Avoiding string + integer addition when using recursive function

Question

I am trying to build a function(which uses function recursion) that scans a number, n, to look for a digit d, & if found, I would like to replace d with a specified number r, as shown in the code below. This code works fine but the output is in a string format. I have tried numerous ways to change it to output an integer but to no avail. Thanks for the help!

def replace_digit(n, d, r):
    number = str(n)
    i = 0
    if len(number) == 1:
        if number == str(d):
            return str(r)
        else:
            return number
    else:
        if number[i] == str(d):
            return number[:i] + str(r) + replace_digit(int(number[i+1:]),d,r)
        else:
            return number[i] + replace_digit(int(number[i+1:]),d ,r)

Is it an exercise? in other words, do you specifically required to use recursion? this is a very trivial task. — DeepSpace
– DeepSpace, Commented Aug 18, 2016 at 10:14
Yeah, I have to specifically write in recursion, hence the problem. Thanks. — jaren
– jaren, Commented Aug 18, 2016 at 10:15

cdlane · Accepted Answer · 2016-08-18 22:43:07Z

1

def replace_digit(number, digit, replacement):

    if number == 0:
        return number  # base case

    quotient, remainder = divmod(number, 10)

    if remainder == digit:
        remainder = replacement

    return replace_digit(quotient, digit, replacement) * 10 + remainder


print(replace_digit(961748941982451653, 9, 2))

OUTPUT

261748241282451653

edited Aug 18, 2016 at 22:43

answered Aug 18, 2016 at 22:36

cdlane

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1 Comment

ElmoVanKielmo Over a year ago

Awesome. I was also thinking about a solution based purely on integer calculations, but I had the feeling that OP is actually struggling with understanding type conversions.

polku · Accepted Answer · 2016-08-18 10:35:47Z

0

If you already have a working function why not just split the problem ?

def replace(n, d, r):
    def replace_digit(n, d, r): # doesn't change
        return ...
    return int(replace_digit(str(n), str(d), str(r))

answered Aug 18, 2016 at 10:35

polku

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1 Comment

jaren Over a year ago

Thanks a lot man! Was getting close to a solution of some sort like this. :)

Sevanteri · Accepted Answer · 2016-08-18 10:47:54Z

0

The solution is to wrap the return values with int() as has been already been stated in the comments and other answers.

But here's a version that doesn't use string manipulation at all. Just for fun.

def replace_digit(n, d, r):
    rest = n // 10  # all but the rightmost digit
    digit = n - rest * 10  # only the rightmost digit
    digit = r if digit == d else digit
    if rest == 0:
        return digit

    return replace_digit(rest, d, r) * 10 + digit

answered Aug 18, 2016 at 10:47

Sevanteri

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Comments

ElmoVanKielmo · Accepted Answer · 2016-08-18 12:56:01Z

0

It's quite simple but you need some type conversions between str and int:

def replace_digit(n, d, r):
    number = str(n)
    rest = str(replace_digit(int(number[1:]), d, r)) if len(number) > 1 else ""
    digit = number[0]
    digit = str(r) if digit == str(d) else digit
    return int(digit + rest)

There's also another possiblity, using a wrapper. This limits number of type conversions.

def replace_digit(n, d, r):
    def replace(n, d, r):
        rest = replace(n[1:], d, r) if len(n) > 1 else ""
        return r + rest if n[0] == d else n[0] + rest

    return int(replace(str(n), str(d), str(r)))

edited Aug 18, 2016 at 12:56

answered Aug 18, 2016 at 10:26

ElmoVanKielmo

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Comments

Carlos Afonso · Accepted Answer · 2016-08-18 13:05:46Z

0

Instead of returning the concatenation try this:

if number[i] == str(d):
    new_number = number[:i] + str(r) + replace_digit(int(number[i+1:]),d,r)
else:
    new_number = number[i] + replace_digit(int(number[i+1:]),d ,r)

return int(new_number)

answered Aug 18, 2016 at 13:05

Carlos Afonso

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