replacing list items in reverse order and skipping every other item python

I am making a program to check whether a card number is potentially valid using the Luhn algorithm.

num = "79927398713" #example num

digits = [int(x) for x in num]

reverse = digits[1:][::2][::-1] #step 1: start from rightmost digit, skip first, skip every other
count = 0
digitsum = 0
print(reverse) #output here is: [1, 8, 3, 2, 9]

for x in (reverse):
    reverse[count] *= 2
    if reverse[count] > 9:
        for x in str(reverse[count]):  #multiply each digit in step 1 by 2, if > 9, add digits to make single-digit number
            digitsum += int(x)
        reverse[count] = digitsum
    count += 1
    digitsum = 0
count = 0
print(reverse) #output here is [2, 7, 6, 4, 9]

basically, I want to input [2, 7, 6, 4, 9] back into the corresponding places in the list digits. It would look like this (changed numbers in asterisks)

[7, **9**, 9, **4**, 7, **6**, 9, **7**, 7, **2**, 3]

the problem is, I would have to read digits backwards, skipping the first (technically last) element, and skipping every other element from there, replacing the values each time.

Am I going about this the wrong way/making it too hard on myself? Or is there a way to index backwards, skipping the first (technically last) element, and skipping every other element?

Solution

You can do this with simple indexing

Once you have the variable reverse, you can index on the left hand side:

# reversed is [2, 7, 6, 4, 9] here
digits[1::2] = reversed(reverse) # will place 9,4,6,7,2 in your example

Note, you can use this trick too for your line where you initialize reverse

reverse = digits[1::2][::-1]

I think you could even use:

reverse = digits[-1 - len(digits) % 2::-2]

which should be even more efficient

Edit

Running timeit, the last solution of digits[-1 - len(digits) % 2::-2] on an array of size 10,000 was 3.6 times faster than the original, I'd highly suggest using this