String index out of range, Python

Question

code for a function which increments a string, to create a new string. If the string already ends with a number, the number should be incremented by 1. If the string does not end with a number the number 1 should be appended to the new string.

The output is correct but it is showing a String index out of range error. Can someone help me on where and how the string index is out of range?

test cases,expected output: (increment_string("foo"), "foo1"),(increment_string("foobar001"), "foobar002"),(increment_string("foobar1"), "foobar2"),(increment_string("foobar00"), "foobar01"),("foobar99"), "foobar100"),("foobar099"), "foobar100"),(increment_string(""), "1")

def increment_string(strng):
   if strng[-1].isdigit():
      exp_strng=strng[::-1]
      new_strng=""
      new_strng1=""
      for i in exp_strng:
         if i.isdigit():
            new_strng+=i
         else:
            break
      new_strng=new_strng[::-1]
      new_strng1=int(new_strng)+1
      new_strng1='{num:{fill}{width}}'.format(num=new_strng1, fill='0', width=len(new_strng))
      return(strng[:-len(new_strng)]+new_strng1)

   else:
      strng+="1"
      return(strng)

Huh, I am not seeing the string index out of range error. Are you sure this is the entirety of your erroneous code? — Jerrybibo
– Jerrybibo, Commented Jul 30, 2017 at 14:13
You're passing a string into new_strng1=int(new_strng)+1. Did you mean len(new_strng) + 1? — Skam
– Skam, Commented Jul 30, 2017 at 14:14
no new_strng1 is to convert string to integer then adding 1 to it @Skam — Yami Danchou
– Yami Danchou, Commented Jul 30, 2017 at 14:16
@Jerrybibo it shows error in codewars website while submitting code. all outputs are passed though. — Yami Danchou
– Yami Danchou, Commented Jul 30, 2017 at 14:18
does it provide you with the input used? if we can't reproduce, then it's hard for us to help you. ¯_(ツ)_/¯ — Skam
– Skam, Commented Jul 30, 2017 at 14:18

Jerrybibo · Accepted Answer · 2017-07-30 14:33:23Z

2

Since you gave us more information on the test cases given, you can bypass the edge case of an empty string by modifying the if statement:

def increment_string(strng):
   # Add it here #
   if strng == "":
      return "1"

   elif strng[-1].isdigit():
      exp_strng = strng[::-1]
      new_strng = ""
      new_strng1 = ""
      for i in exp_strng:
         if i.isdigit():
            new_strng += i
         else:
            break
      new_strng = new_strng[::-1]
      new_strng1 = int(new_strng) + 1
      new_strng1 = '{num:{fill}{width}}'.format(num=new_strng1, fill='0', width=len(new_strng))
      return strng[:-len(new_strng)] + new_strng1

   else:
      strng += "1"
      return strng

answered Jul 30, 2017 at 14:33

Jerrybibo

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Comments

gribvirus74 · Accepted Answer · 2017-07-30 15:03:31Z

2

If think this would be a better solution to your problem:

from re import search

def increment_string(s):
    number = search('\d+$', s)
    if number != None:
        number = number.group()
        first_part = s.split(number)[0]
        return first_part + str(int(number)+1)
    else:
        return s + '1'

I don't know what you want when the number is 9 though: 0 or 10. This code produces 10.

edited Jul 30, 2017 at 15:03

answered Jul 30, 2017 at 14:29

gribvirus74

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2 Comments

Jerrybibo Over a year ago

This does not solve the problem that OP has – the event in which an empty string "" is given as the argument.

gribvirus74 Over a year ago

I've edited it and now it should work perfectly, sorry for the mess.

Yami Danchou · Accepted Answer · 2017-07-30 14:39:31Z

-1

the error was caused when empty string is passed. and I resolved it by adding one more if else:(thanks to Skam)

def increment_string(strng):
if len(strng)>0:
    if strng[-1].isdigit():
        exp_strng=strng[::-1]
        new_strng=""
        new_strng=""
        for i in exp_strng:
            if i.isdigit():
                new_strng+=i
            else:
                break
        new_strng=new_strng[::-1]
        new_strng1=int(new_strng)+1
        new_strng1=str(new_strng1)
        new_strng1=new_strng1.zfill(len(new_strng))
        return(strng[:-len(new_strng)]+new_strng1)
    else:
        strng+="1"
        return(strng)
else:
    strng+="1"
    return(strng)

answered Jul 30, 2017 at 14:39

Yami Danchou

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