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Python – Accessing K element in set without deletion
  • Last Updated : 02 Feb, 2021

In this article given a set(), the task is to write a Python program to access an element K, without performing deletion using pop().

Example:

Input : test_set = {6, 4, 2, 7, 9}, K = 7
Output : 3
Explanation : 7 occurs in 3rd index in set.

Input : test_set = {6, 4, 2, 7, 9}, K = 9
Output : 4
Explanation : 9 occurs in 4th index in set.

Method #1 : Using loop

The most generic method is to perform iteration using loop, and if K is found, print the element, and if required index.

Python3




# Python3 code to demonstrate working of
# Accessing K element in set without deletion
# Using loop
  
# initializing set
test_set = {6, 4, 2, 7, 9}
  
# printing original set
print("The original set is : " + str(test_set))
  
# initializing K
K = 7
  
res = -1
for ele in test_set:
  
    # checking for K element
    res += 1
    if ele == K:
        break
  
# printing result
print("Position of K in set : " + str(res))

Output:



The original set is : {2, 4, 6, 7, 9}
Position of K in set : 3

Method #2 : Using next() + iter()

In this, container is converted to iterator and next() is used to increment the position pointer, when element is found, we break from loop.

Python3




# Python3 code to demonstrate working of
# Accessing K element in set without deletion
# Using next() + iter()
  
# initializing set
test_set = {6, 4, 2, 7, 9}
  
# printing original set
print("The original set is : " + str(test_set))
  
# initializing K
K = 7
  
set_iter = iter(test_set)
for idx in range(len(test_set)):
  
    # incrementing position
    ele = next(set_iter)
    if ele == K:
        break
  
# printing result
print("Position of K in set : " + str(idx))

Output:

The original set is : {2, 4, 6, 7, 9}
Position of K in set : 3

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