Sum of digits in a string

Question

if i just read my sum_digits function here, it makes sense in my head but it seems to be producing wrong results. Any tip?

def is_a_digit(s):
''' (str) -> bool

Precondition: len(s) == 1

Return True iff s is a string containing a single digit character (between
'0' and '9' inclusive).

>>> is_a_digit('7')
True
>>> is_a_digit('b')
False
'''

return '0' <= s and s <= '9'

def sum_digits(digit):
    b = 0
    for a in digit:
        if is_a_digit(a) == True:
            b = int(a)
            b += 1

    return b

For the function sum_digits, if i input sum_digits('hihello153john'), it should produce 9

Answer 1

An equivalent for your code, using list comprehensions:

def sum_digits(your_string):
    return sum(int(x) for x in your_string if '0' <= x <= '9')

It will run faster then a "for" version, and saves a lot of code.

Answer 2

Just a variation to @oscar's answer, if we need the sum to be single digit,

def sum_digits(digit):
    s = sum(int(x) for x in str(digit) if x.isdigit())
    if len(str(s)) > 1:
        return sum_digits(s)
    else:
        return s

Answer 3

Another way of doing it:

def digit_sum(n):
  new_n = str(n)
  sum = 0
  for i in new_n:
    sum += int(i)
  return sum

Answer 4

Notice that you can easily solve this problem using built-in functions. This is a more idiomatic and efficient solution:

def sum_digits(digit):
    return sum(int(x) for x in digit if x.isdigit())

sum_digits('hihello153john')
=> 9

In particular, be aware that the is_a_digit() method already exists for string types, it's called isdigit().

And the whole loop in the sum_digits() function can be expressed more concisely using a generator expression as a parameter for the sum() built-in function, as shown above.

Answer 5

One liner

sum_digits = lambda x: sum(int(y) for y in x if y.isdigit())

Answer 6

Another way of using built in functions, is using the reduce function:

>>> numeric = lambda x: int(x) if x.isdigit() else 0
>>> reduce(lambda x, y: x + numeric(y), 'hihello153john', 0)
9