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Gemasoft
Asked: 2020-12-24 03:31:26 +0800 CST 2020-12-24 03:31:26 +0800 CST

How can I get the digital root of a positive number using Python?

  • 772

I need to get the digital root of a positive number using Python, so far my algorithms don't work.

In theory, the digits of each number should be added until there is only 1 digit left in the number.

For example:

127 = 1+2+7 = 10
1+0 = 1

Therefore the digital root of 127 would be 1.

This was my last algorithm:

def getDigitalRoot(num):
    return num if num==9 else num%9

But it doesn't work well since if I put for example 99 it returns 0 and it would have to be 9.

python

4 Answers

  1. Using a recursive function, you can do something like this:

    def raiz_digital(numero):
    suma = sum(int(digito) for digito in str(numero))
    if suma < 10:
        return suma
    else:
        return raiz_digital(suma)

print raiz_digital(12345) # Resultado: 6
print raiz_digital(98765) # Resultado: 8

    What it does first is calculate the sum of all the digits by converting the integer to a string:

    >>> numero = 12345
>>> [digito for digito in str(numero)]
['1', '2', '3', '4', '5']

    Then each digit is converted back to an integer to be added with the function sum:

    >>> [int(digito) for digito in str(numero)]
[1, 2, 3, 4, 5]
>>> sum([int(digito) for digito in str(numero)])
15

    It is verified that the result is a one-digit number ( suma < 10) and recursion is used if it is greater than one digit to finally obtain the digital root.

    Update

    Using @Darkhogg's suggestion, I've changed the list comprehension to a generator, which only affects the sum line. Changed from:

    suma = sum([int(digito) for digito in str(numero)])

    A:

    suma = sum(int(digito) for digito in str(numero))
    • 6
  2. Best Answer

    I found a pretty elegant way to get what I was looking for:

    def getDigitalRoot(num):
    return num if num == 0 else num % 9 or 9

    In this way I solved the problem I had when putting 99 for example, before it returned me 0 and now it gives me 9 correctly.

    • 3

  3. The problem can be solved without recursion and by using the reduce builtin and the fact that in Python strings are iterable just like lists.

    def digital_root(n):
    n = str(n)
    while(len(n) > 1):
        n = str(reduce(lambda x, y: x + y, map(int, n)))
    return int(n)

print(digital_root(99))
print(digital_root(127))

    Result

    9
1

    The digital root of 127 is 1 and not 10 (appears in the question):

    • 1 + 2 + 7 -> 10 -> 1 + 0 -> 1
    • 2

  4. The digital root of n is equal to the remainder of dividing n by 9 dr(n)= n mod 9

    this is demonstrated in

    https://www.todoexpertos.com/preguntas/63pngi45njkno6f3/cual-es-el-procedimiento-que-permite-expresar-la-raiz-digital-por-medio-de-la-funcion-parte-entera-piso

    The only drawback is that if n is a multiple of 9, n mod 9 = 0 and we want 9 to come out. This is solved with the function

    dr(n) = (n - 1) mod 9 + 1

    In Python this would be def getDigitalRoot(num): return (num-1)%9 +1

    • 1