Python程序删除给定单词的第N次出现

给定Python中的单词列表，任务是删除该列表中给定单词的第 N^次出现。
例子：

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" "?"]

方法#1：通过另一个列表。
制作一个新列表，比如newList 。迭代列表中的元素并检查要删除的单词是否与元素和出现次数匹配，否则，将元素附加到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)

输出 ：

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

方法#2：从列表本身中删除。
与其创建一个新列表，不如从列表本身中删除匹配的元素。迭代列表中的元素并检查要删除的单词是否与元素和出现次数匹配，如果是，则删除该项目并返回true。如果返回 True，则打印 List，否则打印“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")

输出 ：

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

方法 #3：使用pop()从列表中删除。
我们可以使用 pop() 从列表中弹出匹配元素，而不是创建一个新列表并使用 if/else 语句。我们需要使用一个额外的计数器来跟踪索引。
为什么我们需要索引？因为pop() 需要index 来传递，即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))

输出 ：

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