How do I remove the second occurrence from a list in Python?

Python program to remove Nth occurrence of the given word

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

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Python program to remove Nth occurrence of the given word
Introduction
Remove All the Occurrences of an Element From a List in Python
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Python program to remove nth occurrence of the given word list
1. Using list.remove() function

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" "?"]

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

# 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

# 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

# Python3 program to remove Nth 
# occurrence of the given word
# Function to remove nth word

def omit(list1,word,n1):

# for counting the occurrence 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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Introduction

In this section of python programming, we will understand the code to eliminate the nth occurrence of the given element from the list.

on Oct 13 2021 Comment

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Python program to remove nth occurrence of the given word list

To allow the user to input a number of elements in the list and store it in a variable.
Accept the values into the list using a for loop and insert them into the python list.
Use a for loop to traverse through the elements in the list.
Then use an if statement to check if the word to be removed matches the element and the occurrence number and otherwise, it appends the element to another list.
The number of repetitions along with the updated list and distinct elements is printed.

# python program to remove nth occurrence of the given word a=[] n= int(input("Enter the number of elements in list:")) for x in range(0,n): element=input("Enter element" + str(x+1) + ":") a.append(element) print(a) c=[] count=0 b=input("Enter word to remove: ") n=int(input("Enter the occurrence to remove: ")) for i in a: if(i==b): count=count+1 if(count!=n): c.append(i) else: c.append(i) if(count==0): print("Item not found ") else: print("The number of repetitions is: ",count) print("Updated list is: ",c) print("The distinct elements are: ",set(a))

After executing the program, the output will be:

Enter the number of elements in list: 5 Enter element1: test Enter element2: test Enter element3: my Enter element4: world Enter element5: world ['test', 'test', 'my', 'world', 'world'] Enter word to remove: world Enter the occurrence to remove: 4 The number of repetitions is: 2 Updated list is: ['test', 'test', 'my', 'world', 'world'] The distinct elements are: {'test', 'world', 'my'}

1. Using list.remove() function

list.remove(x) removes the first occurrence of value x from the list, but it fails to remove all occurrences of value x from the list.

if __name__ == '__main__':

l = [0, 1, 0, 0, 1, 0, 1, 1]

val = 0

l.remove(val)

print(l)# prints [1, 0, 0, 1, 0, 1, 1]

DownloadRun Code

To remove all occurrences of an item from a list using list.remove(), you can take advantage of the fact that it raises a ValueError when it can’t find the specified item in the list. The idea is to repeatedly call remove() function until it raises a ValueError exception. This is demonstrated below: