CODEWITHSUNDEEP
pythonpython-3.x
M-M
M-M

Reputation: 901

Adding all digits for a positive integer

I am trying to add all the digits of a positive integer. When I test out the number 434, the result was 4, instead of 11.

Here is what I have, I don't understand why is my code not going through all the digits of the number. How to correct it?

def digit_sum(x):
  n=0
  if x>0:
    #get the end digit of the number,add to n
      n+=x%10
      #remove the end digit of the number
      x=x//10
  return n

Upvotes: 0

Views: 610

Answers (4)

jeppoo1
jeppoo1

Reputation: 698

Although this is not a function, it helped me:

digits = "12345678901234567890"
digit_sum = sum(map(int, digits))
print("The equation is: ", '+'.join(digits))
print("The sum is:", digit_sum)

(This prints:

The equation is:  1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9+0
The sum is: 90

)

Upvotes: 1

paxdiablo
paxdiablo

Reputation: 881563

Two main constructs of procedural programming are:

  1. Choosing among different things once. (Selection)
  2. Doing things more that once. (Iteration)

An if statement is a selection statement. Since you want to add up all the digits in the number, your use of if is inappropriate. In other words, using a selection statement will only give you the sum of the first digit you process (the final 4 in 434).

Instead you should use an iteration statement like while:

def digit_sum(number):
  sumOfDigits = 0
  while number > 0:
      sumOfDigits += number % 10
      number = number // 10
  return sumOfDigits

You'll notice I've used more descriptive variable names. That's a good habit to get into because it self-documents your code.

You can also look into more Pythonic ways of doing this, such as:

def digit_sum(number):
    return sum([int(digit) for digit in str(number)])

Whether that can be considered better is open for debate but it's a well-known way of writing Python succinctly.

Upvotes: 2

Anshul Goyal
Anshul Goyal

Reputation: 76887

Instead of the if statement, use a while statement:

def digit_sum(x):
  n=0
  while x>0:
    #get the end digit of the number,add to n
      n+=x%10
      #remove the end digit of the number
      x=x//10
  return n

The if will evaluate only once and return value on n after one iteration, while runs the code in a loop till all digits are added to the sum.

This will give:

In [2]: digit_sum(10)
Out[2]: 1

In [3]: digit_sum(434)
Out[3]: 11

Alternatively, a one liner:

In [4]: digit_sum = lambda x: sum(map(int, str(x)))

In [5]: digit_sum(434)
Out[5]: 11

If instead of 11, you want to sum those digits again as well to get a single digit, just recurse:

In [6]: def digit_sum(x):
   ...:   n=0
   ...:   while x>0:
   ...:     #get the end digit of the number,add to n
   ...:       n+=x%10
   ...:       #remove the end digit of the number
   ...:       x=x//10
   ...:   return n if n < 10 else digit_sum(n)
   ...: 

In [7]: digit_sum(434)
Out[7]: 2

Upvotes: 1

Jake Reece
Jake Reece

Reputation: 1168

I would convert the number to a string, then parse and sum each digit:

def digit_sum(x):
    numList = [int(d) for d in str(x)]
    return sum(numList)

Upvotes: 0

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