Sum the digits of a number

Question

If I want to find the sum of the digits of a number, i.e.:

• Input: 932
• Output: 14, which is (9 + 3 + 2)

Answer 1

def sumOfDigits():

    n=int(input("enter digit:")) 
    sum=0
    while n!=0 :

        m=n%10
        n=n/10
        sum=int(sum+m)

    print(sum)

sumOfDigits()

Answer 2

A base 10 number can be expressed as a series of the form

a × 10^^p + b × 10^^p-1 .. z × 10^⁰

so the sum of a number's digits is the sum of the coefficients of the terms.

Based on this information, the sum of the digits can be computed like this:

import math

def add_digits(n):
    # Assume n >= 0, else we should take abs(n)
    if 0 <= n < 10:
        return n
    r = 0
    ndigits = int(math.log10(n))
    for p in range(ndigits, -1, -1):
        d, n = divmod(n, 10 ** p)
        r += d
    return r

This is effectively the reverse of the continuous division by 10 in the accepted answer. Given the extra computation in this function compared to the accepted answer, it's not surprising to find that this approach performs poorly in comparison: it's about 3.5 times slower, and about twice as slow as

sum(int(x) for x in str(n))

Answer 3

I cam up with a recursive solution:

def sumDigits(num):
#   print "evaluating:", num
  if num < 10:
    return num

  # solution 1
#   res = num/10
#   rem = num%10
#   print "res:", res, "rem:", rem
#   return sumDigits(res+rem)

    # solution 2
  arr = [int(i) for i in str(num)]
  return sumDigits(sum(arr))

# print(sumDigits(1))
# print(sumDigits(49))
print(sumDigits(439230))
# print(sumDigits(439237))

Answer 4

reduce(op.add,map(int,list(str(number))))

Test:

from datetime import datetime
number=49263985629356279356927356923569976549123548126856926293658923658923658923658972365297865987236523786598236592386592386589236592365293865923876592385623987659238756239875692387659238756239875692856239856238563286598237592875498259826592356923659283756982375692835692385653418923564912354687123548712354827354827354823548723548235482735482354827354823548235482354823548235482735482735482735482354823548235489235648293548235492185348235481235482354823548235482354823548235482354823548234



startTime = datetime.now()

for _ in  range(0,100000) :
    out=reduce(op.add,map(int,list(str(number))))

now=datetime.now()
runningTime=(now - startTime)

print ("Running time:%s" % runningTime)
print(out)

Running time:0:00:13.122560 2462

Answer 5

It only works for three-digit numbers, but it works

a = int(input())
print(a // 100 + a // 10 % 10 + a % 10)

Answer 6

Doing some Codecademy challenges I resolved this like:

def digit_sum(n):
arr = []
nstr = str(n)
for x in nstr:
    arr.append(int(x))
return sum(arr)