Python - 在不删除的情况下访问集合中的 K 元素

在本文中，给定一个 set()，任务是编写一个Python程序来访问元素 K，而不使用 pop() 执行删除操作。

例子：

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.

方法#1：使用循环

最通用的方法是使用循环执行迭代，如果找到 K，则打印元素，如果需要，则打印索引。

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))

输出：

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

方法#2：使用next() + iter()

在这种情况下，container 被转换为迭代器，next() 用于增加位置指针，当找到元素时，我们中断循环。

蟒蛇3

# 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))

输出：

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