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Given an Array, derive a sequence of elements in which all duplicates are removed.

There are basically three approaches seen here:

Put the elements into a hash table which does not allow duplicates. The complexity is O(n) on average, and O(n²) worst case. This approach requires a hash function for your type (which is compatible with equality), either built-in to your language, or provided by the user.
Sort the elements and remove consecutive duplicate elements. The complexity of the best sorting algorithms is O(n log n). This approach requires that your type be "comparable", i.e., have an ordering. Putting the elements into a self-balancing binary search tree is a special case of sorting.
Go through the list, and for each element, check the rest of the list to see if it appears again, and discard it if it does. The complexity is O(n²). The up-shot is that this always works on any type (provided that you can test for equality).

ACL2

<lang lisp>(remove-duplicates xs)</lang>

AppleScript

<lang applescript>unique({1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d"})

on unique(x) set R to {} repeat with i in x if i is not in R then set end of R to i's contents end repeat return R end unique</lang>

AutoHotkey

Built in Sort has an option to remove duplicates <lang AutoHotkey>a = 1,2,1,4,5,2,15,1,3,4 Sort, a, a, NUD`, MsgBox % a ; 1,2,3,4,5,15</lang>

We produce an array a with duplicates from a string; then index a second array b with the contents of a, so that duplicates make only one entry; then produce a string with the keys of b, which is finally output. <lang awk>$ awk 'BEGIN{split("a b c d c b a",a);for(i in a)b[a[i]]=1;r="";for(i in b)r=r" "i;print r}' a b c d</lang>

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> DIM list$(15)

     list$() = "Now", "is", "the", "time", "for", "all", "good", "men", \
     \         "to", "come", "to", "the", "aid", "of", "the", "party."
     num% = FNremoveduplicates(list$())
     FOR i% = 0 TO num%-1
       PRINT list$(i%) " " ;
     NEXT
     PRINT
     END
     
     DEF FNremoveduplicates(l$())
     LOCAL i%, j%, n%, i$
     n% = 1
     FOR i% = 1 TO DIM(l$(), 1)
       i$ = l$(i%)
       FOR j% = 0 TO i%-1
         IF i$ = l$(j%) EXIT FOR
       NEXT
       IF j%>=i% l$(n%) = i$ : n% += 1
     NEXT
     = n%</lang>

Output:

Now is the time for all good men to come aid of party.

Bracmat

Here are three solutions. The first one (A) uses a hash table, the second (B) uses a pattern for spotting the elements that have a copy further on in the list and only adds those elements to the answer that don't have copies further on. The third solution (C) utilises an mechanism that is very typical of Bracmat, namely that sums (and also products) always are transformed to a normalised form upon evaluation. Normalisation means that terms are ordered in a unique way and that terms that are equal, apart from a numerical factor, are replaced by a single term with a numerical factor that is the sum of the numerical factors of each term. The answer is obtained by replacing all numerical factors by 1 as the last step.

The list contains atoms and also a few non-atomic expressions. The hash table needs atomic keys, so we apply the str function when searching and inserting elements. <lang bracmat>2 3 5 7 11 13 17 19 cats 222 (-100.2) "+11" (1.1) "+7" (7.) 7 5 5 3 2 0 (4.4) 2:?LIST

(A=

 ( Hashing
 =   h elm list
   .   new$hash:?h
     &   whl
       ' ( !arg:%?elm ?arg
         & ( (h..find)$str$!elm
           | (h..insert)$(str$!elm.!elm)
           )
         )
     & :?list
     &   (h..forall)
       $ (
         = .!arg:(?.?arg)&!arg !list:?list
         )
     & !list
 )

& put$("Solution A:" Hashing$!LIST \n,LIN) );

(B=

 ( backtracking
 =   answr elm
   .     :?answr
       &   !arg
         :   ?
             (   %?`elm
                 ?
                 ( !elm ?
                 | &!answr !elm:?answr
                 )
             & ~
             )
     | !answr
 )

& put$("Solution B:" backtracking$!LIST \n,LIN) );

(C=

 ( summing
 =   sum car LIST
   .   !arg:?LIST
     & 0:?sum
     &   whl
       ' ( !LIST:%?car ?LIST
         & (.!car)+!sum:?sum
         )
     &   whl
       ' ( !sum:#*(.?el)+?sum
         & !el !LIST:?LIST
         )
     & !LIST
 )

& put$("Solution C:" summing$!LIST \n,LIN) );

( !A & !B & !C & )</lang> Only solution B produces a list with the same order of elements as in the input.

Solution A: 19 (4.4) 17 11 13 (1.1) (7.) 222 +11 7 5 3 2 0 cats (-100.2) +7
Solution B: 11 13 17 19 cats 222 (-100.2) +11 (1.1) +7 (7.) 7 5 3 0 (4.4) 2
Solution C: (7.) (4.4) (1.1) (-100.2) cats 222 19 17 13 11 7 5 3 2 0 +7 +11

Brat

<lang brat>some_array = [1 1 2 1 'redundant' [1 2 3] [1 2 3] 'redundant']

unique_array = some_array.unique</lang>

C

O(n^2) version, using linked lists

Since there's no way to know ahead of time how large the new data structure will need to be, we'll return a linked list instead of an array.

<lang c>#include <stdio.h>

include <stdlib.h>

struct list_node {int x; struct list_node *next;}; typedef struct list_node node;

node * uniq(int *a, unsigned alen)

{if (alen == 0) return NULL;
 node *start = malloc(sizeof(node));
 if (start == NULL) exit(EXIT_FAILURE);
 start->x = a[0];
 start->next = NULL;

 for (int i = 1 ; i < alen ; ++i)
    {node *n = start;
     for (;; n = n->next)
        {if (a[i] == n->x) break;
         if (n->next == NULL)
            {n->next = malloc(sizeof(node));
             n = n->next;
             if (n == NULL) exit(EXIT_FAILURE);
             n->x = a[i];
             n->next = NULL;
             break;}}}

 return start;}

int main(void)

  {int a[] = {1, 2, 1, 4, 5, 2, 15, 1, 3, 4};
   for (node *n = uniq(a, 10) ; n != NULL ; n = n->next)
       printf("%d ", n->x);
   puts("");
   return 0;}</lang>

Output:

1 2 4 5 15 3

O(n^2) version, pure arrays

<lang c>#include <stdio.h>

include <stdlib.h>
include <stdbool.h>
include <string.h>

/* Returns `true' if element `e' is in array `a'. Otherwise, returns `false'.

* Checks only the first `n' elements. Pure, O(n).
*/

bool elem(int *a, size_t n, int e) {

   for (size_t i = 0; i < n; ++i)
       if (a[i] == e)
           return true;

   return false;

}

/* Removes the duplicates in array `a' of given length `n'. Returns the number

* of unique elements. In-place, order preserving, O(n ^ 2).
*/

size_t nub(int *a, size_t n) {

   size_t m = 0;

   for (size_t i = 0; i < n; ++i)
       if (!elem(a, m, a[i]))
           a[m++] = a[i];

   return m;

}

/* Out-place version of `nub'. Pure, order preserving, alloc < n * sizeof(int)

* bytes, O(n ^ 2).
*/

size_t nub_new(int **b, int *a, size_t n) {

   int *c = malloc(n * sizeof(int));
   memcpy(c, a, n * sizeof(int));
   int m = nub(c, n);
   *b = malloc(m * sizeof(int));
   memcpy(*b, c, m * sizeof(int));
   free(c);
   return m;

}

int main(void) {

   int a[] = {1, 2, 1, 4, 5, 2, 15, 1, 3, 4};
   int *b;

   size_t n = nub_new(&b, a, sizeof(a) / sizeof(a[0]));

   for (size_t i = 0; i < n; ++i)
       printf("%d ", b[i]);
   puts("");

   free(b);
   return 0;

}</lang>

Output:

1 2 4 5 15 3

Sorting method

Using qsort and return uniques in-place:<lang c>#include <stdio.h>

include <stdlib.h>

int icmp(const void *a, const void *b) {

define _I(x) *(const int*)x

return _I(a) < _I(b) ? -1 : _I(a) > _I(b);

undef _I

}

/* filter items in place and return number of uniques. if a separate

  list is desired, duplicate it before calling this function */

int uniq(int *a, int len) { int i, j; qsort(a, len, sizeof(int), icmp); for (i = j = 0; i < len; i++) if (a[i] != a[j]) a[++j] = a[i]; return j + 1; }

int main() { int x[] = {1, 2, 1, 4, 5, 2, 15, 1, 3, 4}; int i, len = uniq(x, sizeof(x) / sizeof(x[0])); for (i = 0; i < len; i++) printf("%d\n", x[i]);

return 0; }</lang>

Output:

C++

This version uses std::set, which requires its element type be comparable using the < operator. <lang cpp>#include <set>

include <iostream>

using namespace std;

int main() {

   typedef set<int> TySet;
   int data[] = {1, 2, 3, 2, 3, 4};

   TySet unique_set(data, data + 6);

   cout << "Set items:" << endl;
   for (TySet::iterator iter = unique_set.begin(); iter != unique_set.end(); iter++)
         cout << *iter << " ";
   cout << endl;

}</lang>

This version uses hash_set, which is part of the SGI extension to the Standard Template Library. It is not part of the C++ standard library. It requires that its element type have a hash function.

Works with: GCC

<lang cpp>#include <ext/hash_set>

include <iostream>

using namespace std;

int main() {

   typedef __gnu_cxx::hash_set<int> TyHash;
   int data[] = {1, 2, 3, 2, 3, 4};

   TyHash unique_set(data, data + 6);

   cout << "Set items:" << endl;
   for (TyHash::iterator iter = unique_set.begin(); iter != unique_set.end(); iter++)
         cout << *iter << " ";
   cout << endl;

}</lang>

This version uses unordered_set, which is part of the TR1, which is likely to be included in the next version of C++. It is not part of the C++ standard library. It requires that its element type have a hash function.

Works with: GCC

<lang cpp>#include <tr1/unordered_set>

include <iostream>

using namespace std;

int main() {

   typedef tr1::unordered_set<int> TyHash;
   int data[] = {1, 2, 3, 2, 3, 4};

   TyHash unique_set(data, data + 6);

   cout << "Set items:" << endl;
   for (TyHash::iterator iter = unique_set.begin(); iter != unique_set.end(); iter++)
         cout << *iter << " ";
   cout << endl;

}</lang>

Alternative method working directly on the array:

<lang cpp>#include <iostream>

include <iterator>
include <algorithm>

// helper template template<typename T> T* end(T (&array)[size]) { return array+size; }

int main() {

 int data[] = { 1, 2, 3, 2, 3, 4 };
 std::sort(data, end(data));
 int* new_end = std::unique(data, end(data));
 std::copy(data, new_end, std::ostream_iterator<int>(std::cout, " ");
 std::cout << std::endl;

}</lang>

Using sort, unique, and erase on a vector.

Works with: C++11

<lang cpp>#include <algorithm>

include <iostream>
include <vector>

int main() {

 std::vector<int> data = {1, 2, 3, 2, 3, 4};

 std::sort(data.begin(), data.end());
 data.erase(std::unique(data.begin(), data.end()), data.end());

 for(int& i: data) std::cout << i << " ";
 std::cout << std::endl;
 return 0;

}</lang>

C#

Works with: C# version 2+

<lang csharp>int[] nums = { 1, 1, 2, 3, 4, 4 }; List<int> unique = new List<int>(); foreach (int n in nums)

   if (!unique.Contains(n))
       unique.Add(n);</lang>

Works with: C# version 3+

<lang csharp>int[] nums = {1, 1, 2, 3, 4, 4}; int[] unique = nums.Distinct().ToArray();</lang>

Clojure

<lang lisp>user=> (distinct [1 3 2 9 1 2 3 8 8 1 0 2]) (1 3 2 9 8 0) user=></lang>

Common Lisp

To remove duplicates non-destructively:

<lang lisp>(remove-duplicates '(1 3 2 9 1 2 3 8 8 1 0 2)) > (9 3 8 1 0 2)</lang>

Or, to remove duplicates in-place:

<lang lisp>(delete-duplicates '(1 3 2 9 1 2 3 8 8 1 0 2)) > (9 3 8 1 0 2)</lang>

D

<lang d>import std.stdio, std.algorithm;

void main() {

   auto data = [1, 3, 2, 9, 1, 2, 3, 8, 8, 1, 0, 2];
   data.sort();
   writeln(uniq(data));

}</lang> Output:

[0, 1, 2, 3, 8, 9]

Using an associative array: <lang d>import std.stdio;

void main() {

   auto data = [1, 3, 2, 9, 1, 2, 3, 8, 8, 1, 0, 2];

   int[int] hash;
   foreach (el; data)
       hash[el] = 0;
   writeln(hash.keys);

}</lang> Output:

[8, 0, 1, 9, 2, 3]

Delphi

Generics were added in Delphi2009.

<lang Delphi>program RemoveDuplicateElements;

{$APPTYPE CONSOLE}

uses Generics.Collections;

var

 i: Integer;
 lIntegerList: TList<Integer>;

const

 INT_ARRAY: array[1..7] of Integer = (1, 2, 2, 3, 4, 5, 5);

begin

 lIntegerList := TList<Integer>.Create;
 try
 for i in INT_ARRAY do
   if not lIntegerList.Contains(i) then
     lIntegerList.Add(i);

 for i in lIntegerList do
   Writeln(i);
 finally
   lIntegerList.Free;
 end;

end.</lang>

Output:

E

<lang e>[1,2,3,2,3,4].asSet().getElements()</lang>

Erlang

<lang erlang>List = [1, 2, 3, 2, 2, 4, 5, 5, 4, 6, 6, 5]. UniqueList = gb_sets:to_list(gb_sets:from_list(List)). % Alternatively the builtin: Unique_list = lists:usort( List ). </lang>

Euphoria

<lang euphoria>include sort.e

function uniq(sequence s)

   sequence out
   s = sort(s)
   out = s[1..1]
   for i = 2 to length(s) do
       if not equal(s[i],out[$]) then
           out = append(out, s[i])
       end if
   end for
   return out

end function

constant s = {1, 2, 1, 4, 5, 2, 15, 1, 3, 4} ? s ? uniq(s)</lang>

Output:

{1,2,1,4,5,2,15,1,3,4}
{1,2,3,4,5,15}

F#

The simplest way is to build a set from the given array (this actually works for any enumerable input sequence type, not just arrays): <lang fsharp> set [|1;2;3;2;3;4|] </lang> gives: <lang fsharp> val it : Set<int> = seq [1; 2; 3; 4] </lang>

Factor

<lang factor>USING: sets ; V{ 1 2 1 3 2 4 5 } members .

V{ 1 2 3 4 5 }</lang>

Forth

Forth has no built-in hashtable facility, so the easiest way to achieve this goal is to take the "uniq" program as an example.

The word uniq, if given a sorted array of cells, will remove the duplicate entries and return the new length of the array. For simplicity, uniq has been written to process cells (which are to Forth what "int" is to C), but could easily be modified to handle a variety of data types through deferred procedures, etc.

The input data is assumed to be sorted.

<lang forth>\ Increments a2 until it no longer points to the same value as a1 \ a3 is the address beyond the data a2 is traversing.

skip-dups ( a1 a2 a3 -- a1 a2+n )

   dup rot ?do
     over @ i @ <> if drop i leave then
   cell +loop ;

\ Compress an array of cells by removing adjacent duplicates \ Returns the new count

uniq ( a n -- n2 )

  over >r             \ Original addr to return stack
  cells over + >r     \ "to" addr now on return stack, available as r@
  dup begin           ( write read )
     dup r@ <
  while
     2dup @ swap !    \ copy one cell
     cell+ r@ skip-dups
     cell 0 d+        \ increment write ptr only
  repeat  r> 2drop  r> - cell / ;</lang>

Here is another implementation of "uniq" that uses a popular parameters and local variables extension words. It is structurally the same as the above implementation, but uses less overt stack manipulation.

<lang forth>: uniqv { a n \ r e -- n }

   a n cells+ to e
   a dup to r
   \ the write address lives on the stack
   begin
     r e <
   while
     r @ over !
     r cell+ e skip-dups to r
     cell+
   repeat
   a - cell / ;</lang>

To test this code, you can execute:

<lang forth>create test 1 , 2 , 3 , 2 , 6 , 4 , 5 , 3 , 6 , here test - cell / constant ntest

.test ( n -- ) 0 ?do test i cells + ? loop ;

test ntest 2dup cell-sort uniq .test</lang>

output

Fortran

Fortran has no built-in hash functions or sorting functions but the code below implements the compare all elements algorithm.

program remove_dups

 implicit none
 integer :: example(12)         ! The input
 integer :: res(size(example))  ! The output
 integer :: k                   ! The number of unique elements
 integer :: i, j

 example = [1, 2, 3, 2, 2, 4, 5, 5, 4, 6, 6, 5]
 k = 1
 res(1) = example(1)
 outer: do i=2,size(example)
    do j=1,k
       if (res(j) == example(i)) then
          ! Found a match so start looking again
          cycle outer
       end if
    end do
    ! No match found so add it to the output
    k = k + 1
    res(k) = example(i)
 end do outer
 write(*,advance='no',fmt='(a,i0,a)') 'Unique list has ',k,' elements: '
 write(*,*) res(1:k)

end program remove_dups

</lang>

GAP

<lang gap># Built-in, using sets (which are also lists) a := [ 1, 2, 3, 1, [ 4 ], 5, 5, [4], 6 ];

[ 1, 2, 3, 1, [ 4 ], 5, 5, [ 4 ], 6 ]

b := Set(a);

[ 1, 2, 3, 5, 6, [ 4 ] ]

IsSet(b);

true

IsList(b);

true</lang>

Go

Map solution

<lang go>package main import "fmt"

func uniq(list []int) []int {

 unique_set := make(map[int] bool, len(list))
 for _, x := range list {
   unique_set[x] = true
 }
 result := make([]int, len(unique_set))
 i := 0
 for x := range unique_set {
   result[i] = x
   i++
 }
 return result

}

func main() {

 fmt.Println(uniq([]int {1,2,3,2,3,4})) // prints: [3 1 4 2]

}</lang>

Map preserving order

It takes only small changes to the above code to preserver order. Just store the sequence in the map: <lang go>package main

import "fmt"

func uniq(list []int) []int {

   unique_set := make(map[int]int, len(list))
   i := 0
   for _, x := range list {
       if _, there := unique_set[x]; !there {
           unique_set[x] = i
           i++
       }
   }
   result := make([]int, len(unique_set))
   for x, i := range unique_set {
       result[i] = x
   }
   return result

}

func main() {

   fmt.Println(uniq([]int{1, 2, 3, 2, 3, 4})) // prints: [1 2 3 4]

}</lang>

Float64, removing duplicate NaNs.

In solutions above, you just replace "int" with another type to use for a list of another type. (See Associative_arrays/Creation#Go for acceptable types.) Except a weird thing happens with NaNs. They don't compare equal, so you have to special case them if you want to remove duplicates: <lang go>package main

import (

   "fmt"
   "math"

)

func uniq(list []float64) []float64 {

   unique_set := map[float64]int{}
   i := 0
   nan := false
   for _, x := range list {
       if _, exists := unique_set[x]; exists {
           continue
       }
       if math.IsNaN(x) {
           if nan {
               continue
           } else {
               nan = true
           }
       }
       unique_set[x] = i
       i++
   }
   result := make([]float64, len(unique_set))
   for x, i := range unique_set {
       result[i] = x
   }
   return result

}

func main() {

   fmt.Println(uniq([]float64{1, 2, math.NaN(), 2, math.NaN(), 4}))

}</lang>

Groovy

<lang groovy>def list = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] assert list.size() == 12 println " Original List: ${list}"

// Filtering the List list.unique() assert list.size() == 8 println " Filtered List: ${list}"

list = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] assert list.size() == 12

// Converting to Set def set = new HashSet(list) assert set.size() == 8 println " Set: ${set}"

// Converting to Order-preserving Set set = new LinkedHashSet(list) assert set.size() == 8 println "List-ordered Set: ${set}"</lang>

Output:

   Original List: [1, 2, 3, a, b, c, 2, 3, 4, b, c, d]
   Filtered List: [1, 2, 3, a, b, c, 4, d]
             Set: [1, d, 2, 3, 4, b, c, a]
List-ordered Set: [1, 2, 3, a, b, c, 4, d]

Haskell

<lang haskell>values = [1,2,3,2,3,4] unique = List.nub values</lang>

HicEst

<lang hicest>REAL :: nums(12) CHARACTER :: workspace*100

nums = (1, 3, 2, 9, 1, 2, 3, 8, 8, 1, 0, 2) WRITE(Text=workspace) nums ! convert to string EDIT(Text=workspace, SortDelDbls=workspace) ! do the job for a string READ(Text=workspace, ItemS=individuals) nums ! convert to numeric

WRITE(ClipBoard) individuals, "individuals: ", nums ! 6 individuals: 0 1 2 3 8 9 0 0 0 0 0 0 </lang>

Icon and Unicon

This solution preserves the original order of the elements. <lang Icon>procedure main(args)

   every write(!noDups(args))

end

procedure noDups(L)

   every put(newL := [], notDup(set(),!L))
   return newL

end

procedure notDup(cache, a)

   if not member(cache, a) then {
        insert(cache, a)
        return a
        }

end</lang> A sample run is:

->noDups a b c d c a b e
a
b
c
d
e
->

IDL

<lang idl>non_repeated_values = array[uniq(array, sort( array))]</lang>

Inform 7

<lang inform7>To decide which list of Ks is (L - list of values of kind K) without duplicates: let result be a list of Ks; repeat with X running through L: add X to result, if absent; decide on result.</lang>

J

The verb ~. removes duplicate items from any array (numeric, character, or other; vector, matrix, rank-n array). For example: <lang j> ~. 4 3 2 8 0 1 9 5 1 7 6 3 9 9 4 2 1 5 3 2 4 3 2 8 0 1 9 5 7 6

  ~. 'chthonic eleemosynary paronomasiac'

chtoni elmsyarp</lang> Or

<lang j> 0 1 1 2 0 */0 1 2 0 0 0 0 1 2 0 1 2 0 2 4 0 0 0

  ~. 0 1 1 2 0 */0 1 2

0 0 0 0 1 2 0 2 4</lang>

Java

Works with: Java version 1.5

<lang java5>import java.util.Set; import java.util.HashSet; import java.util.Arrays;

Object[] data = {1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d"}; Set<Object> uniqueSet = new HashSet<Object>(Arrays.asList(data)); Object[] unique = uniqueSet.toArray();</lang>

JavaScript

This uses the === "strict equality" operator, which does no type conversions (4 == "4" is true but 4 === "4" is false) <lang javascript>function unique(ary) {

   // concat() with no args is a way to clone an array
   var u = ary.concat().sort();
   for (var i = 1; i < u.length; ) {
       if (u[i-1] === u[i])
           u.splice(i,1);
       else
           i++;
   }
   return u;

}

var ary = [1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d", "4"]; var uniq = unique(ary); for (var i = 0; i < uniq.length; i++)

   print(uniq[i] + "\t" + typeof(uniq[i]));</lang>

1 - number
2 - number
3 - number
4 - number
4 - string
a - string
b - string
c - string
d - string

Or, extend the prototype for Array: <lang javascript>Array.prototype.unique = function() {

   var u = this.concat().sort();
   for (var i = 1; i < u.length; ) {
       if (u[i-1] === u[i])
           u.splice(i,1);
       else
           i++;
   }
   return u;

} var uniq = [1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d"].unique();</lang>

Julia

<lang julia>a = [1,2,3,4,1,2,3,4] unique(a)</lang>

K

(Inspired by the J version.)

<lang K> a:4 5#20?13 / create a random 4 x 5 matrix (12 7 12 4 3

6 3 7 4 7
3 8 3 1 2
2 12 6 4 1)

  ,/a           / flatten to array

12 7 12 4 3 6 3 7 4 7 3 8 3 1 2 2 12 6 4 1

  ?,/a          / distinct elements

12 7 4 3 6 8 1 2

  ?"chthonic eleemosynary paronomasiac"

"chtoni elmsyarp"

  ?("this";"that";"was";"that";"was";"this")

("this"

"that"
"was")

  0 1 1 2 0 *\: 0 1 2

(0 0 0

  ?0 1 1 2 0 *\: 0 1 2

(0 0 0

0 1 2
0 2 4)</lang>

Lang5

<lang lang5>: dip swap '_ set execute _ ;

remove-duplicates

   [] swap do unique? length 0 == if break then loop drop ;

unique?

   0 extract swap "2dup in if drop else append then" dip ;

[1 2 6 3 6 4 5 6] remove-duplicates .</lang> Built-in function: <lang lang5>[1 2 6 3 6 4 5 6] 's distinct [1 2 6 3 6 4 5 6] 's dress dup union .</lang>

Liberty BASIC

LB has arrays, but here the elements are stored in a space-separated string. <lang lb> a$ =" 1 $23.19 2 elbow 3 2 Bork 4 3 elbow 2 $23.19 " print "Original set of elements = ["; a$; "]"

b$ =removeDuplicates$( a$) print "With duplicates removed = ["; b$; "]"

end

function removeDuplicates$( in$)

   o$ =" "
   i  =1
   do
       term$    =word$( in$, i, " ")
       if instr( o$, " "; term$; " ") =0 and term$ <>" " then o$ =o$ +term$ +" "
       i        =i +1
   loop until term$ =""
   removeDuplicates$ =o$

end function </lang>

Original set of elements = [ 1 $23.19 2 elbow 3 2 Bork 4 3 elbow 2 $23.19 ]
With duplicates removed  = [ 1 $23.19 2 elbow 3 Bork 4  ]

Logo

Works with: UCB Logo

<lang logo>show remdup [1 2 3 a b c 2 3 4 b c d] ; [1 a 2 3 4 b c d]</lang>

Lua

<lang Lua>items = {1,2,3,4,1,2,3,4,"bird","cat","dog","dog","bird"} flags = {} io.write('Unique items are:') for i=1,#items do

  if not flags[items[i]] then
     io.write(' ' .. items[i])
     flags[items[i]] = true
  end

end io.write('\n')</lang> Output:

Unique items are: 1 2 3 4 bird cat dog

MAXScript

<lang maxscript>uniques = #(1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d") for i in uniques.count to 1 by -1 do (

   id = findItem uniques uniques[i]
   if (id != i) do deleteItem uniques i

)</lang>

Maple

This is simplest with a list, which is an immutable array. <lang Maple>> L := [ 1, 2, 1, 2, 3, 3, 2, 1, "a", "b", "b", "a", "c", "b" ];

     L := [1, 2, 1, 2, 3, 3, 2, 1, "a", "b", "b", "a", "c", "b"]

> [op]({op}(L));

                       [1, 2, 3, "a", "b", "c"]</lang>

That is idiomatic, but perhaps a bit cryptic; here is a more verbose equivalent: <lang Maple>> convert( convert( L, 'set' ), 'list' );

                       [1, 2, 3, "a", "b", "c"]</lang>

For an Array, which is mutable, the table solution works well in Maple. <lang Maple>> A := Array( L ): > for u in A do T[u] := 1 end: Array( [indices]( T, 'nolist' ) );

                       [1, 2, 3, "c", "a", "b"]</lang>

Note that the output (due to the Array() constructor) is in fact an Array.

Mathematica

Built-in function: <lang Mathematica>DeleteDuplicates[{0, 2, 1, 4, 2, 0, 3, 1, 1, 1, 0, 3}]</lang> gives back: <lang Mathematica>{0, 2, 1, 4, 3}</lang> Custom function (reordering of elements is possible): <lang Mathematica>NoDupes[input_List] := Split[Sort[input]]All, 1 NoDupes[{0, 2, 1, 4, 2, 0, 3, 1, 1, 1, 0, 3}]</lang> gives back: <lang Mathematica>{0, 1, 2, 3, 4}</lang>

MATLAB

MATLAB has a built-in function, "unique(list)", which performs this task.
Sample Usage: <lang MATLAB>>> unique([1 2 6 3 6 4 5 6])

ans =

    1     2     3     4     5     6</lang>

NOTE: The unique function only works for vectors and not for true arrays.

Maxima

<lang maxima>unique([8, 9, 5, 2, 0, 7, 0, 0, 4, 2, 7, 3, 9, 6, 6, 2, 4, 7, 9, 8, 3, 8, 0, 3, 7, 0, 2, 7, 6, 0]); [0, 2, 3, 4, 5, 6, 7, 8, 9]</lang>

MUMPS

We'll take advantage of the fact that an array can only have one index of any specific value. Sorting into canonical order is a side effect. If the indices are strings containing the separator string, they'll be split apart.

<lang MUMPS>REMDUPE(L,S)

;L is the input listing
;S is the separator between entries
;R is the list to be returned
NEW Z,I,R
FOR I=1:1:$LENGTH(L,S) SET Z($PIECE(L,S,I))=""
;Repack for return
SET I="",R=""
FOR  SET I=$O(Z(I)) QUIT:I=""  SET R=$SELECT($L(R)=0:I,1:R_S_I)
KILL Z,I
QUIT R</lang>

Example:

USER>W $$REMDUPE^ROSETTA("1,2,3,4,5,2,5,""HELLO"",42,""WORLD""",",")
1,2,3,4,5,42,"HELLO","WORLD"

Nemerle

<lang Nemerle>using System.Linq; using System.Console;

module RemDups {

   Main() : void
   {
       def nums = [1, 4, 6, 3, 6, 2, 7, 2, 5, 2, 6, 8];
       def unique = $[n | n in nums.Distinct()];
       WriteLine(unique);
   }

}</lang>

NetRexx

This sample takes advantage of the NetRexx built-in Rexx object's indexed string capability (associative arrays). Rexx indexed strings act very like hash tables: <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

-- Note: Task requirement is to process "arrays". The following converts arrays into simple lists of words: -- Putting the resulting list back into an array is left as an exercise for the reader. a1 = [2, 3, 5, 7, 11, 13, 17, 19, 'cats', 222, -100.2, +11, 1.1, +7, '7.', 7, 5, 5, 3, 2, 0, 4.4, 2] a2 = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] a3 = ['Now', 'is', 'the', 'time', 'for', 'all', 'good', 'men', 'to', 'come', 'to', 'the', 'aid', 'of', 'the', 'party.'] x = 0 lists = x = x + 1; lists[0] = x; lists[x] = array2wordlist(a1) x = x + 1; lists[0] = x; lists[x] = array2wordlist(a2) x = x + 1; lists[0] = x; lists[x] = array2wordlist(a3)

loop ix = 1 to lists[0]

 nodups_list = remove_dups(lists[ix])
 say ix.right(4)':' lists[ix]
 say .right(4)':' nodups_list
 say
 end ix

return

-- ============================================================================= method remove_dups(list) public static

 newlist = 
 nodups = '0'
 loop w_ = 1 to list.words()
   ix = list.word(w_)
   nodups[ix] = nodups[ix] + 1 -- we can even collect a count of dups if we want
   end w_
 loop k_ over nodups
   newlist = newlist k_
   end k_

 return newlist.space

-- ============================================================================= method array2wordlist(ra = Rexx[]) public static

 wordlist = 
 loop r_ over ra
   wordlist = wordlist r_
   end r_

 return wordlist.space

</lang> Output:

   1: 2 3 5 7 11 13 17 19 cats 222 -100.2 11 1.1 7 7. 7 5 5 3 2 0 4.4 2
    : 13 2 3 17 19 7. 4.4 5 222 7 -100.2 1.1 cats 0 11

   2: 1 2 3 a b c 2 3 4 b c d
    : c 2 d 3 4 a b 1

   3: Now is the time for all good men to come to the aid of the party.
    : Now aid for men to the party. come time of is all good

NewLISP

<lang NewLISP>(unique '(1 2 3 a b c 2 3 4 b c d))</lang>

Nial

<lang nial>uniques := [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] cull uniques =+-+-+-+-+-+-+-+-+ =|1|2|3|a|b|c|4|d| =+-+-+-+-+-+-+-+-+</lang>

Using strand form <lang nial>cull 1 1 2 2 3 3 =1 2 3</lang>

Objective-C

<lang objc>NSArray *items = [NSArray arrayWithObjects:@"A", @"B", @"C", @"B", @"A", nil];

NSSet *unique = [NSSet setWithArray:items];</lang>

Objeck

<lang objeck> use Structure;

bundle Default {

 class Unique {
   function : Main(args : String[]) ~ Nil {
     nums := [1, 1, 2, 3, 4, 4];
     unique := IntVector->New();

     each(i : nums) {
       n := nums[i];
       if(unique->Has(n) = false) {
         unique->AddBack(n);
       };
     };

     each(i : unique) {
       unique->Get(i)->PrintLine();
     };
   }
 }

} </lang>

OCaml

<lang ocaml>let uniq lst =

 let unique_set = Hashtbl.create (List.length lst) in
   List.iter (fun x -> Hashtbl.replace unique_set x ()) lst;
   Hashtbl.fold (fun x () xs -> x :: xs) unique_set []

let _ =

 uniq [1;2;3;2;3;4]</lang>

Octave

<lang octave> input=[1 2 6 4 2 32 5 5 4 3 3 5 1 2 32 4 4]; output=unique(input); </lang>

ooRexx

<lang ooRexx> data = .array~of(1, 2, 3, "a", "b", "c", 2, 3, 4, "b", "c", "d") uniqueData = .set~new~union(data)~makearray~sort

say "Unique elements are" say do item over uniqueData

  say item

end </lang>

Oz

The following solutions only works if the value type is allowed as a key in a dictionary.

<lang oz>declare

 fun {Nub Xs}
    D = {Dictionary.new}
 in
    for X in Xs do D.X := unit end
    {Dictionary.keys D}
 end

in

 {Show {Nub [1 2 1 3 5 4 3 4 4]}}</lang>

PARI/GP

Sort and remove duplicates. Other methods should be implemented as well. <lang parigp>rd(v)={

 vecsort(v,,8)

};</lang>

Pascal

<lang pascal>Program RemoveDuplicates;

const

 iArray: array[1..7] of integer = (1, 2, 2, 3, 4, 5, 5);

var

 rArray: array[1..7] of integer;
 i, pos, last: integer;
 newNumber: boolean;

begin

 rArray[1] := iArray[1];
 last := 1;
 pos := 1;
 while pos < high(iArray) do
 begin
   inc(pos);
   newNumber := true;
   for i := low(rArray) to last do
     if iArray[pos] = rArray[i] then
     begin
       newNumber := false;

break;

     end;
   if newNumber then
   begin
     inc(last);
     rArray[last] := iArray[pos];
   end;
 end;
 for i := low(rArray) to last do
   writeln (rArray[i]);

end.</lang> Output:

% ./RemoveDuplicates
1
2
3
4
5

Perl

Library: List::MoreUtils MoreUtils

(this version even preserves the order of first appearance of each element) <lang perl>use List::MoreUtils qw(uniq);

my @uniq = uniq qw(1 2 3 a b c 2 3 4 b c d);</lang>

It is implemented like this: <lang perl>my %seen; my @uniq = grep {!$seen{$_}++} qw(1 2 3 a b c 2 3 4 b c d);</lang>

Note: the following two solutions convert elements to strings in the result, so if you give it references they will lose the ability to be dereferenced.

Alternately: <lang perl>my %hash = map { $_ => 1 } qw(1 2 3 a b c 2 3 4 b c d); my @uniq = keys %hash;</lang>

Alternately: <lang perl>my %seen; @seen{qw(1 2 3 a b c 2 3 4 b c d)} = (); my @uniq = keys %seen;</lang>

Perl 6

Works with: Rakudo version #23 "Lisbon"

<lang perl6>sub nub (@a) {

   my @b;
   none(@b) eqv $_ and push @b, $_ for @a;
   return @b;

}

my @unique = nub [1, 2, 3, 5, 2, 4, 3, -3, 7, 5, 6];</lang> Or just use the .uniq builtin.

<lang perl6>my @unique = [1, 2, 3, 5, 2, 4, 3, -3, 7, 5, 6].uniq;</lang>

PHP

<lang php>$list = array(1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'); $unique_list = array_unique($list);</lang>

PicoLisp

There is a built-in function <lang PicoLisp>(uniq (2 4 6 1 2 3 4 5 6 1 3 5))</lang> Output:

-> (2 4 6 1 3 5)

PL/I

  declare t(20) fixed initial (1, 5, 6, 2, 1, 7,
     5, 22, 4, 19, 1, 1, 6, 6, 6, 8, 9, 10, 11, 12);
  declare (i, j, k, n, e) fixed;

  n = hbound(t,1);
  i = 0;

outer:

  do k = 1 to n;
     e = t(k);
     do j = k-1 to 1 by -1;
         if e = t(j) then iterate outer;
     end;
     i = i + 1;
     t(i) = e;
  end;

  put skip list ('Unique elements are:');
  put edit ((t(k) do k = 1 to i)) (skip, f(11));

</lang>

Pop11

<lang pop11>;;; Initial array lvars ar = {1 2 3 2 3 4};

Create a hash table

lvars ht= newmapping([], 50, 0, true);

Put all array as keys into the hash table

lvars i; for i from 1 to length(ar) do

  1 -> ht(ar(i))

endfor;

Collect keys into a list

lvars ls = []; appdata(ht, procedure(x); cons(front(x), ls) -> ls; endprocedure);</lang>

PowerShell

The common array for both approaches: <lang powershell>$data = 1,2,3,1,2,3,4,1</lang> Using a hash table to remove duplicates: <lang powershell>$h = @{} foreach ($x in $data) {

   $h[$x] = 1

} $h.Keys</lang> Sorting and removing duplicates along the way can be done with the Sort-Object cmdlet. <lang powershell>$data | Sort-Object -Unique</lang> Removing duplicates without sorting can be done with the Select-Object cmdlet. <lang powershell>$data | Select-Object -Unique</lang>

PostScript

Library: initlib

[10 8 8 98 32 2 4 5 10 ] dup length dict begin  aload let* currentdict {pop} map end

</lang>

Prolog

<lang prolog>uniq(Data,Uniques) :- sort(Data,Uniques).</lang>

Example usage: <lang prolog>?- uniq([1, 2, 3, 2, 3, 4],Xs). Xs = [1, 2, 3, 4]</lang>

Because sort/2 is GNU prolog and not ISO here is an ISO compliant version: <lang prolog>member1(X,[H|_]) :- X==H,!. member1(X,[_|T]) :- member1_(X,T).

distinct([],[]). distinct([H|T],C) :- member1(H,T),!, distinct(T,C). distinct([H|T],[H|C]) :- distinct(T,C).</lang>

Example usage: <lang prolog>?- distinct([A, A, 1, 2, 3, 2, 3, 4],Xs). Xs = [A, 1, 2, 3, 4]</lang>

PureBasic

Task solved with the built in Hash Table which are called Maps in PureBasic <lang PureBasic>NewMap MyElements.s()

For i=0 To 9 ;Mark 10 items at random, causing high risk of duplication items.

 x=Random(9)
 t$="Number "+str(x)+" is marked"
 MyElements(str(x))=t$   ; Add element 'X' to the hash list or overwrite if already included.

 Debug MyElements()

Next</lang> Output may look like this, e.g. duplicated items are automatically removed as they have the same hash value.

Number 0 is marked
Number 2 is marked
Number 5 is marked
Number 6 is marked

Python

If all the elements are hashable (this excludes list, dict, set, and other mutable types), we can use a set: <lang python>items = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] unique = list(set(items))</lang>

If all the elements are comparable (i.e. <, >=, etc. operators; this works for list, dict, etc. but not for complex and many other types, including most user-defined types), we can sort and group: <lang python>import itertools items = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] unique = [k for k,g in itertools.groupby(sorted(items))]</lang>

If both of the above fails, we have to use the brute-force method, which is inefficient: <lang python>items = [1, 2, 3, 'a', 'b', 'c', 2, 3, 4, 'b', 'c', 'd'] unique = [] for x in items:

   if x not in unique:
       unique.append(x)</lang>

Qi

<lang qi> (define remove-duplicates

 []    -> []
 [A|R] -> (remove-duplicates R) where (element? A R)
 [A|R] -> [A|(remove-duplicates R)])

(remove-duplicates [a b a a b b c d e]) </lang>

R

<lang r>items <- c(1,2,3,2,4,3,2) unique (items)</lang>

Racket

Using the built-in function <lang Racket> -> (remove-duplicates '(2 1 3 2.0 a 4 5 b 4 3 a 7 1 3 x 2)) '(2 1 3 2.0 a 4 5 b 7 x) </lang>

Using a hash-table: <lang Racket> (define (unique/hash lst)

 (hash-keys (for/hash ([x (in-list lst)]) (values x #t))))

</lang>

Using a set: <lang Racket> (define unique/set (compose1 set->list list->set)) </lang>

A definition that works with arbitrary sequences and allows specification of an equality predicate.

<lang Racket> (define (unique seq #:same-test [same? equal?])

 (for/fold ([res '()])
           ([x seq] #:unless (memf (curry same? x) res))
   (cons x res)))

</lang>

-> (unique '(2 1 3 2.0 a 4 5 b 4 3 a 7 1 3 x 2))
'(1 2 3 a b x 4 5 7 2.0)
-> (unique '(2 1 3 2.0 4 5 4.0 3 7 1 3 2) #:same-test =)
'(7 5 4 3 1 2)
-> (unique #(2 1 3 2.0 4 5 4.0 3 7 1 3 2))
'(7 5 4 3 1 2)
-> (apply string (unique "absbabsbdbfbd"))
"fdsba"

REBOL

<lang REBOL>print mold unique [1 $23.19 2 elbow 3 2 Bork 4 3 elbow 2 $23.19]</lang>

Output:

[1 $23.19 2 elbow 3 Bork 4]

Raven

<lang raven>[ 1 2 3 'a' 'b' 'c' 2 3 4 'b' 'c' 'd' ] as items items copy unique print

list (8 items)

0 => 1
1 => 2
2 => 3
3 => "a"
4 => "b"
5 => "c"
6 => 4
7 => "d"</lang>

REXX

Note that in REXX, strings are quite literal.

+7 is different from 7 (but compares numerically equal).
00 is different from 0 (but compares numerically equal).
-0 is different from 0 (but compares numerically equal).
7. is different from 7 (but compares numerically equal).
Ab is different from AB (but can compare equal if made case insensitive).

Note that all three REXX examples below don't care what type of element is used, integer, floating point, character, binary, ...

version 1, using a modified method 3

Instead of discard an element if it's a duplicated, it just doesn't add it to the new list.
This method is faster than method 3. <lang rexx>/*REXX program to remove duplicate elements (items) in a list. */ $= '2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 7 5 5 3 2 0 4.4 2' say 'original list:' $ say right(words($),13) ' words in the original list.'; say

 do j=words($)  by -1  to 1; y=word($,j)  /*process words in the list, */
 _=wordpos(y, $, j+1);  if _\==0  then $=delword($, _, 1) /*del if dup.*/
 end   /*j*/

say 'modified list:' space($) say right(words($),13) ' words in the modified list.'

                                      /*stick a fork in it, we're done.*/</lang>

output

original list: 2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 7 5 5 3 2 0 4.4 2
           23  words in the original list.

modified list: 2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 0 4.4
           17  words in the modified list.

version 2, using method 3

<lang rexx>/*REXX program to remove duplicate elements (items) in a list. */ old= '2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 7 5 5 3 2 0 4.4 2' say 'original list:' old say right(words(old),13) ' words in the original list.'; say new= /*start with a clean slate. */

 do j=1 for words(old); _=word(old,j) /*process the words in old list. */
 if wordpos(_,new)==0  then new=new _ /*Doesn't exist? Then add to list*/
 end   /*j*/

say 'modified list:' space(new) say right(words(new),13) ' words in the modified list.'

                                      /*stick a fork in it, we're done.*/</lang>

output is identical to the 1^st version.

version 3, using method 1 (hash table) via REXX stems

<lang rexx>/* REXX ************************************************************

26.11.2012 Walter Pachl
added: show multiple occurrences
- - - - /

old='2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 7 5 5',

   '3 2 0 4.4 2'

say 'old list='old say 'words in the old list=' words(old) new= found.=0 count.=0 Do While old<>

 Parse Var old w old                                                   
 If found.w=0 Then Do                                                  
   new=new w                                                           
   found.w=1                                                           
   End                                                                 
 count.w=count.w+1                                                     
 End

say 'new list='strip(new) say 'words in the new list=' words(new) Say 'Multiple occurrences:' Say 'occ word' Do While new<>

 Parse Var new w new                                                   
 If count.w>1 Then                                                     
   Say right(count.w,3) w                                              
 End</lang>

Output:

old list=2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 7 5 5 3 2 0 4.4 2 
words in the old list= 23                                                    
new list=2 3 5 7 11 13 17 19 cats 222 -100.2 +11 1.1 +7 7. 0 4.4             
words in the new list= 17                                                    
Multiple occurrences:                                                        
occ word                                                                     
  3 2                                                                        
  2 3                                                                        
  3 5                                                                        
  2 7

Ruby

<lang ruby>ary = [1,1,2,1,'redundant',[1,2,3],[1,2,3],'redundant'] uniq_ary = ary.uniq

=> [1, 2, "redundant", [1, 2, 3]]</lang>

Scala

<lang scala>val list = List(1,2,3,4,2,3,4,99) val l2 = list.distinct // l2: scala.List[scala.Int] = List(1,2,3,4,99)

val arr = Array(1,2,3,4,2,3,4,99) val arr2 = arr.distinct // arr2: Array[Int] = Array(1, 2, 3, 4, 99) </lang>

Scheme

<lang scheme>(define (remove-duplicates l)

 (cond ((null? l)
        '())
       ((member (car l) (cdr l))
        (remove-duplicates (cdr l)))
       (else
        (cons (car l) (remove-duplicates (cdr l))))))

(remove-duplicates '(1 2 1 3 2 4 5))</lang>

Alternative approach: <lang scheme>(define (remove-duplicates l)

 (do ((a '() (if (member (car l) a) a (cons (car l) a)))
      (l l (cdr l)))
   ((null? l) (reverse a))))

(remove-duplicates '(1 2 1 3 2 4 5))</lang>

The function 'delete-duplicates' is also available in srfi-1.

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: main is func

 local
   const array integer: data is [] (1, 3, 2, 9, 1, 2, 3, 8, 8, 1, 0, 2);
   var set of integer: dataSet is (set of integer).value;
   var integer: number is 0;
 begin
   for number range data do
     incl(dataSet, number);
   end for;
   writeln(dataSet);
 end func;</lang>

Output:

{0, 1, 2, 3, 8, 9}

SETL

<lang SETL>items := [0,7,6,6,4,9,7,1,2,3,2]; print(unique(items));</lang> Output in arbitrary order (convert tuple->set then set->tuple): <lang SETL>proc unique(items);

 return [item: item in {item: item in items}];

end proc;</lang>

Preserving source order <lang SETL>proc unique(items);

 seen := {};
 return [item: item in items, nps in {#seen} | #(seen with:= item) > nps];

end proc;</lang>

Slate

<lang slate> [|:s| #(1 2 3 4 1 2 3 4) >> s] writingAs: Set.

"==> {"Set traitsWindow" 1. 2. 3. 4}"

</lang>

Smalltalk

<lang smalltalk>"Example of creating a collection" |a| a := #( 1 1 2 'hello' 'world' #symbol #another 2 'hello' #symbol ). a asSet.</lang> Output:

Set (1 2 #symbol 'world' #another 'hello' )

the above has the disadvantage of loosing the original order (because Sets are unordered, and the hashing shuffles elements into an arbitrary order). When tried, I got:

Set('world' 1 #another 'hello' #symbol 2)

on my system. This can be avoided by using an ordered set (which has also O(n) complexity) as below:

Works with: Smalltalk/X

<lang smalltalk>|a| a := #( 1 1 2 'hello' 'world' #symbol #another 2 'hello' #symbol ). a asOrderedSet.</lang> Output:

OrderedSet(1 2 'hello' 'world' #symbol #another)

Tcl

The concept of an "array" in Tcl is strictly associative - and since there cannot be duplicate keys, there cannot be a redundant element in an array. What is called "array" in many other languages is probably better represented by the "list" in Tcl (as in LISP). With the correct option, the lsort command will remove duplicates. <lang tcl>set result [lsort -unique $listname]</lang>

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT list_old="b'A'A'5'1'2'3'2'3'4" list_sort=MIXED_SORT (list_old) list_new=REDUCE (list_sort) PRINT list_old PRINT list_new </lang> Output (sorted)

b'A'A'5'1'2'3'2'3'4
1'2'3'4'5'A'b

or <lang tuscript> $$ MODE TUSCRIPT list_old="b'A'A'5'1'2'3'2'3'4" DICT list CREATE LOOP l=list_old DICT list LOOKUP l,num IF (num==0) DICT list ADD l ENDLOOP DICT list unload list list_new=JOIN (list) PRINT list_old PRINT list_new </lang> Output:

b'A'A'5'1'2'3'2'3'4
b'A'5'1'2'3'4

UnixPipes

Assuming a sequence is represented by lines in a file. <lang bash>bash$ # original list bash$ printf '6\n2\n3\n6\n4\n2\n' 6 2 3 6 4 2 bash$ # made uniq bash$ printf '6\n2\n3\n6\n4\n2\n'|sort -n|uniq 2 3 4 6 bash$</lang>

or

<lang bash>bash$ # made uniq bash$ printf '6\n2\n3\n6\n4\n2\n'|sort -nu 2 3 4 6 bash$</lang>

Ursala

The algorithm is to partition the list by equality and take one representative from each class, which can be done by letting the built in partition operator, |=, use its default comparison relation. This works on lists of any type including character strings but the comparison is based only on structural equivalence. It's up to the programmer to decide whether that's a relevant criterion for equivalence or else specify a better one. <lang Ursala>#cast %s

example = |=hS& 'mississippi'</lang> output:

'mspi'

       if S(I) = 0 then return I;

func Unique(S); \Remove duplicate bytes from string char S; int I, J, K, L; [L:= StrLen(S); \string length for I:= 0 to L-1 do \for all characters in string...

   for J:= I+1 to L-1 do               \scan rest of string for duplicates
       if S(I) = S(J) then             \if duplicate then
           [for K:= J+1 to L do        \ shift rest of string down (including
               S(K-1):= S(K);          \ terminating zero)
           L:= L-1                     \ string is now one character shorter
           ];

return S; \return pointer to string ];

Text(0, Unique("Pack my box with five dozen liquor jugs."))</lang>

Output:

Pack myboxwithfvedznlqurjgs.