How to remove 2nd occurrence of an item in a list using remove() without removing 1st occurrence in Python

Question

a = [9,8,2,3,8,3,5]

How to remove 2nd occurrence of 8 without removing 1st occurrence of 8 using remove().

Answer 1

remove() removes the first item from the list which matches the specified value. To remove the second occurrence, you can use del instead of remove.The code should be simple to understand, I have used count to keep track of the number of occurrences of item and when count becomes 2, the element is deleted.

a = [9,8,2,3,8,3,5]
item  = 8
count = 0
for i in range(0,len(a)-1):
        if(item == a[i]):
               count =  count + 1
               if(count == 2):
                      del a[i]
                      break
print(a)

Answer 2

Here's a way you can do this using itertools.count along with a generator:

from itertools import count

def get_nth_index(lst, item, n):
    c = count(1)
    return next((i for i, x in enumerate(lst) if x == item and next(c) == n), None)

a = [9,8,2,3,8,3,5]  
indx = get_nth_index(a, 8, 2)
if indx is not None:
    del a[indx]

print(a)
# [9, 8, 2, 3, 3, 5]

Answer 3

It's not clear to me why this specific task requires a loop:

array = [9, 8, 2, 3, 8, 3, 5]

def remove_2nd_occurance(array, value):

    ''' Raises ValueError if either of the two values aren't present '''

    array.pop(array.index(value, array.index(value) + 1))


remove_2nd_occurance(array, 8)

print(array)