Python – 根据第 K 个字符追加

给定一个字符串列表，根据第 K 个索引值附加到字符串 i 或 j 值。

Input : test_list = [“geeksforgeeks”, “best”, “for”, “geeks”], K = 2, N = ‘e’, i, j = “@@”, “..”
Output : [‘geeksforgeeks..’, ‘best@@’, ‘for@@’, ‘geeks..’]
Explanation : geeksforgeeks and geeks having similar 2nd occ. value as ‘e’, hence gets appended by “..”.

Input : test_list = [“giiksforgeeks”, “bst”, “for”, “geeks”], K = 2, N = ‘e’, i, j = “@@”, “..”
Output : [‘giiksforgeeks@@’, ‘best@@’, ‘for@@’, ‘geeks@@’]
Explanation : No values with K value ‘e’, all appended by @@.

方法#1：使用循环

这是解决这个问题的蛮力方法，我们检查每个字符串的第 K 个索引，如果发现是 N，则附加 i 值，否则附加 j。

Python3

# Python3 code to demonstrate working of 
# Append according to Kth character
# Using loop
  
# initializing lists
test_list = ["geeksforgeeks", "best", "for", "geeks"]
  
# printing string
print("The original list : " + str(test_list))
  
# initializing K
K = 2 
  
# initializing N 
N = 'e'
  
# initializing i, j 
i, j = "**", "##"
  
res = []
for sub in test_list:
      
    # checking for Kth index to be N
    if sub[K] == N:
        res.append(sub + i)
    else :
        res.append(sub + j)
  
# printing results 
print("The resultant List : " + str(res))

输出

The original list : ['geeksforgeeks', 'best', 'for', 'geeks']
The resultant List : ['geeksforgeeks**', 'best##', 'for##', 'geeks**']

方法#2：使用列表推导

这以类似的方式解决了这个问题，只是不同的是，它是一种速记，可以用作解决这个问题的一种线性方法。

Python3

# Python3 code to demonstrate working of 
# Append according to Kth character
# Using list comprehension
  
# initializing lists
test_list = ["geeksforgeeks", "best", "for", "geeks"]
  
# printing string
print("The original list : " + str(test_list))
  
# initializing K
K = 2 
  
# initializing N 
N = 'e'
  
# initializing i, j 
i, j = "**", "##"
  
# shorthand to solve this problem
res = [sub + i if sub[K] == N else sub + j for sub in test_list]
  
# printing results 
print("The resultant List : " + str(res))

输出

The original list : ['geeksforgeeks', 'best', 'for', 'geeks']
The resultant List : ['geeksforgeeks**', 'best##', 'for##', 'geeks**']