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up vote 3 down vote favorite

Given a string containing decimal numbers:

teststring134this 123test string54 100

increment every number in this string by one to give the new string

teststring135this 124test string55 101.

The string can be provided as:

  • a command line argument
  • STDIN
  • a hard-coded variable or function argument

Cover all possible positions for a number:

  • as a prefix for a word; 123test ► 124test
  • as a suffix for a word; test123 ► test124
  • inside a word; te123st ► te124st
  • alone test 123 test ► test 124 test

Here's a non-golfed solution in Python:

NUMBERS = '0123456789'

def increment(s):
    out = ''

    number = ''
    for c in s:
        if c in NUMBERS:
            number += c
        else:
            if number != '':
                out += str(int(number) + 1)
                number = ''
            out += c

    if number != '':
        out += str(int(number) + 1)
        number = ''

    return out


print "\"%s\"" % (increment('teststring134this 123test string54 100'))

This is a code-golf question, shortest code wins.

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2  
Fun fact: this can be done with 3 pure regex substitutions (no callbacks) stackoverflow.com/questions/12941362/… (that wouldn't be the golfiest way though) –  Martin Büttner 5 hours ago
4  
You specified input but not output. From your input spec I assume both STDOUT and and return value are fine. But can we also store the result in a hardcoded variable (just as we can take input from it)? –  Martin Büttner 5 hours ago
    
Your code shouldn't work... Don't you need to declare NUMBERS as a global variable inside increment(s) or is it different in Python 2? –  Beta Decay 2 hours ago

9 Answers 9

up vote 6 down vote

Perl, 23

Assumes the input string is assigned to $_

s/\d+/@{[$&+1]}/g;print
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up vote 6 down vote

Perl, 14 bytes

s/\d+/$&+1/ge

Requires the -p switch, which I have counted as one byte.

Example run

$ perl -p <(echo 's/\d+/$&+1/ge') <<< 'teststring134this 123test string54 100'
teststring135this 124test string55 101
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up vote 4 down vote

Python 2 - 59

Supply the string as the variable n

import re
print re.sub('\d+',lambda x:`int(x.group())+1`,n)
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up vote 4 down vote

J (20)

Expects the input to be stored in the variable a.

'\d+'>:&.".rxapply a

Test:

   a=:'teststring134this 123test string54 100'
   '\d+'>:&.".rxapply a
teststring135this 124test string55 101
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up vote 4 down vote

Ruby, 30 24 bytes

$><<s.gsub(/\d+/,&:next)

Expects the input to be stored in s.

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1  
Couldn't $1.next be used in the block? –  August 4 hours ago
    
@August nice, thanks! I didn't know next was that sophisticated. –  Martin Büttner 4 hours ago
up vote 4 down vote

JavaScript (ES6) - 28

H=k=>k.replace(/\d+/g,k=>++k)

Run by using H("test 123 234t").

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up vote 2 down vote

C 191 (if line endings count as 1)

This will never win but it is in a language that doesn't lend itself to golfing...

This is using a hard coded string but the contents of the string are not counted towards the total.

#include<stdlib.h>
#include<string.h>
#include<stdio.h>
i;main(){char o[99],*p=o,*q,*s="test123test999test-1test";for(;*s;*p=0){i=strtol(s,&q,0);if(q>s){sprintf(p,"%d",i+1);p+=strlen(p);s=q;}else*p++=*s++;}puts(o);}

Basically it runs through the string one character at a time checking each to see if it is an integer. If it is then it increments the integer and writes it to the output otherwise it copies the current character to the output.

This leaks the hardcoded string because it increments s.

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1  
I'm sure you can get rid of some of the includes and it will still compile fine with gcc. –  Martin Büttner 4 hours ago
    
Will this work with a string containing e.g. 99? –  anatolyg 23 mins ago
up vote 1 down vote

Lua - 68 characters

d='(%D-)'for k,i,j in s:gmatch(d..'(%d+)'..d)do io.write(k,i+1,j)end

Expects the input to be stored in s.

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up vote 0 down vote

Emacs - 20 characters

C-M-% [0-9]+ RET \,(1+ \#0) RET !

Requires text to be processed to be present in the current buffer. I counted C-M-% as one character here since it can be entered with one keystroke when holding down three modifiers.

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