以相反的顺序替换列表项并跳过所有其他项目 python

Question

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?

Answer 1

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