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   What is the fastest way of doing this  I instinctively did  sum int digit  for digit in str number    and I found this online  sum map int  str number     Which is best to use for speed  and are there any other methods which are even faster

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This might help  def digit sum n       num str   str n      sum   0     for i in range 0  len num str            sum    int num str i       return sum

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It only works for three-digit numbers  but it works a   int input    print a    100   a    10   10   a   10

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If you want to keep summing the digits until you get a single-digit number  one of my favorite characteristics of numbers divisible by 9  you can do   def digital root n       x   sum int digit  for digit in str n       if x  lt  10          return x     else          return digital root x    Which actually turns out to be pretty fast itself      timeit digital root 12312658419614961365   10000 loops  best of 3  22 6   s per loop

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Doing some Codecademy challenges I resolved this like  def digit sum n       digits          nstr   str n      for x in nstr          digits append int x       return sum digits

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def digitsum n       result   0     for i in range len str n             result   result   int str n  i i 1       return result     result  is initialized with 0   Inside the for loop  the number n  is converted into a string to be split with loop index i  and get each digit  ---  str n  i i 1   This sliced digit is converted back to an integer ----  int str n  i i 1    And hence added to result

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Found this on one of the problem solving challenge websites   Not mine  but it works  num   0              replace 0 with whatever number you want to sum up print sum  int k  for k in str num

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Both lines you posted are fine  but you can do it purely in integers  and it will be the most efficient   def sum digits n       s   0     while n          s    n   10         n     10     return s   or with divmod   def sum digits2 n       s   0     while n          n  remainder   divmod n  10          s    remainder     return s     Even faster is the version without augmented assignments   def sum digits3 n      r   0    while n         r  n   r   n   10  n    10    return r      gt   timeit sum digits n  1000000 loops  best of 3  574 ns per loop   gt   timeit sum digits2 n  1000000 loops  best of 3  716 ns per loop   gt   timeit sum digits3 n  1000000 loops  best of 3  479 ns per loop   gt   timeit sum map int  str n    1000000 loops  best of 3  1 42 us per loop   gt   timeit sum  int digit  for digit in str n    100000 loops  best of 3  1 52 us per loop   gt   timeit sum int digit  for digit in str n   100000 loops  best of 3  2 04 us per loop

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num   123 dig   0 sum   0 while num  gt  0     dig   int num 10    sum   sum dig   num   num 10   print sum     make sure to add space above this line

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The best way is to use math  I knew this from school  kinda also from codewars  def digital sum num       return  num   9  or num and 9  Just don t know how this works in code  but I know it s maths If a number is divisible by 9 then  it s digital sum will be 9  if that s not the case then num   9 will be the digital sum

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you can also try this with built in function called divmod      number   int input  enter any integer        sum   0 while number  0       take   divmod number  10       dig   take 1       sum    dig      number   take 0   print sum     you can take any number of digit

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Try this      print sum list map int input  Enter your number

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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

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Here is a solution without any loop or recursion but works for non-negative integers only  Python3    def sum digits n       if n  gt  0          s    n-1     9             return n-9 s     return 0

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Here is the best solution I found  function digitsum n        n   n toString        let result   0      for  let i   0  i  lt  n length  i              result    parseInt n i              return result    console log digitsum 192

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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

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I cam up with a recursive solution   def sumDigits num       print  evaluating    num   if num  lt  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

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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  gt   0  else we should take abs n      if 0  lt   n  lt  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

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n   str input  Enter the number n     list1       for each number in n           list1 append int each number    print sum list1

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You can try this  def sumDigits number       sum   0     while number gt 0           lastdigit   number 10         sum    lastdigit         number   number  10      return sum

[python] Sum the digits of a number

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