1.remove(remove1) line is not getting executed

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Python

#Write a Python program to remove an item from a set if it is present in the set.
set1=set()
num = int(input("How many numbers: "))
for n in range(num):
    numbers = int(input("Enter number "))
    set1.add(numbers)
print("elements in given list is :", set1)
remove1 =input("\n Enter the element you want to remove :")
if remove1 in set1:
    set1.remove(remove1)
    print("The updated set :", set1)
else:
    print("\n The element you have mentioned cannot be found. Please check\n")
    print(set1)

What I have tried:

I have tired with discard and pop () method but the if condition is not getting executed only else part executing.

Posted 31-Dec-21 3:00am

Joshna J U

Updated 31-Dec-21 3:15am

Wendelius

v3

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2 solutions

Solution 2

Which means very simply that the condition is failing: remove1 is not in set1. Why not? Because input always returns a string, an d you are comparing a string with an integer.
Either convert all your inputs to integers:

Python

remove1 = int(input("\n Enter the element you want to remove :"))
if remove1 in set1:
    ...

Or none of them:

Python

for n in range(num):
    numbers = input("Enter number ")
    set1.add(numbers)
print("elements in given list is :", set1)
remove1 = input("\n Enter the element you want to remove :")
if remove1 in set1:
    ...

Posted 31-Dec-21 3:15am

OriginalGriff

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Wendelius · Accepted Answer · 2021-12-31T03:12:00

Solution 1

When you input the numbers you cast them to integer, however you don't do the same when asking for number to delete. So instead of

Python

remove1 =input("\n Enter the element you want to remove :")

try

Python

remove1 =int(input("\n Enter the element you want to remove :"))

Posted 31-Dec-21 3:12am

Wendelius