Tips for cleaning up a challenge's answer? Wei

Question

I've been going through some code challenges. And one struck me in particular. CodeAbbey has a Weighted Sum of Digits challenge. And this was my answer.

# open and read a file, hardcoded numbers, etc. Here is a sample
raw_data = (6, 19, 64, 6527226, 12345146)  

for number in raw_data:
    wsd = 0  # Stores sum
    number_list = list(str(number))  # Converts the number into a list.
    for i, k in enumerate(number_list):  # Enumerates each number
        wsd += (int(i+1) * int(k))  # Multiplies and adds product to wsd
    print(wsd)

output >>> 6, 19, 14, 114, 137

Anyone with more experience able to see a better way of getting the sum?

Answer 1

for number in raw_data:
    print(sum(i * int(d) for i, d in enumerate(str(number), 1)))

Answer 2

This can be done entirely using math operations. Look how the sum is generated, e.g. for 12345146:

                6
              4+6
            1+4+6
          5+1+4+6
        4+5+1+4+6
      3+4+5+1+4+6
    2+3+4+5+1+4+6
+ 1+2+3+4+5+1+4+6

So firs you add last digit, then last and previous, then last, previous and previous etc. until you add all digits. And this can be simply implemented by following algorithm:

raw_data = (6, 19, 64, 6527226, 12345146)  
for d in raw_data:
    wsd = 0
    lst = 0 # this is for holding last 'sum row'
    while d > 0:
        lst = lst + d % 10 # append another digit
        wsd += lst
        d //= 10
    #
    print wsd
#

On my computer this is about two times faster than algorithms using enumeration and string conversion (I multiplied raw_data 100000 times, thus getting 500000 elements for comparison).

Answer 3

If using map:

>>> map(lambda d: sum([ (i+1) * int(v) for i,v in enumerate(list(str(d)))]), raw_data)
[6, 19, 14, 114, 137]