Indent a string using given parentheses

Question

Given the following input to the program:

List of block start characters
List of block end characters
A string to format

format the string with the blocks delimited by the two character sets indented.

Formatting is done with two spaces per level and the parentheses are placed as shown in the example below. You may assume the sets of opening and closing characters to be disjoint.

E.g. for {[(< and }])> as the opening and closing character sets and the following string:

abc{xyz{text[note{comment(t{ex}t)abc}]}}

the following output would be expected:

abc
{
  xyz
  {
    text
    [
      note
      {
        comment
        (
          t
          {
            ex
          }
          t
        )
        abc
      }
    ]
  }
}

You may not hard-code the list of “parentheses” characters. How input is given is not specified, though; this could be either command-line arguments or via standard input, as you wish.

Can we assume that for each parenthesis there's a closing one, and in the same order? — Champo, Jul 5, 2011 at 1:22
Does the program have to support any parenthesis characters given as arguments? e.g. ./program 'p' 'q' <<< '1p23p45q67q8' Or does it need only support {[(< and }])> ? — Joey Adams, Jul 5, 2011 at 3:27
@Joey, I'd assume not, though that would be all the more impressive. — Neil, Jul 5, 2011 at 7:39
joey : input are 1. open parenthesis characters 2.close parenthesis chars 3. string to indent. Juan : we can assume that, though code need not rely on that, what I mean is if delim is part of opening parenthesis chars, increase indent, else if part of closing parenthesis chars decrease indent. — Prashant Bhate, Jul 5, 2011 at 11:25

Lowjacker · Accepted Answer · 2011-07-09 23:20:08Z

7

Ruby, 106 101 96 95

s,e,i=$*
i.scan(/[#{z=Regexp.quote s+e}]|[^#{z}]*/){|l|puts'  '*(s[l]?~-$.+=1:e[l]?$.-=1:$.)+l}

Input is provided via the command line.

edited Jul 9, 2011 at 23:20

answered Jul 6, 2011 at 22:56

Lowjacker

4,8691 gold badge19 silver badges19 bronze badges

1

\$\begingroup\$ You can save 4 characters by using ~-j+=1 instead of (j+=1;j-1). Additionally, using $. everywhere instead of j allows you to remove the j=0, which saves another character. \$\endgroup\$
– Ventero
Jul 7, 2011 at 11:37

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DCharness · Accepted Answer · 2011-08-02 06:42:52Z

Perl - 131 96 94 chars

$i="";for$_(split/([\Q$ARGV[0]$ARGV[1]\E])/,$ARGV[2]){$i=~s/..// if/[\Q$ARGV[1]\E]/;print "$i$_\n"if$_;$i.='  'if/[\Q$ARGV[0]\E]/;}

Seems like there should be room for eliminating common expressions, at least, but it's a quick take that handles the example, as well as Joey Adams's hypothetical about arbitrary brackets.

There was, indeed, plenty of room for improvement:

$_=pop;($s,$e)=map"[\Q$_\E]",@ARGV;for(split/($s|$e)/){print"  "x($i-=/$e/),"$_\n"if$_;$i+=/$s/}

...and still a little more:

$_=pop;($s,$e)=map"[\Q$_\E]",@ARGV;map{print"  "x($i-=/$e/),"$_\n"if$_;$i+=/$s/}split/($s|$e)/

Ry- · Accepted Answer · 2013-10-01 00:02:07Z

3

JavaScript, 255 227 205 characters

~~Hey, its length fits perfectly in a byte! :D~~

function(s,e,t){R=eval.bind(0,"Array(n).join(' ')");for(i=n=0,b=r='';c=t[i++];)~s.indexOf(c)?(r+=b,b='\n'+R(++n)+c+'\n '+R(++n)):~e.indexOf(c)?b+='\n'+((n-=2)?R()+' ':'')+c+'\n'+(n?R()+' ':''):b+=c;return r+b}

It's a function, pass it the start characters, the end characters, then the text.

edited Oct 1, 2013 at 0:02

answered Jul 5, 2011 at 17:23

Ry-

5,3831 gold badge23 silver badges32 bronze badges

\$\begingroup\$ Your own edit comment has been used against you. :D \$\endgroup\$
– Doorknob
Sep 30, 2013 at 23:38
\$\begingroup\$ @Doorknob: I… I thought I had never done that. D: I am so sorry. (Were you hunting?) \$\endgroup\$
– Ry-
Sep 30, 2013 at 23:52
\$\begingroup\$ @Doorknob: And thanks for reminding me about this; shortened :) \$\endgroup\$
– Ry-
Oct 1, 2013 at 0:02
\$\begingroup\$ No I wasn't hunting, just stumbled upon this question, but I decided to, and I found this :O :P \$\endgroup\$
– Doorknob
Oct 1, 2013 at 2:03

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Python – 162 chars

i=f=0
s=""
l,r,z=[raw_input()for c in'   ']
o=lambda:s+("\n"+"  "*i)*f+c
for c in z:
 if c in l:f=1;s=o();i+=1
 elif c in r:i-=1;f=1;s=o()
 else:s=o();f=0
print s

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 5, 2011 at 3:41

Champo

3,1681 gold badge17 silver badges19 bronze badges

\$\begingroup\$ Note that the task calls for the two sets of parentheses to be part of the input, not hardcoded. \$\endgroup\$
– Joey
Jul 5, 2011 at 12:51
\$\begingroup\$ @Joey noted, I'll get around to fixing that in a while. Thanks \$\endgroup\$
– Champo
Jul 5, 2011 at 13:40

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Python 2.7.X - 136 chars

import sys
a,c=sys.argv,0
for i in a[3]:
 if not(i in a[2]):print ' '*c+i
 else:print ' '*(c-4)+i
 if i in a[1]:c+=4
 if i in a[2]:c-=4

Usage : $ ./foo.py '(' ')' '(ab(cd(ef)gh)ij)'

Resulting Output:

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 2, 2011 at 16:01

arrdem

8465 silver badges15 bronze badges

\$\begingroup\$ Do you need the spaces after the print statements? \$\endgroup\$
– Adalynn
Dec 4, 2016 at 0:25

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FireFly · Accepted Answer · 2013-10-13 13:19:00Z

C - 213 209

^{I hate stupid mistakes... >.<}

#include<stdio.h>
#include<string.h>
int main(int i,char**s){for(char q,r,c,t,a=0;~(c=getchar());t=q|r){q=!!strchr(s[1],c);a-=r=!!strchr(s[2],c);for(i=0;t|q|r&&i<2*a+1;putchar(i++?' ':'\n'));a+=q;putchar(c);}}

Reads left-parens from first command-line argument, right-parens from second argument, and input to indent on stdin.

Pretty-printed & commented:

int main(int i, char **s) {
  for (char q, r, /* is left-paren? is right-paren? */
            c,    /* character read from input */
            t,    /* last char was a paren-char */
            a=0;  /* indentation */
       ~(c = getchar());
       t = q|r) {
         q = !!strchr(s[1],c);
    a -= r = !!strchr(s[2],c);
    for (i=0; t|q|r && i<2*a+1; putchar(i++? ' ' : '\n'));
    a += q;
    putchar(c);
  }
}

Harry K. · Accepted Answer · 2011-07-13 11:51:28Z

C (159 225 chars)

#define q(s,c)strchr(s,c)
#define p(i,j,k)printf("\n%*s%c%c%*s",i,"",*s,k,j,"")
g(char*b,char*e,char*s){int i;for(i=0;*s;s++)q(b,*s)?p(i-2,i+=2,'\n'):q(e,*s)?q(b,*(s+1))||q(e,*(s+1))?p(i-=2,i-2,0):p(i-=2,i-2,'\n'):putchar(*s);}

It cost me 66 extra characters just to fix the bug with the empty lines :( Frankly, I need a fresh approach, but I'll call it a day for now.

#define p(i,j)printf("\n%*s%c\n%*s",i,"",*s,j,"")
f(char*b,char*e,char*s){int i;for(i=0;*s;s++){strchr(b,*s)?p(i-2,i+=2):strchr(e,*s)?p(i-=2,i-2):putchar(*s);}}

A rather quick & dirty approach. It has a bug of producing empty lines between consecutive closing parenthesis, but otherwise it does the job (or so I think). I will revisit it for a better & cleaner solution, sometime this week.

char *b is the opening parenthesis set, char *e is the closing parenthesis set and char *s is the input string.

Gareth · Accepted Answer · 2011-07-31 09:55:53Z

1

Scala(2.9), 211 characters

object P extends App{def x(j:Int)={"\n"+"  "*j}
var(i,n)=(0,"")
for(c<-args(2)){if(args(0).exists(_==c)){print(x(i)+c)
i+=1
n=x(i)}else{if(args(1).exists(_==c)){i-=1
print(x(i)+c)
n=x(i)}else{print(n+c)
n=""}}}}

answered Jul 31, 2011 at 9:55

Gareth

11.4k1 gold badge36 silver badges85 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Perl - 89 85 bytes

A version of Hojung Youn's answer which accepts the block characters via two arguments.

#!perl -p
BEGIN{$b=pop;$a=pop}s/([$a])|([$b])|\w+/"  "x($1?$t++:$2?--$t:$t)."$&
"/ge

Called like:

perl golf.pl<<<'abc{xyz{text[note{comment(t{ex}t)abc}]}}' '[{(<' ']})>'

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 2, 2011 at 14:24

Sorpigal

1113 bronze badges

\$\begingroup\$ Very nice concept, @Hojung and Sorpigal. It is a bit fragile, though. For example, swap the ] and } in the close-paren argument, and the ] closes the character class, leading to an unmatched paren error. Similarly, suppose the open set starts with ^, perhaps to match v in the close set; you'll get the complement of the intended [$a] class. That's why I used \Q...\E in my answer. The \w+ for non-paren characters works for the example, but what about input like 'x(foo-bar)y' '(' ')'? Of course, it isn't clear the code needs to handle something like that. \$\endgroup\$
– DCharness
Aug 2, 2011 at 15:23

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heinrich5991 · Accepted Answer · 2013-04-03 17:37:11Z

1

Python3, 184 182 chars

import sys
_,p,q,t=sys.argv
i,f,x=0,1,print
for e in t:
 if e in p:f or x();x(' '*i+e);i+=2;f=1
 elif e in q:f or x();i-=2;f=1;x(' '*i+e)
 else:not f or x(' '*i,end='');f=x(e,end='')

Example:

$ python3 ./a.py '{[(<' '}])>' 'abc{xyz{text[note{comment(t{ex}t)abc}]}}'
abc
{
  xyz
  {
    text
    [
      note
      {
        comment
        (
          t
          {
            ex
          }
          t
        )
        abc
      }
    ]
  }
}

edited Apr 3, 2013 at 17:37

heinrich5991

1035 bronze badges

answered Jul 5, 2011 at 17:58

Alexandru

6,1544 gold badges30 silver badges50 bronze badges

\$\begingroup\$ heinrich5991 suggested saving two characters by changing the second line to _,p,q,t=sys.argv \$\endgroup\$
– Peter Taylor
Apr 3, 2013 at 8:53

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pinco · Accepted Answer · 2013-10-10 19:17:26Z

1

Groovy, 125

p=args;i=0;s={a,b->"\n"+"\t"*(b?i++:--i)+a+"\n"+"\t"*i};p[0].each{c->print p[1].contains(c)?s(c,1):p[2].contains(c)?s(c,0):c}

You can save the script in a file indent.groovy and try it with:
groovy indent.groovy "abc{xyz{text[note{comment(t{ex}t)abc}]}}" "{[(" ")]}"

answered Oct 10, 2013 at 19:17

pinco

3031 gold badge3 silver badges12 bronze badges

\$\begingroup\$ I tried around in groovy for an hour before seeing your answer, I used a similar aproach, but mine is much longer than yours so I won't even bother to post.. Good job! :) \$\endgroup\$
– Fels
Oct 11, 2013 at 17:33

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golfer9338 · Accepted Answer · 2013-10-14 09:02:14Z

Python - 407

from sys import*;o=argv[1];c=argv[2];t=argv[3];p=0;n=False;a=lambda:h not in e;b=lambda s:print(s+(" "*p)+h);r="";e=o+c
for h in t:
 for k in o:
  if h==k:
   if(r in e)and(r!=""):b("")
   else:b("\n")
   p+=2;n=True;break
 for k in c:
  if h==k:
   p-=2
   if(r in e)and(r!=""):b("")
   else:b("\n")
   n=True;break
 if a()and n:print((" "*p)+h,end="");n=False
 elif a():print(h,end="")
 r=h

An ungolfed version of the program:

import sys

open_set = sys.argv[1]
close_set = sys.argv[2]
text = sys.argv[3]
spaces = 0
newline = False
a = lambda : char not in b_set
b = lambda s: print(s + (" " * spaces) + char)
prev = ""
b_set = open_set + close_set

for char in text:
    for bracket in open_set:
        if char == bracket:
            if (prev in b_set) and (prev != ""):
                b("")
            else:
            b("\n")
        spaces += 2
        newline = True
        break
    for bracket in close_set:
        if char == bracket:
            spaces -= 2
            if (prev in b_set) and (prev != ""):
                b("")
            else:
                b("\n")
            newline = True
            break
    if a() and newline:
        print((" " * spaces) + char, end="")
        newline = False
    elif a():
        print(char, end="")
    prev = char

The arguments to the program are (in order): the opening parentheses, the closing parentheses, and the text to indent.

Example ($ is command line prompt):

$ python indent.py "{[(<" "}])>" "abc{xyz{text[note{comment(t{ex}t)abc}]}}"
abc
{
  xyz
  {
    text
    [
      note
      {
        comment
        (
          t
          {
            ex
          }
          t
        )
        abc
      }
    ]
  }
}

Mihai Bişog · Accepted Answer · 2011-08-04 21:56:04Z

0

C 284 Non White space Characters

I'm no fan of obfuscation but well...

#include<cstdio>
#include<cstring>
#define g printf
#define j char
int main(int a,j**b){int c=0;for(j*f=b[3];*f!='\0';++f){if(strchr(b[1],*f)!=0){g("\n%*c\n%*c",c,*f,c+2,'\0');c+=2;}else if(strchr(b[2],*(f))!=0){c-=2;g("\n%*c",c,*f);if(strchr(b[2],*(f+1))==0)g("\n%*c",c,'\0');}else putchar(*f);}}

Usage: ./program start_brackets end_brackets string_to_parse

answered Aug 4, 2011 at 21:56

Mihai Bişog

1012 bronze badges

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migimaru · Accepted Answer · 2011-08-08 13:56:41Z

0

php (187) (153)

function a($s,$o,$e){while(''!=$c=$s[$i++]){$a=strpbrk($c,$o)?2:0;$b=strpbrk($c,$e)?2:0;echo ($a+$b||$r)?"\n".str_pad('',$t-=$b):'',$c;$t+=$a;$r=$a+$b;}}

Function takes string, opening delimiters, ending delimiters as arguments.

edited Aug 8, 2011 at 13:56

answered Aug 6, 2011 at 15:41

migimaru

1,1008 silver badges13 bronze badges

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tkf · Accepted Answer · 2013-04-02 21:51:26Z

C, 256

Parameters:

e is the ending char,
n is the indentation,
b the opening brackets,
d the closing brackets.

I broke the code up to avoid the horizontal scrollbar.

#define r char
#define P(c) putchar(c);
#define N P(x)
#define W printf("%*s",n,"");
r*s,x='\n';i(r e,int n,r*b,r*d){r*t=s,*p;int l=0;W while(*s!=e)    
{if(p=strchr(b,*s)){if(s!=t){N W}P(*s++)N i(d[p-b],n+2,b,d); N W 
P(*s++);l=1;}else{if(l){N W l=0;}P(*s++)}}}

Complete program is 363 characters.

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#define r char
#define P(c) putchar(c);
#define N P(x)
#define W printf("%*s",n,"");
r*s,x='\n';i(r e,int n,r*b,r*d)
{r*t=s,*p;int l=0;W while(*s!=e)
{if(p=strchr(b,*s)){if(s!=t){N W}
P(*s++)N i(d[p-b],n+2,b,d); N W
P(*s++);l=1;}else{if(l){N W l=0;}
P(*s++)}}}main(int c,r*v[]){s =
v[3];i('\0',0,v[1],v[2]);}

Community · Accepted Answer · 2020-06-17 09:04:33Z

Powershell, 146 Bytes

param([char[]]$s,[char[]]$e,[char[]]$f)$f|%{}{if($_-in$s){$o;'  '*$i+$_;$o='  '*++$i;}elseif($_-in$e){$o;'  '*--$i+$_;$o='  '*$i}else{$o+=$_}}{$o}

Ungolfed Explanation

param([char[]]$start,             # Cast as array of Chars
      [char[]]$end,
      [char[]]$string)
$string | foreach-object { } {    # For every char in string. Empty Begin block
    if ( $_ -in $start ) {        # If char is in start
        $o                        # Print stack ($o)
        '  ' * $i + $_            # Newline, indent, insert start char
        $o = '  ' * ++$i          # Set stack to ident (incremented)
    } elseif ( $_ -in $end ) {    # If char is in end
        $o                        # Print stack
        '  ' * --$i + $_          # Newline, decrement indent, insert end char
        $o = '  ' * $i            # Set stack to indent
    } else {
        $o+ = $_                  # Otherwise add character to stack
    }
} { $o }                          # Print remaining stack (if any)

Cole Cameron · Accepted Answer · 2015-10-08 14:30:04Z

0

C, 181 characters

#define d(m,f)if(strchr(v[m],*s)){puts("");for(j=f;j--;)printf("  ");}
i;main(j,v,s)char**v,*s;{for(s=v[3];*s;s++){d(1,i++)d(2,--i)putchar(*s);d(1,i)if(!strchr(v[2],*(s+1)))d(2,i)}}

Pretty much the most straightforward approach imaginable. Iterate through the string (v[3]), if it's a left brace (as defined in v[1]), increase the indent level, if it's a right brace (as defined in v[2]), decrease the indent level.

answered Oct 8, 2015 at 14:30

Cole Cameron

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Stack Exchange Network

Indent a string using given parentheses

17 Answers 17

Ruby, 106 101 96 95

Perl - 131 96 94 chars

JavaScript, 255 227 205 characters

Python – 162 chars

Python 2.7.X - 136 chars

C - 213 209

C (159 225 chars)

Scala(2.9), 211 characters

Perl - 89 85 bytes

Python3, 184 182 chars

Groovy, 125

Python - 407

C 284 Non White space Characters

php (187) (153)

C, 256

Powershell, 146 Bytes

C, 181 characters

Your Answer

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string
whitespace
balanced-string
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Indent a string using given parentheses

17 Answers 17

Ruby, 106 101 96 95

Perl - 131 96 94 chars

JavaScript, 255 227 205 characters

Python – 162 chars

Python 2.7.X - 136 chars

C - 213 209

C (159 225 chars)

Scala(2.9), 211 characters

Perl - 89 85 bytes

Python3, 184 182 chars

Groovy, 125

Python - 407

C 284 Non White space Characters

php (187) (153)

C, 256

Powershell, 146 Bytes

C, 181 characters

Your Answer

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