How to skip specific indexes in an array?

Question

I have a list of the form L = [1, 2, 3, 4, 1, 2, 1, 3, 0, 4] and I want to remove the third index where 1 occurs and replace it with 0. My code is this, but this removes the previous indexes of 1(1st and 2nd also) which I want in my List. My code is this

counter=0
index = 2
L = list([1, 2, 3, 4, 1, 2, 1, 3, 0, 4])
print("Value before is\n", L)
for i in range(len(L)):
    y=L.index(1)
    print(y)
    if(counter==index):
        L[y]=0
        break
    else:
        counter=counter+1
        L[y]
        print("Value of list in else\n",L)
        print("Value of counter\n",counter)

print("After the value is\n",L)

So output comes as

 [2, 3, 4, 2, 0, 3, 0, 4]

but I want it as

L = [1, 2, 3, 4, 1, 2, 0, 3, 0, 4]

and remember that I will not be given directly the index which I want to change So I could Do L[7]=0 Thanks in advance

Answer 1

There are a few issues with your algorithm, but it boils down to this: by doing y = L.index(1) you find the first index where a 1 appears. So by doing L[y] = 0, all you can do is update the first occurence of a 1.

Finding the nth index

There is no builin to find the nth appearance and so you will have to write it.

To be consistent with list.index, I made the following index function raise a ValueError when the item is not found.

Code

def index(lst, obj, n=1):
    count = 0
    for index, item in enumerate(lst):
        if item == obj:
            count += 1
        if count == n:
            return index
    raise ValueError('{} is not in list at least {} times'.format(obj, n))

L = [1, 2, 3, 4, 1, 2, 1, 3, 0, 4]

index = index(L, 1, n=3)

L[index] = 0

print(L)

Output

[1, 2, 3, 4, 1, 2, 0, 3, 0, 4]

Using list-comprehension

Alternatively, if all you want to do is replace the nth occurence, but do not care about its actual index, you can generate a new list with a list-comprehension and an itertools.count object.

Code

from itertools import count

def replace(lst, obj, repl, n=1):
    counter = count(1)
    return [repl if x == obj and next(counter) == n else x for x in lst]


L = [1, 2, 3, 4, 1, 2, 1, 3, 0, 4]

new_list = replace(L, 1, 0, n=3)
print(new_list)

Output

[1, 2, 3, 4, 1, 2, 0, 3, 0, 4]