Removing unique elements from string

Question

I came upon this question, because it seems to be very common use-case to find unique characters in string. But what if we want to get rid of them?

Input contains only lower case alphabets. Only letters from a to z are used. Input length may be from 1 to 1000 characters.

Example:
input: helloworld
output: llool

Objective: Shortest code wins
Language: Any of the top 20 of TIOBE languages

Answer 1

(GolfScript, 15 13 characters)

:;{.;?);>?)},

GolfScript is not one of the top 20, but a codegolf without GolfScript... (run it yourself)

Previous Version: (run script)

1/:;{;\-,;,(<},

Answer 2

Perl, 28 24 characters (includes 1 for 'p' option)

s/./$&x(s!$&!$&!g>1)/eg

Usage:

> perl -pe 's/./$&x(s!$&!$&!g>1)/eg'
helloworld
llool

At first I thought I could do this with negative look-ahead and negative look-behind, but it turns out that negative look-behinds must have a fixed length. So I went for nested regexes instead. With thanks to mob for the $& tip.

Answer 3

J, 12 characters

Having entered a valid Perl answer, here's an invalid (language not in the TIOBE top 20) answer.

a=:#~1<+/@e.

Usage:

   a 'helloworld'
llool

Declares a verb a which outputs only non unique items.

Answer 4

GolfScript (14 chars)

:x{{=}+x\,,(},

Online demo

Might not qualify to win, but it's useful to have a yardstick.

Answer 5

Ruby 46 40 36

gets.chars{|c|$><<c if$_.count(c)>1}

Answer 6

Japt, 6 5 bytes

ÆèX É

-1 byte thanks to @Oliver

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Answer 7

Perl 44

$l=$_;print join"",grep{$l=~/$_.*$_/}split""

Execution:

perl -lane '$l=$_;print join"",grep{$l=~/$_.*$_/}split""' <<< helloworld
llool

Answer 8

Mathematica 72 63

Ok, Mathematica isn't among the top 20 languages, but I decided to join the party anyway.

x is the input string.

"" <> Select[y = Characters@x, ! MemberQ[Cases[Tally@y, {a_, 1} :> a], #] &]

Answer 9

K, 18

{x@&x in&~1=#:'=x}

Answer 10

Python 2.7 (52 51), Python 3 (52)

I didn't expect it to be so short.

2.7: a=raw_input();print filter(lambda x:a.count(x)>1,a)

3.0: a=input();print''.join(i for i in a if a.count(x)>1)

raw_input(): store input as a string (input() = eval(raw_input()))
(Python 3.0: input() has been turned into raw_input())

filter(lambda x:a.count(x)>1,a): Filter through all characters within a if they are found in a more than once (a.count(x)>1).

Answer 11

sed and coreutils (128)

Granted this is not part of the TIOBE list, but it's fun (-:

<<<$s sed 's/./&\n/g'|head -c -1|sort|uniq -c|sed -n 's/^ *1 (.*)/\1/p'|tr -d '\n'|sed 's:^:s/[:; s:$:]//g\n:'|sed -f - <(<<<$s)

De-golfed version:

s=helloworld
<<< $s sed 's/./&\n/g'        \
| head -c -1                  \
| sort                        \
| uniq -c                     \
| sed -n 's/^ *1 (.*)/\1/p'   \
| tr -d '\n'                  \
| sed 's:^:s/[:; s:$:]//g\n:' \
| sed -f - <(<<< $s)

Explanation

The first sed converts input into one character per line. The second sed finds characters that only occur once. Third sed writes a sed script that deletes unique characters. The last sed executes the generated script.

Answer 12

Brachylog (v2), 8 bytes

⊇.oḅlⁿ1∧

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Function submission. Technically noncompeting because the question has a limitation on what langauges are allowed to compete (however, several other answers have already ignored the restriction).

Explanation

⊇.oḅlⁿ1∧
⊇         Find {the longest possible} subset of the input
  o       {for which after} sorting it,
   ḅ        and dividing the sorted input into blocks of identical elements,
    lⁿ1     the length of a resulting block is never 1
 .     ∧  Output the subset in question.

Answer 13

Python (56)

Here's another (few chars longer) alternative in Python:

a=raw_input();print''.join(c for c in a if a.count(c)>1)

---

If you accept output as a list (e.g. ['l', 'l', 'o', 'o', 'l']), then we could boil it down to 49 characters:

a=raw_input();print[c for c in a if a.count(c)>1]

Answer 14

Perl (55)

@x=split//,<>;$s{$_}++for@x;for(@x){print if($s{$_}>1)}

Reads from stdin.

Answer 15

C# – 77 characters

Func<string,string>F=s=>new string(s.Where(c=>s.Count(d=>c==d)>1).ToArray());

If you accept the output as an array, it boils down to 65 characters:

Func<string,char[]>F=s=>s.Where(c=>s.Count(d=>c==d)>1).ToArray();

Answer 16

Ocaml, 139 133

Uses ExtLib's ExtString.String

open ExtString.String
let f s=let g c=fold_left(fun a d->a+Obj.magic(d=c))0 s in replace_chars(fun c->if g c=1 then""else of_char c)s

Non-golfed version

open ExtString.String
let f s =
  let g c =
    fold_left
      (fun a c' -> a + Obj.magic (c' = c))
      0
      s
  in replace_chars
  (fun c ->
    if g c = 1
    then ""
    else of_char c)
  s

The function g returns the number of occurences of c in the string s. The function f replaces all chars either by the empty string or the string containing the char depending on the number of occurences. Edit: I shortened the code by 6 characters by abusing the internal representation of bools :-)

Oh, and ocaml is 0 on the TIOBE index ;-)

Answer 17

PHP - 70

while($x<strlen($s)){$c=$s[$x];echo substr_count($s,$c)>1?$c:'';$x++;}

with asumption $s = 'helloworld'.

Answer 18

Java 8, 90 bytes

s->{for(char c=96;++c<123;s=s.matches(".*"+c+".*"+c+".*")?s:s.replace(c+"",""));return s;}

Explanation:

Try it online.

s->{                         // Method with String as both parameter and return-type
  for(char c=96;++c<123;     //  Loop over the lowercase alphabet
    s=s.matches(".*"+c+".*"+c+".*")?
                             //   If the String contains the character more than once
       s                     //    Keep the String as is
      :                      //   Else (only contains it once):
       s.replace(c+"",""));  //    Remove this character from the String
  return s;}                 //  Return the modified String

Answer 19

PowerShell, 59 bytes

"$args"-replace"[^$($args|% t*y|group|?{$_.Count-1}|% n*)]"

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Less golfed:

$repeatedСhars=$args|% toCharArray|group|?{$_.Count-1}|% name
"$args"-replace"[^$repeatedСhars]"

Note: $repeatedChars is an array. By default, a Powershell joins array elements by space char while convert the array to string. So, the regexp contains spaces (In this example, [^l o]). Spaces do not affect the result because the input string contains letters only.

Answer 20

APL (Dyalog Extended), 8 bytes^SBCS

Anonymous tacit prefix function.

∊⊢⊆⍨1<⍧⍨

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⍧⍨ count-in selfie (count occurrences of argument elements in the argument itself)

1< Boolean mask where one is less than that

⊢⊆⍨ partition the argument by that mask (beginning a new partition on 1s and removing on 0s)

∊ ϵnlist (flatten)

Answer 21

JavaScript, 45 bytes

s=>[...s].filter(c=>s.match(c+'.*'+c)).join``

Answer 22

R, 70 bytes

a=utf8ToInt(scan(,''));intToUtf8(a[!a%in%names(table(a)[table(a)<2])])

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A poor attempt, even from a TIOBE top 20 language. I know something can be done about the second half, but at the moment, any golfs escape me.

Answer 23

JavaScript, 34 bytes

Input as a string, output as a character array.

s=>[...s].filter(x=>s.split(x)[2])

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Answer 24

JavaScript (Node.js), 82 bytes

p=>[...p].map((v,i,a)=>a.filter(f=>f==v).length).reduce((a,c,i)=>c>1?a+=p[i]:a,[])

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Answer 25

C++, 139 bytes

string s;cin>>s;string w{s}; auto l=remove_if(begin(s),end(s),[&w](auto&s){return count(begin(w),end(w),s)==1;});s.erase(l,end(s));cout<<s;

ungolfed:

#include <algorithm>
#include <string>
#include <iostream>

int main() {
  using namespace std;
  string s;
  cin >> s;
  const string w{s};
  auto l = remove_if(begin(s), end(s), [&w](auto& s) {
                                         return count(begin(w), end(w), s) == 1;
                                       });
  s.erase(l, end(s));
  cout << s;
  return 0;
}

Answer 26

PHP - 137

Code

implode('',array_intersect(str_split($text),array_flip(array_filter(array_count_values(str_split($text)),function($x){return $x>=2;}))));

Normal Code

$text   = 'helloworld';
$filter = array_filter(array_count_values(str_split($text)), function($x){return $x>=2;});
$output = implode('',array_intersect(str_split($text),array_flip($filter)));

echo $output;

Answer 27

PHP - 83 78

<?for($a=$argv[1];$i<strlen($a);$r[$a[$i++]]++)foreach($ras$k=>$c)if($c>1)echo$k

Improved version:

<?for($s=$argv[1];$x<strlen($s);$c=$s[$x++]) echo substr_count($s,$c)>1?$c:'';

Of course this needs notices to be turned off

Edit: Improvement inspired by @hengky mulyono

I am so bad at codegolf :)