Bingkai ASCII-Art yang bagus ini

30

pengantar

Saya pikir semua orang setuju bahwa gambar yang bagus harus memiliki bingkai yang bagus. Tetapi sebagian besar tantangan di situs ini tentang ASCII-Art hanya ingin gambaran mentah dan tidak peduli tentang pelestariannya.
Bukankah lebih baik jika kita memiliki program yang membutuhkan ASCII-Art dan mengelilinginya dengan bingkai yang bagus?

Tantangan

Tulis sebuah program yang menggunakan ASCII-Art sebagai input dan outputnya dikelilingi oleh frame yang bagus.

Contoh:

*****
 ***
  *
 ***
*****

menjadi

╔═══════╗
║ ***** ║
║ *** ║
║ * ║
║ *** ║
║ ***** ║
╚═══════╝
  • Anda harus menggunakan karakter yang sama persis untuk bingkai seperti dalam contoh: ═ ║ ╔ ╗ ╚ ╝
  • Bagian atas dan bawah bingkai dimasukkan sebelum baris pertama dan setelah input terakhir.
  • Bagian kiri dan kanan frame harus memiliki satu bantalan ruang yang tepat untuk garis input terluas.
  • Mungkin tidak ada spasi putih memimpin atau tertinggal dalam output. Hanya baris tambahan yang diizinkan.
  • Anda dapat berasumsi bahwa input tersebut tidak memiliki spasi putih terkemuka yang tidak perlu.
  • Anda dapat mengasumsikan bahwa input tidak memiliki spasi spasi tambahan pada garis apa pun.
  • Anda tidak harus menangani input kosong.
  • Input hanya akan berisi karakter ASCII yang dapat dicetak dan baris baru.

Aturan

  • Fungsi atau program lengkap diizinkan.
  • Aturan default untuk input / output.
  • Celah standar berlaku.
  • Ini adalah , sehingga byte-count terendah akan menang. Tiebreak adalah pengiriman sebelumnya.

Selamat Coding!

Menggunakan beberapa ASCII-Art yang hebat, yang dihasilkan dalam setiap tantangan di situs ini, sebagai masukan untuk program Anda dan menunjukkannya dengan kerangka yang bagus sangat dianjurkan!

Denker
sumber
29
Bingkai non-ASCII untuk seni ASCII? Bidaah!
Dennis
5
Sangat terkait erat. Tantangan yang sama, tetapi hanya menggunakan karakter tunggal (ASCII) untuk bingkai.
Martin Ender
13
(I should clarify I don't think it's a dupe. Having to use 6 different characters makes this a lot trickier. The other challenge can be solved by rotating the grid and appending # four times. Adapting such an approach here will be tricky at best, and not viable at worst.)
Martin Ender
6
@IsmaelMiguel I have won the previous contest and don't see how I could adapt my old answer at all.
Martin Ender
2
I suspect that DenkerAffe is assuming CP437 or something where the frame chars are also one byte.
Joshua

Jawaban:

6

CJam, 45 chars / 52 bytes

qN/_z,)[_)'═*N]2*C,3%'╔f+.\4/@@f{Se]'║S@2$N}*

Trying to avoid those expensive 3-byte chars was... interesting.

Try it online

Explanation

qN/                   Split input by newline
_z,                   Zip and get length L, i.e. length of longest line
)                     Increment -> L+1
[_)'═*N]              Make two-element array of "═"*(L+2) and newline
2*                    Double the array, giving ["═"*(L+2) "\n" "═"*(L+2) "\n"]

C,                    range(12), i.e. [0 1 2 ... 11]
3%                    Every third element, i.e. [0 3 6 9]
'╔f+                  Add "╔" to each, giving "╔╗╚╝"
.\                    Vectorised swap with the previous array, giving
                      ["╔" "═"*(L+2) "╗" "\n" "╚" "═"*(L+2) "╝" "\n"]
4/                    Split into chunks of length 4

@@                    Move split input and L+1 to top
f{...}                Map with L+1 as extra parameter...
  Se]                   Pad line to length L+1, with spaces
  '║S                   Put "║" and space before it
  2$N                   Put "║" and newline after it

*                     Join, putting the formatted lines between the top and bottom rows
Sp3000
sumber
16

Haskell, 139 bytes

q=length
g x|l<-lines x,m<-maximum$q<$>l,s<-[-1..m]>>"═"='╔':s++"╗\n"++(l>>= \z->"║ "++z++([q z..m]>>" ")++"║\n")++'╚':s++"╝"

As an example I'm framing snowman "12333321".

*Main> putStrLn $ g " _===_\n (O.O)\n/(] [)\\\n ( : )"
╔═════════╗
║  _===_  ║
║  (O.O)  ║
║ /(] [)\ ║
║  ( : )  ║
╚═════════╝

How it works:

bind
  l: input split into lines
  m: maximum line length
  s: m+2 times ═

build top line
prepend left frame to each line, pad with spaces, append right frame
build bottom line.
nimi
sumber
9

JavaScript (ES6), 138 bytes

This is 138 bytes in the IBM866 encoding, which at time of writing is still supported in Firefox, but 152 in UTF-8.

s=>`╔${t='═'.repeat(w=2+Math.max(...(a=s.split`
`).map(s=>s.length)))}╗
${a.map(s=>('║ '+s+' '.repeat(w)).slice(0,w+1)).join`║
`}║
╚${t}╝`
Neil
sumber
1
Can you actually encode Javascript using CP437 and still run it? If not, then this isn't actually 138 bytes.
Mama Fun Roll
@ӍѲꝆΛҐӍΛПҒЦꝆ Although I couldn't find anything supporting CP437, Firefox currently supports IBM866 which also has these box drawing characters, so I've updated my answer.
Neil
Okay, cool. Have an upvote!
Mama Fun Roll
6

Bash, 173 171 150 148 147 bytes, 157 136 134 133 characters

q(){((n=${#2}>n?${#2}:n));};mapfile -tc1 -C q v;for((p=++n+1;p;--p));do z+=═;done;echo ╔$z╗;printf "║ %-${n}s║\n" "${v[@]}";echo ╚$z╝

Multiline:

q() {
    (( n = ${#2} > n ? ${#2} : n))
}
mapfile -tc1 -C q v

for((p=++n+1;p;--p))
do 
    z+=═
done

echo ╔$z╗
printf "║ %-${n}s║\n" "${v[@]}"
echo ╚$z╝

Example execution:

bash -c 'q(){((n=${#2}>n?${#2}:n));};mapfile -tc1 -C q v;for((p=++n+1;p;--p));do z+=═;done;echo ╔$z╗;printf "║ %-${n}s║\n" "${v[@]}";echo ╚$z╝'< bear.txt

Sample run from script:

$ cat bear2.txt 
     (()__(()
     /       \
    ( /    \  \
     \ o o    /
     (_()_)__/ \
    / _,==.____ \
   (   |--|      )
   /\_.|__|'-.__/\_
  / (        /     \
  \  \      (      /
   )  '._____)    /
(((____.--(((____/mrf
$ ./frame< bear2.txt 
╔═══════════════════════╗
║      (()__(()         ║
║      /       \        ║
║     ( /    \  \       ║
║      \ o o    /       ║
║      (_()_)__/ \      ║
║     / _,==.____ \     ║
║    (   |--|      )    ║
║    /\_.|__|'-.__/\_   ║
║   / (        /     \  ║
║   \  \      (      /  ║
║    )  '._____)    /   ║
║ (((____.--(((____/mrf ║
╚═══════════════════════╝
Runium
sumber
1
Your example has an emtpy line between the bottom frame and the input which is invalid. The top and bottom frames have to be inserted directly before and after the input (your previous version was fine btw).
Denker
1
Nice!, But you could save approx 5 char if ...?${#2}+2:n)) instead of +1, drop 2 spaces and printf -v z %${n}s; instead of printf -v z " %*.s" $n.
F. Hauri
@Sukminder Ok, was already assumnig that, but wanted to make that sure since the input you show does not contain an empty line. I did not demand that you clear the input of leading or traliing empty lines, so you program is perfectly fine.
Denker
5

AWK, 159 bytes

{a[NR]=$0
x=length($0)
m=m<x?x:m
a[NR,1]=x}
END{for(;i<m+2;i++)t=t"═"
print"╔"t"╗"
for(j=1;j<NR;j++){f="║ %-"m"s ║\n"
printf f,a[j]}print"╚"t"╝"}

Apparently awk can print Unicode if you can figure out how to get it in the code.

Robert Benson
sumber
I'm having so many ideas for awesome pipes now ...
Sebb
@Sebb That does seem like fun. :)
Robert Benson
5

Perl, 111 characters

(score includes +5 for the interpreter flags)

#!/usr/bin/perl -n0 -aF\n
$n=(sort{$b<=>$a}map length,@F)[0];$l="═"x$n;
print"╔═$l═╗\n",(map{sprintf"║ %-${n}s ║\n",$_}@F),"╚═$l═╝";

First, we find the longest line length $n, by numerically sorting the lengths of all lines.

We set $l to be the header/footer bar to be $n repetitions of the horizontal frame character.

Then we print each line formatted to left-align in a field of width $n, sandwiched in between the frame characters.

Result:

╔═══════════╗
║   |\_/|   ║
║  / @ @ \  ║
║ ( > * < ) ║
║  `>>x<<'  ║
║  /  O  \  ║
╚═══════════╝
Toby Speight
sumber
4

Pyth, 44 chars (58 bytes)

++\╔K*JhheSlR.z\═\╗jbm+\║+.[+;d;J\║.z++\╚K\╝

Explanation

++\╔K*JhheSlR.z\═\╗                          - print out the first line
           lR.z                              -        map(len, all_input())
          S                                  -       sorted(^)
         e                                   -      ^[-1]
       hh                                    -     ^+2
      J                                      -    autoassign J = ^
     *         \═                            -   ^*"═"
    K                                        -  autoassign K = ^
++\╔             \╗                          - imp_print("╔"+^+"╗")

                   jbm+\║+.[+;d;J\║.z        - print out the middle
                   jb                        - "\n".join(V)
                     m             .z        -  [V for d in all_input()]
                      +\║+       \║          -   "║"+V+"║"
                          .[   ;J            -    pad(V, " ", J)
                            +;d              -     " "+d

                                     ++\╚K\╝ - print out the end
                                     ++\╚K\╝ - imp_print("╚"+K+"╝")

Try it here.

Blue
sumber
4

PHP 5.3, 209 bytes

This only works using the encoding OEM 860. It is an Extended ASCII superset, used in Portuguese DOS versions. Since I'm Portuguese (and I used to love doing these "frames" in Pascal) and this is a standard encoding, I went ahead with this:

<?foreach($W=explode('
',$argv[1])as$v)$M=max($M,strlen($v)+2);printf("É%'Í{$M}s»
º%1\${$M}sº
%2\$s
º%1\${$M}sº
È%1\$'Í{$M}s¼",'',join('
',array_map(function($v)use($M){return str_pad(" $v ",$M);},$W)));

Here's the base64:

PD9mb3JlYWNoKCRXPWV4cGxvZGUoJwonLCRhcmd2WzFdKWFzJHYpJE09bWF4KCRNLHN0cmxlbigkdikrMik7cHJpbnRmKCLilZQlJ+KVkHskTX1z4pWXCuKVkSUxXCR7JE19c+KVkQolMlwkcwrilZElMVwkeyRNfXPilZEK4pWaJTFcJCfilZB7JE19c+KVnSIsJycsam9pbignCicsYXJyYXlfbWFwKGZ1bmN0aW9uKCR2KXVzZSgkTSl7cmV0dXJuIHN0cl9wYWQoIiAkdiAiLCRNKTt9LCRXKSkpOw==

This answer was based on my answer on: https://codegolf.stackexchange.com/a/57883/14732 (the heavy lifting was all made there, just had to twitch a bit).

Ismael Miguel
sumber
Impressive to say the least :)
MonkeyZeus
The codes is 209 bytes/characters. 22+58+11+5+11+24+66+12=209 The last 12 is newlines and as it is DOS that means CRLF, or two bytes per newline. The charactercountonline site does not count newlines. Each of the non-ASCII glyphs are 1 byte in OEM 860.
Runium
@Sukminder Don't forget that (at least) Windows converts \n into \r\n, when opening the file in ASCII/text mode.
Ismael Miguel
2

Python 3, 119 Bytes

def f(x): 
 n='\n';s="║ ";e=" ║";h=(x.find(n)+2)*"═";return"╔"+h+"╗"+n+s+x.replace(n,e+n+s)+e+n+"╚"+h+"╝"

126 bytes

import sys
o=["║ %s ║\n"%j[:-1] for j in sys.stdin]
h="═"*(len(o[0])-3)
print("╔"+h+"╗\n"+"".join(o)+"╚"+h+"╝")

Input:

hello
there
  !  

Output:

╔═══════╗
 hello 
 there 
   !   
╚═══════╝
SumnerHayes
sumber
Welcome to Progamming Puzzles & Code Golf! Nice first answer! You can always write functions instead of full programs (unless explicitly forbidden in the challenge) which might allow you to save some bytes by taking the input as argument. Also you might want to use Python 2, so you can save 2 bytes by going with print"╔"+h+"╗\n"+"".join(o)+"╚"+h+"╝".
Denker
Thanks. I couldn't figure out how to get the high-bytes working in Python2 (probably setting the codec environment variable would work but I'm not sure how that plays into golf byte counts). The function approach eliminates python2/3 differences but adds a byte in my best approach.
SumnerHayes
Okay, I got it down to 119 characters as a function; takes input as a string. My mini-markdown is obviously not up to snuff; Line 1 is the def, the rest (after the colon) is line 2, with a leading space. def f(x): n='\n';s="║ ";e=" ║";h=(x.find(n)+2)*"═";return"╔"+h+"╗"+n+s+x.replace(n,e+n+s)+e+n+"╚"+h+"╝"
SumnerHayes
Just update your post with the new version and the new score (struck the old score out with <s>...</s>). Also you can add <!-- language-all: lang-python --> before your code-block to add syntax highlighting to your code.
Denker
This doesn't work if the input is non-rectangular, while the question says that no line will have trailing whitespace.
Dennis
2

Python 2, 115 Bytes

def f(i):w='═'*(i.find('\n')+2);return'╔%s╗\n║ %s ║\n╚%s╝'%(w,' ║\n║ '.join(i.split('\n')),w)

Looks shorter than 115 here, but working file includes 3-byte UTF-8 BOM mark signature, bumping it up to 115 bytes. If you were to run it in Python 3 you wouldn't need the BOM and it'd get down to 112 bytes.

Jenny Miller
sumber
Welcome to Programming Puzzles & Code Golf! Unfortunately, your code appears to assume that the input is rectangular, while the question says that no line will have trailing whitespace.
Dennis
I count 107 bytes. I don't think you need to include the "UTF-8 BOM mark signature".
CalculatorFeline
@CatsAreFluffy Are you using Python2? In Python3 all strings are unicode, but it's trickier with Python2.
Jenny Miller
Oops, I counted pipes as 2 bytes, but even after using an actual bytecounter, still only 111 bytes. Tell me where those 5 bytes came from.
CalculatorFeline
The UTF-8 BOM is 3 bytes (en.wikipedia.org/wiki/Byte_order_mark). My count was one high because my text editor was adding a trailing newline, so my solution is really only 115 bytes. You could leave off the leading BOM bytes and get it down to 112 if you were using Python3 (which counts all strings as unicode). But I don't know how you're seeing only 111 bytes. btw, here's how I added the BOM: sed -i '1s/^\(\xef\xbb\xbf\)\?/\xef\xbb\xbf/' codeGolf.py
Jenny Miller
1

C, 290 bytes

Golfed function B, with dependencies; takes input as null-terminated char*

#define l(s) strlen(s)
p(char*s,int n){while(n--)printf(s);}
B(char*s){char*t=strtok(s,"\n");int x=l(t),z=1;while(t=strtok(0,"\n"))z++,x=l(t)>x?l(t):x;p("╔",1);p("=",x+2);p("╗\n",1);while(z--)printf("║ %s", s),p(" ",x-l(s)),p(" ║\n",1),s+=l(s)+1;p("╚",1);p("=",x+2);p("╝\n",1);}

Somewhat-ungolfed function in full program

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 1024

// GOLF-BEGIN =>
#define l(s) strlen(s)
// since multibyte chars don't fit in char: use char* instead
void p (char*s,int n){ while(n--)printf(s); } 
void B (char *s){
    char *t = strtok(s,"\n");
    int x=l(t), z=1;
    while(t=strtok(0,"\n"))z++,x=l(t)>x?l(t):x;  
    // x is l(longest line), z is #lines
    p("╔",1);p("=",x+2);p("╗\n",1);
    while(z--)printf("║ %s", s),p(" ",x-l(s)),p(" ║\n",1),s+=l(s)+1;
    p("╚",1);p("=",x+2);p("╝\n",1);       
}
// <= GOLF-END

int main(int argc, char **argv) {
    char buffer[MAX];
    memset(buffer, 0, MAX);
    FILE *f = fopen(argv[1], "rb");
    fread(buffer, 1, MAX, f); 
    B(buffer);
    return 0;
}

input

     _.,----,._
   .:'        `:.
 .'              `.
.'                `.
:                  :
`    .'`':'`'`/    '
 `.   \  |   /   ,'
   \   \ |  /   /
    `\_..,,.._/'
     {`'-,_`'-}
     {`'-,_`'-}
     {`'-,_`'-}
      `YXXXXY'
        ~^^~

output

╔======================╗
║      _.,----,._      ║
║    .:'        `:.    ║
║  .'              `.  ║
║ .'                `. ║
║ :                  : ║
║ `    .'`':'`'`/    ' ║
║  `.   \  |   /   ,'  ║
║    \   \ |  /   /    ║
║     `\_..,,.._/'     ║
║      {`'-,_`'-}      ║
║      {`'-,_`'-}      ║
║      {`'-,_`'-}      ║
║       `YXXXXY'       ║
║         ~^^~         ║
╚======================╝

C golfing tips appreciated!

tucuxi
sumber