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Python program to remove Nth occurrence of the given word
  • Difficulty Level : Easy
  • Last Updated : 24 Nov, 2020

Given a list of words in Python, the task is to remove the Nth occurrence of the given word in that list.

Examples:

Input: list - ["geeks", "for", "geeks"]
       word = geeks, N = 2
Output: list - ["geeks", "for"]

Input: list - ["can", "you",  "can", "a", "can" "?"]
       word = can, N = 1
Output: list - ["you",  "can", "a", "can" "?"]

 

Approach #1: By taking another list.

Make a new list, say newList. Iterate the elements in the list and check if the word to be removed matches the element and the occurrence number, otherwise, append the element to newList.






# Python3 program to remove Nth 
# occurrence of the given word
  
# Function to remove Ith word
def RemoveIthWord(lst, word, N):
    newList = []
    count = 0
  
    # iterate the elements
    for i in lst:
        if(i == word):
            count = count + 1
            if(count != N):
                newList.append(i)
        else:
            newList.append(i)
              
    lst = newList
      
    if count == 0:
        print("Item not found")
    else:
        print("Updated list is: ", lst)    
      
    return newList
  
# Driver code
list = ["geeks", "for", "geeks"]
word = "geeks"
N = 2
  
RemoveIthWord(list, word, N)

Output :

Updated list is:  ['geeks', 'for']

 
Approach #2: Remove from the list itself.

Instead of making a new list, delete the matching element from the list itself. Iterate the elements in the list and check if the word to be removed matches the element and the occurrence number, If yes delete that item and return true. If True is returned, print List otherwise, print “Item not Found”.




# Python3 program to remove Nth 
# occurrence of the given word
  
# Function to remove Ith word
def RemoveIthWord(list, word, N):
    count = 0
      
    for i in range(0, len(list)):
        if (list[i] == word):
            count = count + 1
              
            if(count == N):
                del(list[i])
                return True
                  
    return False
  
# Driver code
list = ['geeks', 'for', 'geeks']
word = 'geeks'
N = 2
  
flag = RemoveIthWord(list, word, N)
  
if (flag == True):
    print("Updated list is: ", list)
else:
    print("Item not Updated"

Output :

Updated list is:  ['geeks', 'for']

Approach #3: Remove from the list using pop().

Instead of creating a new list and using if/else statement we can pop the matching element from the list using pop( ). We need to use an additional counter to keep track of the index.
Why we need an index ? because pop( ) needs index to pass inside i.e pop(index).




# Python3 program to remove Nth 
# occurrence of the given word
  
# Function to remove nth word
def omit(list1,word,n1):
      
    # for counting the occurence of word
    count=0
      
    # for counting the index number
    # where we are at present             
    index=0   
            
    for i in list1:
        index+=1
        if i==word:
            count+=1
            if count==n1:
                  
                # (index-1) because in list
                # indexing start from 0th position
                list1.pop(index-1)  
    return list1
  
# Driver code
list1 = ["he", "is", "ankit", "is",
         "raj", "is","ankit raj"]
  
word="is"
n1=3
  
print("new list is :",omit(list1,word,n1))

Output :

new list is : ['he', 'is', 'ankit', 'is', 'raj', 'ankit raj']

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