Python – 后第 K 个元素

给定一个列表，任务是从后端提取所有第 K 个元素。

Input : test_list = [3, 4, 5, 2, 1], K = 2
Output : [1, 5, 3]
Explanation : Every 2nd elements are extracted from rear starting from 1.

Input : test_list = [3, 4, 5], K = 1
Output : [5, 4, 3]
Explanation : Every elements are extracted from rear starting from 1.

方法#1：使用循环
解决这个问题的方法很粗暴。在此，我们反转然后执行迭代以获取每个 Kth 多个元素。

# Python3 code to demonstrate working of 
# Rear Kth elements
# Using loop
  
# initializing list
test_list = [3, 5, 7, 9, 10, 2, 8, 6] 
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 3
  
# Rear Kth elements
res = []
test_list.reverse()
for idx, ele in enumerate(test_list):
      
    # Extracting elements divisible by K
    if idx % K == 0:
        res.append(ele)
  
# printing result 
print("Rear Kth elements : " + str(res))

输出 ：

The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]

方法#2：使用列表切片
这是此问题的速记版本解决方案。我们使用 list slicing skip 的力量来跳过第 K 个元素，并将其反转。

# Python3 code to demonstrate working of 
# Rear Kth elements
# Using list slicing
  
# initializing list
test_list = [3, 5, 7, 9, 10, 2, 8, 6] 
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 3
  
# Rear Kth elements
# Starting slicing from Rear (-1) and extracting all Kth elements
res = test_list[-1::-K]
  
# printing result 
print("Rear Kth elements : " + str(res))

输出 ：

The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]