用于获取给定元素的下一个未显示元素的Python程序

给定一个数字列表和列表中的一个特定元素，任务是编写一个Python程序来获取下一个未显示的元素到给定元素的第一次出现，即在给定元素之前不会出现的元素。

例子：

Input : test_list = [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6], nex_to_ele = 6

Output : 2

Explanation : In the above list there are 2, 3, 4, 5, and 6 are elements, as nex_to_ele is 6, whose first occurrence is at index 2, before that only 3 and 4 occurs, so the unrevealed elements are 2 and 5, out of which unrevealed element 2 is next after the first occurrence of 6. Hence 2 is the next unrevealed element of 6 in the list. 

Input : test_list = [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6], nex_to_ele = 3

Output : 4

Explanation : Next different unrevealed element to 3 is 4 as 3 occurs at index 0. So, none of the elements in the list have occurred before 3 and 4 is the next element after 3. 

方法 #1：使用set() + sorted() + index()

在这种情况下，使用 set() 和 sorted() 将列表转换为保留顺序的唯一元素列表。然后 index() 用于获取 number 的索引，结果列表中它的下一个元素是所需的元素。

Python3

# Python3 code to demonstrate working of
# Next different element in list
# Using set() + sorted() + index()
 
# initializing list
test_list = [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing element
nex_to_ele = 6
 
# sorting removing duplicated keeping order
temp = sorted(set(test_list), key=lambda sub: test_list.index(sub))
 
# getting next index
num_idx = temp.index(nex_to_ele)
 
# checking for last index for overflow
if num_idx == len(temp) - 1:
    res = None
 
# getting next index as result
else:
    res = temp[num_idx + 1]
 
# printing result
print("Next different element : " + str(res))

输出：

The original list is : [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6]

Next different element : 2

方法 #2：使用next() + iter() + sorted() + set()

在这种情况下，我们不使用 index() 来获取元素索引，而是使用用户迭代器方法递增到下一个迭代器，直到找到元素。它的下一个元素是结果。

蟒蛇3

# Python3 code to demonstrate working of
# Next different element in list
# Using next() + iter() + sorted() + set()
 
# initializing list
test_list = [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing element
nex_to_ele = 6
 
# sorting removing duplicated keeping order
temp = sorted(set(test_list), key=lambda sub: test_list.index(sub))
 
temp_itr = iter(temp)
flag = 0
for idx in range(0, len(temp)):
 
    # if last element is element to find
    if idx == len(temp) - 1:
        flag = 1
 
    # breaking when element is found
    if next(temp_itr) == nex_to_ele:
        break
 
if flag:
    res = None
else:
 
    # next element is answer
    res = next(temp_itr)
 
# printing result
print("Next different element : " + str(res))

输出：

The original list is : [3, 4, 6, 6, 3, 2, 3, 5, 6, 3, 4, 5, 5, 3, 6]

Next different element : 2