How to sum the digits of each value in an array to create a new array?

Question

• Starting array: [1555, 1116, 221, 997]
• Ending array: [16, 9, 5, 25]
• Where each number in the ending array is the sum of the digits of the corresponding number in the starting array.
• How do I get from the starting array to the ending array?

Answer 1

You can use list comprehension where each number is converted to string -> splitted to list of chars -> converted to digits(numbers) back -> summed:

start = [1555, 1116, 221, 997]
end = [sum(map(int, list(str(x)))) for x in start]
print(end) # output: [16, 9, 5, 25]

Answer 2

I believe what you are asking is to create an array whose elements are the sum of the digits of each number in the original array. The number digit summation is shamelessly stolen from Sum the digits of a number - python , and putting it all together you get:

def sum_digits(n):
   r = 0
   while n:
       r, n = r + n % 10, n // 10
   return r

in_array = [1555, 1116, 221, 997]
out_array = [sum_digits(n) for n in in_array]
print(out_array)

Hope this helps!

Answer 3

• Using sum and a list-comprehension
• Without map & list

t = [1555, 1116, 221, 997]

result = [sum(int(d) for d in str(v)) for v in t]

print(result)
[16, 9, 5, 25]

• Corresponding for-loop

result = list()
for v in t:
    r = list()
    for d in str(v):
        r.append(int(d))
    result.append(sum(r))

%%timeit comparison

• Given the following test list

import numpy as np

np.random.seed(123)
t = [np.random.randint(100, 4000) for _ in range(4000000)]

• This answer

%%timeit
[sum(int(d) for d in str(v)) for v in t]

5.27 s ± 60.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

• Answer from Gadip

%%timeit
[sum(map(int, list(str(x)))) for x in t]

4.63 s ± 37 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

• Answer from minerharry

def sum_digits(n):
    r = 0
    while n:
        r, n = r + n % 10, n // 10
    return r

%%timeit
[sum_digits(n) for n in t]

1.72 s ± 10.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)