Exploded view of a string

Question

Don't you love those exploded-view diagrams in which a machine or object is taken apart into its smallest pieces?

Let's do that to a string!

The challenge

Write a program or function that

1. inputs a string containing only printable ASCII characters;
2. dissects the string into groups of non-space equal characters (the "pieces" of the string);
3. outputs those groups in any convenient format, with some separator between groups.

For example, given the string

Ah, abracadabra!

the output would be the following groups:

!
,
A
aaaaa
bb
c
d
h
rr

Each group in the output contains equal characters, with spaces removed. A newline has been used as separator between groups. More about allowed formats below.

Rules

The input should be a string or an array of chars. It will only contain printable ASCII chars (the inclusive range from space to tilde). If your language does not support that, you can take the input in the form of numbers representing ASCII codes.

You can assume that the input contains at least one non-space character.

The output should consist of characters (even if the input is by means of ASCII codes). There has to be an unambiguous separator between groups, different than any non-space character that may appear in the input.

If the output is via function return, it may also be an array or strings, or an array of arrays of chars, or similar structure. In that case the structure provides the necessary separation.

A separator between characters of each group is optional. If there is one, the same rule applies: it can't be a non-space character that may appear in the input. Also, it can't be the same separator as used between groups.

Other than that, the format is flexible. Here are some examples:

• The groups may be strings separated by newlines, as shown above.

• The groups may be separated by any non-ASCII character, such as ¬. The output for the above input would be the string:

  !¬,¬A¬aaaaa¬bb¬c¬d¬h¬rr
  

• The groups may be separated by n>1 spaces (even if n is variable), with chars between each group separated by a single space:

  !  ,    A   a a a a a    b b  c       d   h  r r
  

• The output may also be an array or list of strings returned by a function:

  ['!', 'A', 'aaaaa', 'bb', 'c', 'd', 'h', 'rr']
  

• Or an array of char arrays:

  [['!'], ['A'], ['a', 'a', 'a', 'a', 'a'], ['b', 'b'], ['c'], ['d'], ['h'], ['r', 'r']]
  

Examples of formats that are not allowed, according to the rules:

• A comma can't be used as separator (!,,,A,a,a,a,a,a,b,b,c,d,h,r,r), because the input may contain commas.
• It's not accepted to drop the separator between groups (!,Aaaaaabbcdhrr) or to use the same separator between groups and within groups (! , A a a a a a b b c d h r r).

The groups may appear in any order in the output. For example: alphabetical order (as in the examples above), order of first appearance in the string, ... The order need not be consistent or even deterministic.

Note that the input cannot contain newline characters, and A and a are different characters (grouping is case-sentitive).

Shortest code in bytes wins.

Test cases

In each test case, first line is input, and the remaining lines are the output, with each group in a different line.

• Test case 1:

  Ah, abracadabra!
  !
  ,
  A
  aaaaa
  bb
  c
  d
  h
  rr
  

• Test case 2:

  \o/\o/\o/
  ///
  \\\
  ooo
  

• Test case 3:

  A man, a plan, a canal: Panama!
  !
  ,,
  :
  A
  P
  aaaaaaaaa
  c
  ll
  mm
  nnnn
  p
  

• Test case 4:

  "Show me how you do that trick, the one that makes me scream" she said
  ""
  ,
  S
  aaaaa
  cc
  dd
  eeeeeee
  hhhhhh
  ii
  kk
  mmmm
  n
  ooooo
  rr
  ssss
  tttttt
  u
  ww
  y
  

Answer 1

Jelly, 5 bytes

ḟ⁶ṢŒg

Try it online!

It does return an array, just that when it is printed to STDOUT, the separator is gone.

This is indeed a function that can be called as such (in Jelly, each line is a function).

ḟ⁶ṢŒg
ḟ⁶      filter out spaces
  Ṣ     sort
   Œg   group

Answer 2

Python 3.5+, 77 46 44 41 bytes

lambda s:[a*s.count(a)for a in{*s}-{' '}]

Pretty simple. Goes through the unique characters in the string by converting it to a set (using Python 3.5's extended iterable unpacking), then uses a list comprehension to construct the exploded diagrams by counting the number of times each character occurs in the string with str.count. We filter out spaces by removing them from the set.

The order of the output may vary from run to run; sets are unordered, so the order in which their items are processed, and thus this answer outputs, cannot be guaranteed.

This is a lambda expression; to use it, prefix lambda with f=.

Try it on Ideone! Ideone uses Python 3.4, which isn't sufficient.

Usage example:

>>> f=lambda s:[a*s.count(a)for a in{*s}-{' '}]
>>> f('Ah, abracadabra!')
[',', 'A', 'aaaaa', 'd', '!', 'bb', 'h', 'c', 'rr']

Saved 3 bytes thanks to @shooqie!

Answer 3

Retina, 13 bytes

O`.
!`(\S)\1*

The sorting is very easy (it's a builtin), it's separating the letters that takes 9 bytes. Try it online!

The first line sOrts all matches of the regex . (which is every character), giving us !,Aaaaaabbcdhrr.

Match is the default stage for the last line of a program, and ! makes it print a linefeed-separated list of matches of the regex. The regex looks for one or more instances of a non-space character in a row.

Answer 4

Perl 6, 28 bytes

*.comb(/\S/).Bag.kv.map(*x*)

Note that Bag like a Hash or Set is unordered so the order of results is not guaranteed.

Explanation:

# Whatever lambda ｢*｣


# grab the characters
*.comb(
  # that aren't white-space characters
  /\S/
)
# ("A","h",",","a","b","r","a","c","a","d","a","b","r","a","!")


# Turn into a Bag ( weighted Set )
.Bag
# {"!"=>1,","=>1,"A"=>1,"a"=>5,"b"=>2,"c"=>1,"d"=>1,"h"=>1,"r"=>2}


# turn into a list of characters and counts
.kv
# ("!",1,",",1,"A",1,"a",5,"b",2,"c",1,"d",1,"h",1,"r",2)


# map over them 2 at a time
.map(
  # string repeat the character by the count
  * x *
)
# ("!",",","A","aaaaa","bb","c","d","h","rr")

Answer 5

Vim, 50, 46 bytes

i <esc>:s/./&\r/g
:sor
qq:%s/\v(.)\n\1/\1\1
@qq@qD

Explanation/gif will come later.

Answer 6

Pyth, 6

.gk-zd

Try it here or run a Test Suite.

Pretty simple, -zd removes spaces from the input, and .gk groups each remaining element by its value. Unfortunately, I haven't found a way to make use of auto-fill variables. Note that the output is shown as Python strings, so certain characters (read: backslashes) are escaped. If you want it to be more readable, add a j to the beginning of the code.

Answer 7

Haskell, 38 bytes

f s=[filter(==c)s|c<-['!'..],elem c s]

Basically nimi's solution, but explicitly checking only letters appearing in the string.

Answer 8

Brachylog, 14 7 bytes

7 bytes thanks to Fatalize.

:@Sxo@b

Try it online!

:@Sxo@b
:@Sx     remove spaces
    o    sort
     @b  group

Answer 9

JavaScript (ES6), 41 bytes

s=>[...s].sort().join``.match(/(\S)\1*/g)

Answer 10

2sable, 7 bytes

Code:

Úð-vyÃ,

Explanation:

Ú       # Uniquify the string, aabbcc would result into abc
 ð-     # Remove spaces
   vy   # For each character...
     Ã  #   Keep those in the string, e.g. 'aabbcc', 'a' would result into 'aa'
      , #   Pop and print with a newline

Uses the CP-1252 encoding. Try it online!

Answer 11

Haskell, 40 bytes

f x=[v:w|d<-['!'..],v:w<-[filter(==d)x]]

Usage example: f "Ah, abracadabra!"-> ["!",",","A","aaaaa","bb","c","d","h","rr"].

The pattern v:w matches only list with at least one element, so all characters not in the input are ignored.

Also 40 bytes:

import Data.List
group.sort.filter(>' ')

Answer 12

C# 125 98 Bytes

using System.Linq;s=>s.GroupBy(c=>c).Where(g=>g.Key!=' ').Select(g=>new string(g.Key,g.Count())));

Explanation

//Using anonymous function to remove the need for a full signature 
//And also allow the implicit return of an IEnumerable
s =>

    //Create the groupings
    s.GroupBy(c => c)

    //Remove spaces
    .Where(g=> g.Key!=' ')

    //Generate a new string using the grouping key (the character) and repeating it the correct number of times
    .Select(g => new string(g.Key, g.Count()));

• Thanks to @TheLethalCoder who suggested the use of an anonymous function, which also allowed me to remove the ToArray call and just implicitly return an IEnumerable which collectively saves 27 bytes

Answer 13

R, 198 189 96 95 bytes

for(i in unique(a<-strsplit(gsub(" ","",readline()),"")[[1]]))cat(rep(i,sum(a==i)),"\n",sep="")

Ungolfed :

a<-strsplit(gsub(" ","",readline()),"")[[1]] #Takes the input from the console

for(i in unique(a)) #loop over unique characters found in variable a

cat(rep(i,sum(a==i)),"\n",sep="") # print that character n times, where n was the number of times it appeared

This solution is currently not entirely working, when \ are involved.
Now it is !

Thank a lot you to @JDL for golfing out 102 bytes !

Answer 14

Ruby, 41 + 1 = 42 bytes

+1 byte for -n flag.

gsub(/(\S)(?!.*\1)/){puts$1*$_.count($1)}

Takes input on stdin, e.g.:

$ echo 'Ah, abracadabra!' | ruby -ne 'gsub(/(\S)(?!.*\1)/){puts$1*$_.count($1)}'
A
h
,
c
d
bb
rr
aaaaa
!

Answer 15

Swift, 105 91 bytes

Thanks to @NobodyNada for 14 bytes :)

Yeah, I'm pretty new to Swift...

func f(a:[Character]){for c in Set(a){for d in a{if c==d && c != " "{print(c)}}
print("")}}

Characters within a group are separated by a single newline. Groups are separated by two newlines.

Answer 16

Mathematica, 36 bytes

Built-in functions Gather and Characters do most of the work here.

Gather@Select[Characters@#,#!=" "&]&

Answer 17

><>, 49 bytes

i:0(?v
84}0~/&{!*
v!?: <}/?=&:&:<
>&1+&}aol1-?!;^

Very spaciously wasteful in the output, but i assume is still allowed given the lenience of the rules

Explanation:

i:0(?v           Collects text from input
84}0~/&{!*       adds 32 (first ascii starting at space) to register and 0 to stack
v!?: <}/?=&:&:<  checks all characters to the current register, if equal:
       o         prints the character and continues looping
>&1+&}aol1-?!;^  when all characters are checked, adds 1 to register, prints a newline,
                 checks the stack length to halt the program if 0, and starts looping again

fit some things in pretty tight, even using jumps to get around some functions so i could run the pointer vertically.

Basically this puts each ASCII character on its own newline, and if none of that character exists, the line will be blank

Try it online

Edit: i was wrong there was an error in the code what would cause it to never complete if there was a space in the input

Answer 18

Pyth, 5 bytes

.gksc

Try it here!

Takes input as a Python string (i.e. wrapped in quotes, escaped quotes and slashes as necessary).

Explanation:

    c    Split (implied) input on whitespace
   s     Sum together
.gk      Group by value

If you guarantee at least one space in the input, there's a 4-byte solution:

t.gk

Try it here!

Explanation:

 .gk (Q)  groups the characters in the string by their value
           this sorts them by their value, which guarantees that spaces are first
t         Remove the first element (the spaces)

Answer 19

Dyalog APL, 11 bytes

Function returning list of strings.

(⊂∩¨∪)~∘' '

(⊂∩¨∪) the intersection of the entirety and its unique characters

~∘' ' except spaces

TryAPL online!

Answer 20

PowerShell v2+, 44 bytes

[char[]]$args[0]-ne32|group|%{-join$_.Group}

Takes input $args[0] as a command-line argument literal string. Casts that as a char-array, and uses the -not equal operator to pull out spaces (ASCII 32). This works because casting has a higher order precedence, and when an array is used as the left-hand operator with a scalar as the right-hand, it acts like a filter.

We pass that array of characters to Group-Object, which does exactly what it says. Note that since we're passing characters, not strings, this properly groups with case-sensitivity.

Now, we've got a custom object(s) that has group names, counts, etc. If we just print that we'll have a host of extraneous output. So, we need to pipe those into a loop |%{...} and each iteration -join the .Group together into a single string. Those resultant strings are left on the pipeline, and output is implicit at program completion.

Example

PS C:\Tools\Scripts\golfing> .\exploded-view-of-substrings.ps1 'Programming Puzzles and Code Golf'
PP
rr
ooo
gg
aa
mm
i
nn
u
zz
ll
ee
s
dd
C
G
f

Answer 21

Processing, 109 bytes

void s(char[] x){x=sort(x);char l=0;for(char c:x){if(c!=l)println();if(c!=' '&&c!='\n'&&c!='\t')print(c);l=c;}}

Its the brute force approach, sort the array, then loop through it. If it doesn't match the last character printed, print a newline first. If it is whitespace, skip the printing step.

Answer 22

Octave, 61 bytes

@(x)mat2cell(y=strtrim(sort(x)),1,diff(find([1 diff(+y) 1])))

This is an anoymous function that takes a string as input and outputs a cell arrray of strings.

Try at Ideone.

How it works

• sort sorts the input string. In particular, spaces will be at the beginning.
• strtrim removes leading spaces.
• diff(+y) computes consecutive differences between characters (to detect group boundaries)...
• ... so diff(find([1 diff(+y) 1]) gives a vector of group sizes.
• mat2cell then splits the sorted string into chunks with those sizes.

Answer 23

Javascript (using external Library - Enumerable) (78 67 bytes)

 n=>_.From(n).Where(y=>y!=' ').GroupBy(x=>x).WriteLine(y=>y.Write())

Link to lib: https://github.com/mvegh1/Enumerable

Code explanation: This is what Enumerable was made to do! Load the string into the library, which converts it to a char array. Filter out the white space entries. Group by char. Write each group to a line, according to the specified predicate. That predicate says to join all the elements of the current group into a string, without a delimiter.

Answer 24

Perl6, 48 47 45

slurp.comb.Bag.kv.map:{$^a.trim&&say $a x$^b}

Thanks to manatwork for the improvements.

Answer 25

MATL, 7 bytes

Xz!t7XQ

MATL Online Demo

Answer 26

Ruby, 35 bytes

->s{(?!..?~).map{|x|s.scan x}-[[]]}

Tried to turn the problem upside down to save some bytes. Instead of starting with the string, start with the possible character set.

Answer 27

Ruby, 46 bytes

Try it online!

->s{(s.chars-[' ']).uniq.map{|c|c*s.count(c)}}

My original full program version, 48 bytes after adding the n flag:

p gsub(/\s/){}.chars.uniq.map{|c|c*$_.count(c)}

Answer 28

Python, 107

Could be shortened by lambda, but later

x=sorted(input())
i=0
while i<len(x):x[i:]=['  '*(x[i]!=x[i-1])]+x[i:];i-=~(x[i]!=x[i-1])
print("".join(x))

Answer 29

CJam, 10 bytes

{S-$e`::*}

An unnamed block that expects the string on top of the stack and replaces it with a list of strings.

Try it online!

Explanation

S-  Remove spaces.
$   Sort.
e`  Run-length encode, gives pairs [R C], where R is the run-length and
    C is the character.
::* Repeat the C in each pair R times.

Answer 30

Pyke, 16 9 bytes

d-}F/i*d+

Try it here!

Answer 31

Common Lisp, 123

(lambda(s &aux(w(string-trim" "(sort s'char<))))(princ(elt w 0))(reduce(lambda(x y)(unless(char= x y)(terpri))(princ y))w))

Ungolfed:

(lambda (s &aux (w (string-trim " " (sort s 'char<))))
  (princ (elt w 0))
  (reduce
    (lambda (x y) 
      (unless (char= x y) (terpri))
      (princ y))
  w))

Not the most golf friendly language. This could probably be modified to return list of lists instead of printing string.

Answer 32

Emacs, 36 keystrokes

C-SPACE C-EM-xsort-rTABRETURN.RETURN.RETURNC-AC-M-S-%\(\(.\)\2*\)RETURN\1C-QC-JRETURN!

Result

A man, a plan, a canal: Panama! -->

!
,,
:
A
P
aaaaaaaaa
c
ll
mm
nnnn
p

Explanation

1. C-SPACE C-E
2. M-x sort-rTAB RETURN .RETURN .RETURN
3. C-A
4. C-M-S-% \(\(.\)\2*\)RETURN\1 C-Q C-JRETURN !

1. Select the input line;
2. Call sort-regexp-fields with arguments . and .;
   • Argument #1: Regexp scpecifying records to sort
   • Argument #2: Regexp scpecifying key within records
3. Return at line start;
4. Apply regexp substitution \(\(.\)\2*\) -> \1\n on all matches.

Answer 33

PHP, 67 bytes

Space as separator

PHP5 using deprecated ereg_replace function.

for(;++$i<128;)echo ereg_replace("[^".chr($i)."]","",$argv[1])." ";

PHP7, 73 bytes

for(;++$i<128;)echo str_repeat($j=chr($i),substr_count($argv[1],$j))." ";

Using built in is worse :(

foreach(array_count_values(str_split($argv[1]))as$a=>$b)echo str_repeat($a,$b)." ";

Answer 34

Perl, 31 bytes

Includes +1 for -p

Run with input on STDIN:

explode.pl <<< "ab.ceaba.d"

explode.pl:

#!/usr/bin/perl -p
s%.%$&x s/\Q$&//g.$/%eg;y/
//s

If you don't care about spurious newlines inbetween the lines the following 24 bytes version works too:

#!/usr/bin/perl -p
s%.%$&x s/\Q$&//g.$/%eg

Answer 35

Oracle SQL 11.2, 123 bytes

SELECT LISTAGG(c)WITHIN GROUP(ORDER BY 1)FROM(SELECT SUBSTR(:1,LEVEL,1)c FROM DUAL CONNECT BY LEVEL<=LENGTH(:1))GROUP BY c;

Un-golfed

SELECT LISTAGG(c)WITHIN GROUP(ORDER BY 1)
FROM (SELECT SUBSTR(:1,LEVEL,1)c FROM DUAL CONNECT BY LEVEL<=LENGTH(:1))
GROUP BY c

Answer 36

S.I.L.O.S 265

The (non competing) code better input format is at the bottom, feel free to try it online!

def : lbl
loadLine :
a = 256
:a
:s
x = get a
z = x
z - 32
z |
if z u
a + 1
GOTO s
:u
if x c
GOTO e
:c
b = x
b + 512
c = get b
c + 1
set b c
a + 1
if x a
:e
a = 512
b = 768
l = 10
:Z
j = a
j - b
if j z
z = get a
c = a
c - 512
:3
if z C
printChar l
a + 1
GOTO Z
:C
printChar c
z - 1
GOTO 3
:z

Input for the above is a series of command line arguments representing ascii values, terminated with a zero.Try it online!.

---

For a more reasonable method of input we must up the byte count (and use features that were nonexistent before the writing of this challenge).
For 291 bytes we get the following code.

\
def : lbl
loadLine :
a = 256
:a
:s
x = get a
z = x
z - 32
z |
if z u
a + 1
GOTO s
:u
if x c
GOTO e
:c
b = x
b + 512
c = get b
c + 1
set b c
a + 1
if x a
:e
a = 512
b = 768
l = 10
:Z
j = a
j - b
if j z
z = get a
c = a
c - 512
:3
if z C
printChar l
a + 1
GOTO Z
:C
printChar c
z - 1
GOTO 3
:z

Feel free to test this version online!. The backslash is unnecessary but is there to show the important leading line feed.

Answer 37

ARM machine code on Linux, 60 bytes

Hex Dump:

b590 4c0d f810 2b01 b11a 58a3 3301 50a3 e7f8 6222 2704 2001 f1ad 0101 2201 237f 5ce5 b135 700b df00 3d01 d8fc 250a 700d df00 3b01 d8f4 bd90 000200bc

This function basically creates an array of size 128, and whenever it reads a character from the input string it increment's the value at that characters position. Then it goes back through the array and prints each character array[character] times.

Ungolfed Assembly (GNU syntax):

.syntax unified
.bss @bss is zero-initialized data (doesn't impact code size)
countArray:
    .skip 128 @countArray[x] is the number of occurances of x

.text
.global expStr
.thumb_func
expstr:
    @Input: r0 - a NUL-terminated string.
    @Output: Groups of characters to STDOUT
    push {r4,r7,lr}
    ldr r4,=countArray @Load countArray into r4
    readLoop:
        ldrb r2,[r0],#1 @r2=*r0++
        cbz r2,endReadLoop @If r2==NUL, break
        ldr r3,[r4,r2] @r3=r4[r2]
        adds r3,r3,#1 @r3+=1
        str r3,[r4,r2] @r4[r2]=r3
        b readLoop @while (true)
    endReadLoop:
    @Now countArray[x] is the number of occurances of x.
    @Also, r2 is zero
    str r2,[r4,#' ] @'<character> means the value of <character>
    @What that just did was set the number of spaces found to zero.
    movs r7,#4 @4 is the system call for write
    movs r0,#1 @1 is stdout
    sub r1,sp,#1 @Allocate 1 byte on the stack
    @Also r1 is the char* used for write
    movs r2,#1 @We will print 1 character at a time
    movs r3,#127 @Loop through all the characters
    writeLoop:
        ldrb r5,[r4,r3] @r5=countArray[r3]
        cbz r5,endCharacterLoop @If we're not printing anything, go past the loop
        strb r3,[r1] @We're going to print byte r3, so we store it at *r0
        characterLoop:
            swi #0 @Make system call
            @Return value of write is number of characters printed in r0
            @Since we're printing one character, it should just return 1, which
            @means r0 didn't change.
            subs r5,r5,#1
            bhi characterLoop
        @If we're here then we're done printing our group of characters
        @Thus we just need to print a newline.
        movs r5,#10 @r5='\n' (reusing r5 since we're done using it as a loop counter
        strb r5,[r1]
        swi #0 @Print the character

        endCharacterLoop:
        subs r3,r3,#1
        bhi writeLoop @while (--r3)
    pop {r4,r7,pc}
.ltorg @Store constants here

Testing script (also assembly):

.global _start
_start:
    ldr r0,[sp,#8] @Read argv[1]
    bl expstr @Call function
    movs r7,#1 @1 is the system call for exit
    swi #0 @Make system call

Answer 38

Scala, 53 bytes

def?(s:String)=s.filter(_!=' ').groupBy(identity).map{_._2.mkString}

When run with REPL:

scala> ?("Ah, abracadabra!")
res2: scala.collection.immutable.Iterable[String] = List(!, A, aaaaa, ,, bb, c, h, rr, d)

scala> print(res2.mkString("\n"))
!
A
aaaaa
,
bb
c
h
rr
d

EDIT: Forgot to filter spaces

Answer 39

Clojure, 46

#(partition-by identity(sort(re-seq #"\S" %)))

Long and descriptive function names for simplest of things, yeah.

Answer 40

PowerShell, 65

@TimmyD has the shorter answer but I don't have enough rep to comment. Here's my answer in 65 bytes.

I didn't think to use group and I didn't know that you could stick -join in front of something instead of -join"" on the end and save two characters (using that would make my method 63).

([char[]]$args[0]-ne32|sort|%{if($l-ne$_){"`n"};$_;$l=$_})-join""

My method sorts the array, then loops through it and concatenates characters if they match the preceding entry, inserting a newline if they do not.

Answer 41

Bash + coreutils, 53 50 45 bytes

sed 's: ::g;s:\(.\)\1*:& :g'<<<`fold -1|sort`

There is no separator between the characters of a group and the groups are separated by space. There is one trailing group separator, but as I understood that's acceptable.

Run:

./explode_view.sh <<< '\o /\ o /\ o k'

Output:

// \\\ k ooo 

Answer 42

Clojure (55 bytes)

(defn f[s](map(fn[[a b]](repeat b a))(frequencies s)))

Test Cases:

(f "Ah, abracadabra!")
;;=> ((\space) (\!) (\A) (\a \a \a \a \a) (\b \b) (\c) (\d) (\h) (\,) (\r \r))

(f "\\o/\\o/\\o/")
;;=> ((\\ \\ \\) (\o \o \o) (\/ \/ \/))

(f "A man, a plan, a canal: Panama!")
;;=> ((\space \space \space \space \space \space) (\!) (\A) (\a \a \a \a \a \a \a \a \a) (\c) (\, \,) (\l \l) (\m \m) (\n \n \n \n) (\P) (\p) (\:))

(f "\"Show me how you do that trick, the one that makes me scream\" she said")
;;=> ((\space \space \space \space \space \space \space \space \space \space \space \space \space \space) (\a \a \a \a \a) (\" \") (\c \c) (\d \d) (\e \e \e \e \e \e \e) (\h \h \h \h \h \h) (\i \i) (\k \k) (\,) (\m \m \m \m) (\n) (\o \o \o \o \o) (\r \r) (\S) (\s \s \s \s) (\t \t \t \t \t \t) (\u) (\w \w) (\y))

Answer 43

Javascript (ES5): 50 48 bytes

As a function:

function(i){return i.split("").sort().join(" ")}

The important code, replace the empty string with the search string (29 bytes):

"".split("").sort().join(" ")

Answer 44

Java 8, 121 120 110 103 Bytes

s->s.chars().distinct().filter(c->c!=32)
    .forEach(c->out.println(s.replaceAll("[^\\Q"+(char)c+"\\E]","")))

Above lambda can be consumed with Java-8 Consumer. It fetches distinct characters and replaces other characters of the String to disassemble each character occurrence.

Answer 45

Elixir, 70 bytes

def f(s), do: '#{s}'|>Enum.group_by(&(&1))|>Map.delete(32)|>Map.values

Answer 46

R, 67 bytes

Even though there's already an R-answer by @Frédéric I thought my solution deserves it's own answer because it's conceptually different.

for(i in unique(x<-strsplit(readline(),"")[[1]]))cat(x[x%in%i]," ")

The program prints the ascii characters in order of appearence in the string where groups are separated by two white spaces and chars within groups by one white space. A special case is if the string has whitespaces in itself, then at one specific place in the output there will be 4+number of white spaces in string white spaces separating two groups E.g:

Ah, abracadabra! => A h , a a a a a b b r r c d !

Ungolfed

Split up the code for clarity even though assignment is done within the unique functions and changed to order of evaluation:

x<-strsplit(readline(),"")[[1]]) # Read string from command line and convert into vector

for(i in unique(x){              # For each unique character of the string, create vector 
  cat(x[x%in%i]," ")             # of TRUE/FALSE and return the elements in x for which 
}                                # these are matched

Answer 47

C, 178

#define F for
#define i8 char
#define R return
z(i8*r,i8*e,i8*a){i8*b,c;F(;*a;++a)if(!isspace(c=*a)){F(b=a;*b;++b)if(r>=e)R-1;else if(*b==c)*b=' ',*r++=c;*r++='\n';}*r=0;R 0;}

z(outputArray,LastPointerOkinOutputarray,inputArray) return -1 on error 0 ok Note:Modify its input array too...

#define P printf
main(){char a[]="A, 12aa99dd333aA,,<<", b[256]; z(b,b+255,a);P("r=%s\n", b);}

/*
 178
 r=AA
 ,,,
 1
 2
 aaa
 99
 dd
 333
 <<
 */

Answer 48

Racket 243 bytes

(let*((l(sort(string->list s)char<?))(g(λ(k i)(list-ref k i))))(let loop((i 1)(ol(list(g l 0))))(cond((= i(length l))
(reverse ol))((equal?(g l(- i 1))(g l i))(loop(+ 1 i)(cons(g l i)ol)))(else(loop(+ 1 i)(cons(g l i)(cons #\newline ol)))))))

Ungolfed:

(define (f s)
  (let*((l (sort (string->list s) char<?))
        (g (λ (k i)(list-ref k i))) )
    (let loop ((i 1)
               (ol (list (g l 0))))
      (cond
        ((= i (length l)) (reverse ol))
        ((equal? (g l (- i 1)) (g l i))
         (loop (+ 1 i) (cons (g l i) ol)))
        (else (loop (+ 1 i) (cons (g l i) (cons #\newline ol))))
        ))))

Testing:

(display (f "Ah, abracadabra!"))

Output:

(  
 ! 
 , 
 A 
 a a a a a 
 b b 
 c 
 d 
 h 
 r r)

Answer 49

PowerShell - 86

I imagine I could get this smaller, but I'll work on it later. This assumes that $d contains the string to explode.

$s="";$i=0;[Array]::Sort(($a=$d-split""));$a|%{$s+=$_;if($a[++$i]-cne$_){$s+="`n"}};$s