Sort a string, sort of

Question

If you sort a string you'll typically get something like:

         ':Iaaceeefggghiiiiklllllmnnooooprrssstttttuuyyyy

_{Yes, that was the first sentence sorted.}

As you can see, there are a lot of repeated characters, aa, eee, ttttt, 9 spaces and so on.

If we add 128 to the ASCII-value of the first duplicate, 256 to the second, 384 to the third and so on, sort it again and output the new string (modulus 128 to get the same characters back) we get the string:

 ':Iacefghiklmnoprstuy aegilnorstuy egilosty iloty lt    

_{(Note the single leading space and the 4 trailing spaces).}

The string is "sequentially sorted" <space>':I....uy, <space>aeg....uy, <space>egi....ty, <space>iloty, <space>lt, <space>, <space>,<space>, <space>.

It might be easier to visualize this if we use a string with digits in it. The string 111222334 will when "sorted" be: 123412312.

Challenge:

To no surprise, the challenge is to write a code that sorts a string according to the description above.

You can assume that the input string will contain only printable ASCII-characters in the range 32-126 (space to tilde).

---

Test cases:

**Test cases:**
 *:Tacest*es*s*

If you sort a string you'll typically get something like:
 ':Iacefghiklmnoprstuy aegilnorstuy egilosty iloty lt    

Hello, World!
 !,HWdelorlol

#MATLAB, 114 bytes
 #,14ABLMTbesty 1A

f=@(s)[mod(sort(cell2mat(cellfun(@(c)c+128*(0:nnz(c)-1),mat2cell(sort(s),1,histc(s,unique(s))),'un',0))),128),''];
'()*+,-0128:;=@[]acdefhilmnoqrstuz'(),0128@acefilmnorstu'(),12celmnostu'(),12celnstu(),clnst(),cls(),cs(),()()()()

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This is code-golf, so the shortest code in each language counted in bytes will win^ref.

Answer 1

Pyth, 5 bytes

s.T.g

Test suite

Very straightforward: Group and sort, transpose, concatenate.

s.T.g
s.T.gkQ    Implicit variables
   .gkQ    Group the input input lists of elements whose values match when the
           identity function is applied, sorted by the output value.
 .T        Transpose, skipping empty values. This puts all first characters into
           a list, then all second, etc.
s          Concatenate.

Answer 2

Jelly, 3 bytes

ĠZị

Try it online!

How it works

Oh boy, this challenge was all but made for Jelly.

The group atom (Ġ) takes an array¹ as input and groups indices that correspond to identical elements of the array. The array of index groups is sorted with the corresponding elements as keys, which is precisely the order we require for this challenge.

Next, the zip atom (Z) transposes rows and columns of the generated (ragged) matrix of indices. This simply consists of reading the columns of the matrix, skipping elements that are not present in that column. As a result, we get the first index of the character with the lowest code point, followed by the first index of the character with the second lowest code point, … followed by the second index of the character with the lowest code point, etc.

Finally, the unindex atom (ị) retrieves the elements of the input array at all of its indices in the generated order. The result is a 2D character array, which Jelly flattens before printing it.

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¹ _{Jelly doesn't have a string type, just arrays of characters.}

Answer 3

Python 3, 109 105 104 103 99 93 90 88 81 79 69 bytes

2 bytes saved thanks to FlipTack

7 bytes saved because flornquake caught my dumb error

2 bytes saved thanks to xnor

10 bytes saved thanks to Dennis

a=[*input()]
while a:
    for c in sorted({*a}):print(end=c);a.remove(c)

Explanation

We start by converting our string to a list using a splat and storing that list in a variable a. Then while our a is not the empty list we go through each unique member of a in sorted order, print it and remove a copy of that character from the list.

Each iteration prints thus prints one copy of each character present in a.

Answer 4

Haskell, 44 bytes

import Data.List
concat.transpose.group.sort

Usage example:

Prelude Data.List> concat.transpose.group.sort $ "If you sort a string you'll typically get something like:"
" ':Iacefghiklmnoprstuy aegilnorstuy egilosty iloty lt    "

Sort, group equal chars to a list of strings (e.g. "aabbc" -> ["aa","bb","c"]), transpose and flatten into a single string, again.

Answer 5

Python 2, 75 bytes

lambda s:`zip(*sorted((s[:i].count(c),c)for i,c in enumerate(s)))[1]`[2::5]

Try it online!

Answer 6

C, 109 106 105 104 102 100 97 98 96 91 Bytes

Back up to 98 Bytes, needed to initialize j to make f(n) re-useable

Down to 96 Bytes using puts in place of strlen B-)

It's strange I had to back to strlen but I got rid of the for(;i++;) loop so now it's down to 91 Bytes. Apparently the man page for puts reads;

"RETURNS
   If successful, the result is a nonnegative integer; otherwise, the result is `EOF'."

... I was lucky it was working in the first place

char*c,i,j;f(m){for(j=strlen(m);j;++i)for(c=m;*c;c++)if(*c==i){*c=7,putchar(i),j--;break;}}

test code...

main(c,v)char**v;
{
    char test[] = "If you sort a string you'll typically get something like: ";
    char test2[] = "Hello, World!";

    f(test);puts("");    
    f(test2);puts("");    
}

Here are a few test cases, now it's time to golf this down

C:\eng\golf>a.exe
 ':Iacefghiklmnoprstuy aegilnorstuy egilosty iloty lt
 !,HWdelorlo

Answer 7

Dyalog APL, 21 chars = 39 bytes

t[0~⍨∊⍉(⊢⌸t)[⍋∪t←⍞;]]

t[...] index t (to be defined shortly) with...

 0~⍨ zeros removed from

 ∊ the enlisted (flattened)

 ⍉ transposed

 (⊢⌸t)[...;] keyed* t, row-indexed by...

  ⍋ the indices which would sort

  ∪ the unique letters of

  t← t, which has the value of

  ⍞ prompted text input

TryAPL online!

---

* ^{⊢⌸t creates a table where the rows (padded with zeros for a rectangular table) list each unique letters' indices in t.}

Answer 8

Mathematica, 68 60 59 bytes

Split[Characters@#~SortBy~ToCharacterCode]~Flatten~{2}<>""&

Accepts a String. Outputs a String.

If list of characters were allowed (46 bytes):

Split[#~SortBy~ToCharacterCode]~Flatten~{2,1}&

Version using Sort (40 bytes):

Split@Sort@Characters@#~Flatten~{2}<>""&

This version cannot be my answer because Sort cannot be used here; Sort sorts by canonical order, not by character code.

Answer 9

JavaScript (ES6), 79 bytes

f=s=>s&&(a=[...new Set(s)]).sort().join``+f(a.reduce((s,e)=>s.replace(e,``),s))

<input oninput=o.textContent=f(this.value)><pre id=o>

Works by extracting the set of unique characters, sorting it, removing them from the original string, and recursively calculating the sort of the rest of the string. 81 byte solution that I found interesting:

f=s=>s&&(s=[...s].sort().join``).replace(r=/(.)(\1*)/g,"$1")+f(s.replace(r,"$2"))

Answer 10

Python 2, 77 76 bytes

d={}
def f(c):d[c]=r=d.get(c,c),;return r
print`sorted(input(),key=f)`[2::5]

Takes a quoted string as input from stdin.

Try it online!

Answer 11

J, 16 15 bytes

/:+/@(={:)\;"0]

This is a verb that takes and returns one string. Try it online!

Miles saved a byte, thanks!

Explanation

Nothing too fancy here: sort primarily by order of occurrence, secondarily by char value.

/:+/@(={:)\;"0]  Input is y.
          \      Map over prefixes:
  +/              Sum
    @(   )        of
      =           bit-array of equality
       {:         with last element.
                 This gives an array of integers whose i'th element is k
                 if index i is the k'th occurrence of y[i].
           ;     Pair this array
            "0   element-wise
              ]  with y
/:               and sort y using it as key.

Answer 12

Mathematica, 55 bytes, non-competing

(Sort@Characters@#//.{a___,b_,b_,c___}:>{a,b,c,b})<>""&

Edit: Unfortunately, Mathematica's sort is not by character codes, but by alphabetical order, where uppercase immediatly follows lowercase (ie Hi There is sorted to { , e, e, h, H, i, r, T}).

This works using patterns:

//.{a___,b_,b_,c___}:>{a,b,c,b}
    a___       c___              (Three _) a zero or more members, named a and c
         b_,b_                   exactly one member, repeated twice (since they have the same name)
                    :>           Delayed Rule (replace left hand side with right hand side.)
                                 Delayed Rule evaluate on each substitution, avoiding conflicts with predefined variables
                      {a,b,c,b}  put one of the b-named member after all other sequences
//.                              repeat until no change (aka Replace Repeated)

Answer 13

Brainf*ck, 458 226 bytes

,[>>>>>>,]<<<<<<[[-<<<+<<<]>>>[>>>[>>>>>>]<<<[>>--[<->--]<-<[>->+<[>]>[<+>-]<<[<]>-]>>.[[-]<]<<<[[>>>>>>+<<<<<<-]<<<]>>>>>>]>>>[>>>[>>>>>>]<<<[>>--[<->--]<-<[>->+<[>]>[<+>-]<<[<]>-]>>[-<+<+>>]<[->>+<<]<[<<<<<<]>>>]>>>]]<<<<<<]

Try it online! - BF

Numberwang, 262 226 bytes

8400000087111111442111911170004000400000071114002241202271214020914070419027114170270034427171114400000091111112711170000007000400040000007111400224120227121402091407041902711417027004219190071420091171411111170007000771111117

Try it online! - NW

I put both of these here because they are identical code.

Answer 14

PHP, 83 bytes

for($s=count_chars($argv[1]);$s=array_filter($s);$c%=128)echo$s[++$c]--?chr($c):'';

Unfortunately you can't have unset in a ternary so I need to use the annoyingly long array_filter.
Use like:

php -r "for($s=count_chars($argv[1]);$s=array_filter($s);$c%=128)echo$s[++$c]--?chr($c):'';" "If you sort a string you'll typically get something like:"

Answer 15

V, 37 36 bytes

Thanks @DJMcMayhem for the byte!

Í./&ò
dd:sor
Íî
òͨ.©¨±«©±À!¨.«©/±³²

Try it online!

Not sure I like the regex at the end, but I needed to make the ò break somehow.

Explain

Í./&ò                    #All chars on their own line
dd:sor                   #Delete empty line, sort chars
Íî                       #Join all lines together s/\n//
òͨ.©¨±«©±À!¨.«©/±³² #until breaking s/\v(.)(\1+)\1@!(.+)/\3\2\1

Answer 16

Python 2, 70 bytes

f=lambda s,i=0,c='':s[i>>7:]and(s.count(c)>i>>7)*c+f(s,i+1,chr(i%128))

Try it online

This is very inefficient. The test link changes the i>>7 to i>>5 and sets the recursion limit to 10000. Assumes the inputs only has ASCII values up to 126.

Uses the div-mod trick to iterate through two loops: minimum counts i/128 in the outer loop and ASCII values i%128 in the inner loop. Includes a character c with the given ASCII value if the number of times it appears in the string is at least its minimum count.

The code uses a trick to simulate the assignment c=chr(i%128) so that it can be referenced in the expression (s.count(c)>i>>7)*c. Python lambdas do not allow assignment because they only take expressions. Converting to a def or full program is still a net loss here.

Instead, the function pushes forward the value chr(i%128) to the next recursive call as an optional input. This is off by one because i has been incremented, but doesn't matter as long as the string doesn't contain special character '\x7f' (we could also raise 128 to 256). The initial c='' is harmless.

Answer 17

Perl 6, 68 bytes

{my \a=.comb.sort;[~] flat roundrobin |a.squish.map({grep *eq$_,a})}

I was a little surprised to find that there's no built-in way to group like elements in a list. That's what the squish-map bit does.

Answer 18

JavaScript (ES6), 77 75 bytes

s=>(a=[],x={},[...s].sort().map(c=>a[x[c]=n=-~x[c]]=(a[n]||'')+c),a).join``

Stable sorts the lexicographically sorted string by n^th occurence

F=s=>(a=[],x={},[...s].sort().map(c=>a[x[c]=n=-~x[c]]=(a[n]||'')+c),a).join``

const update = () => {
  console.clear();
  console.log(F(input.value));
};
input.oninput = update;
update();

#input {
  width: 100%;
  box-sizing: border-box;
}

<input id="input" type="text" value="         ':Iaaceeefggghiiiiklllllmnnooooprrssstttttuuyyyy" length=99/>
<div id="output"></div>

Answer 19

FSharp, 194 190 170 140 133 bytes

let f=Seq.map
let(@)=(>>)
f [e-mail] id@f([e-mail]((*)128@(+)))@[e-mail]@f((%)@(|>)128@byte)@Array.ofSeq@f char

Using Seq instead of Array saves a couple of bytes

Defining a shorter name, and using another maps to avoid a (fun ->) block

It turns out F# can map a char to an in, so removing the shortened name of System.Text.Encoding.ASCII, and adding in another map saves me 20 bytes!

Returning a char array instead of a string, saves me 30 bytes!

I no longer need to make sure it's a string, saves me 7 bytes

Answer 20

Perl 6, 54 bytes

{[~] flat roundrobin |.comb.classify(~*){*}.sort»[*]}

Explanation:

• { }: A lambda that takes one argument -- e.g. 21211.
• .comb: Split the input argument into a list of characters -- e.g. (2,1,2,1,1).
• .classify(~*): Group the characters using string comparison as the grouping condition, returning an unordered Hash -- e.g. { 2=>[2,2], 1=>[1,1,1] }.
• {*}: Return a list of all values of the Hash -- e.g. [2,2], [1,1,1].
• .sort: Sort it -- e.g. [1,1,1], [2,2].
• »[*]: Strip the item containers the arrays were wrapped in due to being in the hash, so that they won't be considered as a single item in the following step -- e.g. (1,1,1), (2,2).
• roundrobin |: Zip the sub-lists until all are exhausted -- e.g. (1,2), (1,2), (1).
• flat: Flatten the result -- e.g. 1, 2, 1, 2, 1.
• [~]: Concatenate it to get a string again -- e.g. 12121.

(Credit for the roundrobin approach goes to Sean's answer.)

Answer 21

05AB1E, 15 bytes

{.¡"ä"©¹g׫øJ®K

Try it online! or as a Test suite

Explanation

{                # sort input
 .¡              # group by equal elements
   "ä"©          # push "ä" and store a copy in the register
       ¹g×       # repeat the "ä" input-nr times
          «      # concatenate the result to each string in the grouped input
           ø     # zip
            J    # join to string
             ®K  # remove all instances of "ä" in the string

10 of the 15 bytes are for getting around 05AB1E's way of handling zipping strings of different length.

Answer 22

JavaScript (ES6), 114 bytes

Separated with newline for clarity, not part of byte count:

s=>[...s].map(a=>(m[a]=-~m[a])*128+a.charCodeAt(),m={})
.sort((a,b)=>a-b).map(a=>String.fromCharCode(a%128)).join``

Demo

`**Test cases:**
 *:Tacest*es*s*

If you sort a string you'll typically get something like:
 ':Iacefghiklmnoprstuy aegilnorstuy egilosty iloty lt    

Hello, World!
 !,HWdelorlol

#MATLAB, 114 bytes
 #,14ABLMTbesty 1A

f=@(s)[mod(sort(cell2mat(cellfun(@(c)c+128*(0:nnz(c)-1),mat2cell(sort(s),1,histc(s,unique(s))),'un',0))),128),''];
'()*+,-0128:;=@[]acdefhilmnoqrstuz'(),0128@acefilmnorstu'(),12celmnostu'(),12celnstu(),clnst(),cls(),cs(),()()()()`.split`\n\n`.map(s=>(p=s.split`\n`,console.log(`${p[0]}\n\n${r=f(p[0])}\n\nmatch: ${r==p[1]}`)),
f=s=>[...s].map(a=>(m[a]=-~m[a])*128+a.charCodeAt(),m={}).sort((a,b)=>a-b).map(a=>String.fromCharCode(a%128)).join``)

Answer 23

Clojure, 79 bytes

#(for[V[(group-by(fn[s]s)%)]i(range 1e9)k(sort(keys V))c[(get(V k)i)]:when c]c)

An anonymous function, returns a sequence of characters. Supports up-to 10^9 repetitions of any characters, which should be plenty.

Answer 24

Retina, 24 bytes

O`.
O$#`(.)(?=(\1)*)
$#2

Try it online!

Answer 25

Ruby, 59+1 = 60 bytes

Adds one byte for the -n flag. Port of @PatrickRoberts' dictionary solution.

d={};print *$_.chars.sort_by{|c|d[c]||=0;c.ord+128*d[c]+=1}