Special String reformatting

Question

This challenge is to take an alphabetical string as input and to apply the following conversion:

The first of each type of character of the string must stay, and must be immediately followed by an integer representing how many of these characters were in the original string. Any repeating characters must be omitted.

All inputs will be entirely lower-case letters (no spaces). Outputs must be ordered in the same way as the inputs (input hi must give output of h1i1, not i1h1)

Examples

Input: potato Output: p1o2t2a1

Input: pqwertyuiop Output: p2q1w1e1r1t1y1u1i1o1

Input: thisisanexample Output: t1h1i2s2a2n1e2x1m1p1l1

Input: oreganoesque Output: o2r1e3g1a1n1s1q1u1

Input: aaaaaaabaaaaaa Output: a13b1

Scoring

This is code-golf. Shortest answer wins!

Answer 1

APL (Dyalog Unicode), 7 bytes (SBCS)

Anonymous tacit prefix function.

,,∘⍕∘≢⌸

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⌸ apply the following function between each unique character and the indices where it occurs:

 , concatenate the character

  ∘ to

   ⍕ the stringification

    ∘ of

     ≢ the indices' count

, flatten

Answer 2

PHP, 63 bytes

foreach(array_count_values(str_split($argn))as$a=>$b)echo$a.$b;

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Built-ins and a whole lot of glue. Input via STDIN, output to STDOUT.

Answer 3

Python 3, ^{58 \$\cdots\$ 50} 57 bytes

Added 3 bytes and switched to Python 3 to fix bugs.
Added 7 bytes to make output a string.

lambda s:''.join(c+f'{s.count(c)}'for c in{}.fromkeys(s))

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Outputs a string.

Outputting a list of strings is 50 bytes.

Answer 4

Wolfram Language (Mathematica), 34 bytes

This is the dreaded obvious chain-the-built-ins. Print can accept multiple arguments and will print them without any separators.

Print@@Flatten@Tally@Characters@#&

More scary version, for the same number of bytes:

Print@@(##&)@@@Tally@Characters@#&

I suspect that a better hack consisting entirely of the characters #&/() can be used to get rid of Flatten, but I couldn't come up with one.

Of course, Mathematica has both LetterCounts and CharacterCounts, and there's also Counts instead of Tally, but all of these seem to return association objects that seem excessively complicated to work with.

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Answer 5

Python 3, 49 bytes

lambda s:''.join({c+str(s.count(c)):1for c in s})

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Outputs a string.

Based on Noodle9's solution, which uses a nice idea of a dictionary to deduplicate while preserving order, which they do in Python 3 but not Python 2. Note that set doesn't preserve order.

The idea is to make the character-with-count strings be keys of a dictionary so that only the first instance is used. The values in the dictionary don't matter, since dictionaries iterate over keys by default.

I had thought at first that the deduplication must be applied to the characters of the input string, but realized that it also works on the strings to be joined in the output, since a given character is always attached to the same count.

Outputting a list of strings without joining takes 43 bytes.

Answer 6

K (oK), 23 18 bytes

{,/t,'$+/x=\:t:?x}

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Answer 7

J, 15 17 bytes

Thanks to Jonah for finding and fixing a bug!

[:;~.<@,&":"0#/.~

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Answer 8

05AB1E, 6 bytes

Ùε¢yì?

Try it online or verify all test cases.

Explanation:

Ù       # Uniquify the (implicit) input-string
 ε      # For-each over each character in this string:
  ¢     #  Count the amount of times the current character occurs in the (implicit) input
   yì   #  Prepend the current character before this count
     ?  #  Print it without newline

Answer 9

Jelly, 5 bytes

Lƙż@Q

A full program printing the result (or a monadic Link yielding a list of pairs of characters and integers).

Try it online!

There are loads of ways to achieve this in 6 bytes (e.g. Qżċ@€¥).

How?

Lƙż@Q - Main Link: list of characters, S          e.g. "cabbage"
 ƙ    - for groups of identical elements (of S):       (c aa bb g e)
L     -   length                                       [1,2,2,1,1]
    Q - de-duplicate S                                 "cabge"
   @  - with swapped arguments:
  ż   -   zip                                          [['c',1],['a',2],['b',2],['g',1],['e',1]]
      - implicit, smashing, print                      c1a2b2g1e1

Answer 10

Python 3, 62 bytes

f=lambda s:s and s[0]+str(s.count(s[0]))+f(s.replace(s[0],''))

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Answer 11

JavaScript (ES6), 55 bytes

s=>[...new Set(s)].map(c=>c+~-s.split(c).length).join``

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Commented

s =>                    // s = input string
  [...new Set(s)]       // create a set from the input string and split it,
                        // resulting in an array of characters arranged in
                        // order of first appearance
  .map(c =>             // for each character c in this array:
    c +                 //   append c
    ~-s.split(c).length //   followed by the number of occurrences in the
                        //   original string
  )                     // end of map()
  .join``               // join everything

Answer 12

perl -p, 52 50 bytes

s/./$&1/g;1while s/(\D)\K(\d+)(.*)\1\d/($2+1).$3/e

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Starts off by adding 1 to each character. Then, as often as possible, find a letter followed by a number, with the same letter elsewhere in the string followed by a digit (which has to be 1). Increment the count, and remove the same letters followed by its 1.

Initial solution, following a very different technique:

perl -nF//, 52 bytes

$F{$_}++for@F;$F{$_}&&print$_,$F{$_}xor$F{$_}=0for@F

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Reads a line from STDIN, assuming it's not newline terminated.

Splits the input into characters, available (in order), in @F (due to the -F//). Counts the occurrence of each character in the hash %F. Then loops over @F: if present in %F with a true value, print the character and its count, then set the corresponding entry in %F to 0. This ensures each character is only outputted once.

The TIO code has some header and footer code so we can handle multiple test inputs. They're not needed if we just have one line of input.

Answer 13

Brachylog, 5 bytes

ọ∋∋w⊥

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Full program, or alternatively a predicate which prints the output then fails. Could be one byte shorter if it could generate the characters as a mix of strings and integers, but that seems like a bit too far out there of an output format.

   w     Print (without a newline)
  ∋      an element of
 ∋       an element of
ọ        a list of pairs [element, # of occurrences] in order of first appearance,
    ⊥    then try it again some other way.

Answer 14

C (gcc), ^{105 \$\cdots\$93} 90 bytes

Saved 2 5 bytes from the man himself Arnauld!!!

d;c;f(char*s){for(;d=1,c=*s;c-1&&printf("%c%d",c,d))for(char*q=s++;*++q;d+=*q==c?*q=1:0);}

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Answer 15

AWK + -F '', 65 + 4 = 69 bytes

{while(i++<NF)if(!a[$i]++)b[i]=$i;while(j++<i)printf b[j]a[b[j]]}

Try it at awk.js.org

A shorter 64 59+4 byte program, which runs on GNU awk with -F '', is this:

{while(i++<NF)if(!a[$i]++)b[i]=1;for(i in b)printf$i a[$i]}

Annoyingly, though, while 'Try it online' links to the GNU awk manual page, it doesn't seem to use GNU awk, and refuses the -F '' command-line option.

The alternative link above (to awk .js.org) accepts the command-line option, but then outputs in a different order, which costs a frustrating additional one six bytes to correct (which I have included above as the price of verifiability).

Answer 16

CJam, 12 bytes

Port of the Pyth answer.

q:A_&{_Ae=}%

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Explanation

q            Take the whole input
 :A          Assign to a variable
   _&        Set union w/ itself
     {    }% Map:
      _          Join the uniquified character
       Ae=       With the count of the character in the input string.

CJam, 16 bytes

CJam has the built-in, so I guess it simplifies the question. Unfortunately the built-in does it in the wrong order...

q:A{A#}$e`{-1%}%

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Answer 17

Perl 5 -p, 28 bytes

s|.|($b=s/$&//g)?$&.$b:''|ge

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Answer 18

Ruby 2.7 -paF|, 17 bytes

$_=[*$F.tally]*''

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This is more or less built in to the recent Ruby release as the tally method.

Answer 19

K4 17 16 bytes

Solution:

{,/(?x),'$#:'=x}

Examples:

q)k){,/(?x),'$#:'=x}"potato"
"p1o2t2a1"
q)k){,/(?x),'$#:'=x}"pqwertyuiop"
"p2q1w1e1r1t1y1u1i1o1"
q)k){,/(?x),'$#:'=x}"thisisanexample"
"t1h1i2s2a2n1e2x1m1p1l1"
q)k){,/(?x),'$#:'=x}"oreganoesque"
"o2r1e3g1a1n1s1q1u1"
q)k){,/(?x),'$#:'=x}"aaaaaaabaaaaaa"
"a13b1"

Explanation:

{,/(?x),'$#:'=x} / the solution
{              } / lambda function taking implicit 'x' argument
             =x  / group x (dict of unique chars => indices)
          #:'    / count length of each group
         $       / cast to string
       ,'        / join each-both
   (  )          / do this together
    ?x           / distinct x
 ,/              / flatten

Answer 20

R, 72 71 bytes

cat(rbind(z<-unique(y<-el(strsplit(scan(,""),""))),table(y)[z]),sep="")

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Answer 21

SNOBOL4 (CSNOBOL4), 96 bytes

	I =INPUT
N	I LEN(1) . X	:F(O)
	N =
S	I X =	:F(B)
	N =N + 1	:(S)
B	O =O X N	:(N)
O	OUTPUT =O
END

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	I =INPUT		;* Read input, set to I
N	I LEN(1) . X	:F(O)	;* Get the first character of I as X; if I is empty then goto O
	N =			;* set N to empty string (evaled as 0 in arithmetic)
S	I X =	:F(B)		;* remove the first occurrence of X from I. If none exist, goto B
	N =N + 1	:(S)	;* increment N and goto S
B	O =O X N	:(N)	;* Add to the output string and goto N to get the Next character
O	OUTPUT =O		;* print the result
END

Answer 22

T-SQL, 111 bytes

Added a line change to make it readable

WHILE @+@ like'_[a-z]%'
SELECT @=concat(s,left(@,1),len(@)-len(s))FROM(SELECT
replace(@,left(@,1),'')s)s
PRINT @

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Answer 23

JavaScript (V8), 106 102 bytes

e=>{for(o="",i=0;i<e.length;i++)o.includes(e[i])||(o+=e[i]+e.match(RegExp(e[i],"g")).length);return o}

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Answer 24

Pyth, 8 7 bytes

-1 byte thanks to @isaacg

s+R/Qd{

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s+R/Qd{
      {  Deduplicate: keep the first occurrence of each unique character
  R      For each of these unique characters:
 +        - append the character
   /Qd    - to its count in the original input
s        Join the resulting strings

Answer 25

Stax, 7 bytes

ô!Ω;òá☺

Run and debug it

Answer 26

Retina 0.8.2, 31 bytes

+`(.)(.+)\1
$1$1$2
(.)\1*
$1$.&

Try it online! Link includes test cases, unusually without even needing a header. Explanation:

+`(.)(.+)\1
$1$1$2

Collect all repeated characters into a single run at the first appearance.

(.)\1*
$1$.&

Replace each run with its first character and its length.

Answer 27

Charcoal, 14 bytes

⭆Φθ¬№…θκι⁺ι№θι

Try it online! Link is to verbose version of code. Explanation:

  θ             Input string
 Φ              Filter over characters
    №           Count of
        ι       Current character in
      θ         Input string
     …          Truncated to length
       κ        Current index
   ¬            Is zero
⭆               Map over unique characters and join
          ι     Current character
         ⁺      Concatenated with
           №    Count of
             ι  Current character in
            θ   Input string
                Implicitly print

Answer 28

Kotlin, 69 67 bytes

fun String.f()=groupBy{it}.map{(a,b)->"$a"+b.size"}.joinToString("")

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Answer 29

C# (Visual C# Interactive Compiler), 57 bytes

s=>s.GroupBy(c=>c).Aggregate("",(r,g)=>r+g.Key+g.Count())

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Answer 30

SimpleTemplate, 54 47 bytes

This was a very easy, but incredibly fun challenge!

The code simply loops through every character and counts how many times it shows, and presents it all again.
Nothing fancy at all...

{@eachargv.0}{@incR.[_]}{@/}{@eachR}{@echo__,_}

Hey, I didn't say the code was readable!

---

Here's an ungolfed and readable version:

{@each argv.0 as char}
    {@inc by 1 result.[char]}
{@/}
{@each result as times key char}
    {@echo char, times}
{@/}

Should be easy to understand ...
{@inc} increments the value or creates a new one, if it doesn't exist. (this keeps key ordering).

---

You can try this on http://sandbox.onlinephpfunctions.com/code/a180782e659c29674fbb0d77dc82d90d238c6e08
Older version: http://sandbox.onlinephpfunctions.com/code/6ee5077eaf38ec445d84086cc07966026ca7c565

(There, you have an example on how to use this in a function, with multiple tests.)