Filter

From Rosetta Code
Task
Filter
You are encouraged to solve this task according to the task description, using any language you may know.

Select certain elements from an Array into a new Array in a generic way. To demonstrate, select all even numbers from an Array.

ActionScript

<lang actionscript>var arr:Array = new Array(1, 2, 3, 4, 5); var evens:Array = new Array(); for (var i:int = 0; i < arr.length(); i++) {

   if (arr[i] % 2 == 0)
       evens.push(arr[i]);

}</lang>

Ada

<lang ada>with Ada.Integer_Text_Io; use Ada.Integer_Text_Io; with Ada.Text_Io; use Ada.Text_Io;

procedure Array_Selection is

  type Array_Type is array (Positive range <>) of Integer;
  Null_Array : Array_Type(1..0);
  
  function Evens (Item : Array_Type) return Array_Type is
  begin
     if Item'Length > 0 then
        if Item(Item'First) mod 2 = 0 then
           return Item(Item'First) & Evens(Item((Item'First + 1)..Item'Last));
        else
           return Evens(Item((Item'First + 1)..Item'Last));
        end if;
     else
        return Null_Array;
     end if;
  end Evens;
  
  procedure Print(Item : Array_Type) is
  begin
     for I in Item'range loop
        Put(Item(I));
        New_Line;
     end loop;
  end Print;
   
  Foo : Array_Type := (1,2,3,4,5,6,7,8,9,10);

begin

  Print(Evens(Foo));

end Array_Selection;</lang> Here is a non-recursive solution: <lang ada>with Ada.Text_IO; use Ada.Text_IO;

procedure Array_Selection is

  type Array_Type is array (Positive range <>) of Integer;
 
  function Evens (Item : Array_Type) return Array_Type is
     Result : Array_Type (1..Item'Length);
     Index  : Positive := 1;
  begin
     for I in Item'Range loop
        if Item (I) mod 2 = 0 then
           Result (Index) := Item (I);
           Index := Index + 1;
        end if;
     end loop;
     return Result (1..Index - 1);
  end Evens;
 
  procedure Put (Item : Array_Type) is
  begin
     for I in Item'range loop
        Put (Integer'Image (Item (I)));
     end loop;
  end Put;

begin

  Put (Evens ((1,2,3,4,5,6,7,8,9,10)));
  New_Line;

end Array_Selection;</lang>

ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386

<lang algol68>MODE TYPE = INT;

PROC select = ([]TYPE from, PROC(TYPE)BOOL where)[]TYPE: BEGIN

 FLEX[0]TYPE result;
 FOR key FROM LWB from TO UPB from DO
   IF where(from[key]) THEN
     [UPB result+1]TYPE new result;
     new result[:UPB result] := result;
     new result[UPB new result] := from[key];
     result := new result
   FI
 OD;
 result

END;

[]TYPE from values = (1,2,3,4,5,6,7,8,9,10); PROC where even = (TYPE value)BOOL: NOT ODD value;

print((select(from values, where even), new line));

  1. Or as a simple one line query #

print((select((1,4,9,16,25,36,49,64,81,100), (TYPE x)BOOL: NOT ODD x ), new line))</lang> Output:

         +2         +4         +6         +8        +10
         +4        +16        +36        +64       +100

AmigaE

<lang amigae>PROC main()

 DEF l : PTR TO LONG, r : PTR TO LONG, x
 l := [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
 r := List(ListLen(l))
 SelectList({x}, l, r, `Mod(x,2)=0)
 ForAll({x}, r, `WriteF('\d\n', x))

ENDPROC</lang>

AppleScript

<lang applescript>set array to {1, 2, 3, 4, 5, 6} set evens to {} repeat with i in array if (i mod 2 = 0) then set end of evens to i's contents end repeat

return evens</lang>Result is (a list):

{2, 4, 6}

AutoHotkey

<lang autohotkey>array = 1,2,3,4,5,6,7 loop, parse, array, `, { if IsEven(A_LoopField) evens = %evens%,%A_LoopField% } stringtrimleft, evens, evens, 1 msgbox % evens return

IsEven(number) { return !mod(number, 2) }</lang>

AWK

In this example, an array is filled with the numbers 1..9. In a loop, even elements are collected into the string r. Note that sequence is not necessarily maintained. <lang awk>$ awk 'BEGIN{split("1 2 3 4 5 6 7 8 9",a);for(i in a)if(!(a[i]%2))r=r" "a[i];print r}' 4 6 8 2</lang>

C

<lang c>int arr[5] = {1,2,3,4,5}; int *result; int memoryReqd = 0,length = sizeof(arr)/sizeof(arr[0]), i,j; for (i=0; i<length; i++) {

       if(0 == arr[i]%2)
       {
               memoryReqd++;
       }

} result = malloc(memoryReqd*sizeof(int)); for(i=0,j=0; i<length; i++) {

       if(0 == arr[i]%2)
       {
               result[j++]=arr[i];
       }

}</lang>

Test our resultant array: <lang c>for(i=0; i<memoryReqd; i++) {

       printf("%d\n",result[i]);

}</lang>

C++

<lang cpp>#include <vector>

  1. include <algorithm>
  2. include <functional>
  3. include <iterator>
  4. include <iostream>

int main() {

 std::vector<int> ary;
 for (int i = 0; i < 10; i++)
   ary.push_back(i);
 std::vector<int> evens;
 std::remove_copy_if(ary.begin(), ary.end(), back_inserter(evens),
                     std::bind2nd(std::modulus<int>(), 2)); // filter copy
 std::copy(evens.begin(), evens.end(),
           std::ostream_iterator<int>(std::cout, "\n"));
 return 0;

}</lang>

C#

Works with: .NET version 1.1

<lang csharp>ArrayList array = new ArrayList( new int[] { 1, 2, 3, 4, 5 } ); ArrayList evens = new ArrayList(); foreach( int i in array ) {

       if( (i%2) == 0 )
               evens.Add( i );

} foreach( int i in evens )

      System.Console.WriteLine( i.ToString() );</lang>
Works with: .NET version 2.0

<lang csharp>List<int> array = new List<int>( new int[] { 1, 2, 3, 4, 5 } ); List<int> evens = array.FindAll( delegate( int i ) { return (i%2)==0; } ); foreach( int i in evens )

      System.Console.WriteLine( i.ToString() );</lang>
Works with: .NET version 3.5

<lang csharp>IEnumerable<int> array = new List<int>( new int[] { 1, 2, 3, 4, 5 } ); IEnumerable<int> evens = array.Where( delegate( int i ) { return (i%2)==0; } ); foreach( int i in evens )

      System.Console.WriteLine( i.ToString() );</lang>

Replacing the delegate with the more concise lambda expression syntax. <lang csharp>int[] array = { 1, 2, 3, 4, 5 }; int[] evens = array.Where(i => (i % 2) == 0).ToArray();

foreach (int i in evens)

   Console.WriteLine(i);</lang>

Clean

The standard environment is required for list and array comprehensions. We specify the types of the functions because array comprehensions are overloaded. Clean provides lazy, strict, and unboxed arrays.

<lang clean>module SelectFromArray

import StdEnv</lang>

Create a lazy array where each element comes from the list 1 to 10.

<lang clean>array :: {Int} array = {x \\ x <- [1 .. 10]}</lang>

Create (and print) a strict array where each element (coming from another array) is even.

<lang clean>Start :: {!Int} Start = {x \\ x <-: array | isEven x}</lang>

Clojure

<lang lisp>;; range and filter create lazy seq's (filter even? (range 0 100))

vec will convert any type of seq to an array

(vec (filter even? (vec (range 0 100))))</lang>

Common Lisp

Common Lisp has many ways of accomplishing this task. Most of them involve higher-order sequence functions that take a predicate as the first argument and a list as the second argument. A predicate is a function that returns a boolean. The higher-order functions call the predicate for each element in list, testing the element.

In this example, the goal is to find the even numbers. The most straight-forward function is to use remove-if-not, which removes elements from the list that does not pass the predicate. The predicate, in this case, tests to see if an element is even. Therefore, the remove-if-not acts like a filter:

<lang lisp>(remove-if-not #'evenp '(1 2 3 4 5 6 7 8 9 10)) > (2 4 6 8 10)</lang>

However, this function is non-destructive, meaning the function creates a brand new list. This might be too prohibitive for very large lists. There is a destructive version that modifies the list in-place:

<lang lisp>(delete-if-not #'evenp '(1 2 3 4 5 6 7 8 9 10)) > (2 4 6 8 10)</lang>

D

Library: Tango

<lang d>import tango.core.Array; import tango.io.Stdout;

void main() {

   auto array = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];
   // removeIf places even elements at the beginnig of the array and returns number of found evens
   auto evens = array.removeIf( ( typeof(array[0]) i ) { return (i % 2) == 1; } );
   Stdout("Evens - ")( array[0 .. evens] ).newline; // The order of even elements is preserved
   Stdout("Odds - ")( array[evens .. $].sort ).newline; // Unlike odd elements

}</lang> Output:

Evens - [ 2, 4, 6, 8, 10 ]
Odds - [ 1, 3, 5, 7, 9 ]

E

There are several ways this could be done.

<lang e>pragma.enable("accumulator") accum [] for x ? (x %% 2 <=> 0) in [1,2,3,4,5,6,7,8,9,10] { _.with(x) }</lang>

<lang e>var result := [] for x ? (x %% 2 <=> 0) in [1,2,3,4,5,6,7,8,9,10] {

   result with= x

} result</lang>

<lang e>def makeSeries := <elang:control.makeSeries> makeSeries([1,2,3,4,5,6,7,8,9,10]).filter(fn x,_{x %% 2 <=> 0}).asList()</lang>

Erlang

<lang erlang>Numbers = lists:seq(1, 5). EvenNumbers = lists:filter(fun (X) -> X rem 2 == 0 end, Numbers).</lang>

Or using a list comprehension:

<lang erlang>EvenNumbers = [X || X <- Numbers, X rem 2 == 0].</lang>

F#

<lang fsharp>let evenList = List.filter (fun i -> i%2 = 0 ) let evenList2 = fun list -> [for i in list do if i%2=0 then yield i] let evenArray = Array.filter (fun i -> i%2 = 0 ) let evenSeq = fun list -> seq { for i in list do if i%2=0 then yield i}</lang>

Factor

<lang factor>10 iota >array [ even? ] filter .</lang>

Forth

<lang forth>: sel ( dest 0 test src len -- dest len )

 cells over + swap do   ( dest len test )
   i @ over execute if
     i @ 2over cells + !
     >r 1+ r>
   then
 cell +loop drop ;

create nums 1 , 2 , 3 , 4 , 5 , 6 , create evens 6 cells allot

.array 0 ?do dup i cells + @ . loop drop ;
even? ( n -- ? ) 1 and 0= ;

evens 0 ' even? nums 6 sel .array \ 2 4 6</lang>


Fortran

<lang fortran>module funcs

 implicit none

contains

 pure function iseven(x)
   logical :: iseven
   integer, intent(in) :: x
   iseven = mod(x, 2) == 0
 end function iseven

end module funcs</lang>

<lang fortran>program Filter

 use funcs
 implicit none
 integer, parameter                 :: N = 100
 integer, dimension(N)              :: array
 integer, dimension(:), pointer     :: filtered
 integer :: i
 forall(i=1:N) array(i) = i
 filtered => filterwith(array, iseven)
 print *, filtered

contains

 function filterwith(ar, testfunc)
   integer, dimension(:), pointer        :: filterwith
   integer, dimension(:), intent(in)     :: ar
   interface
      elemental function testfunc(x)
        logical :: testfunc
        integer, intent(in) :: x
      end function testfunc
   end interface
   integer :: i, j, n
   n = count( testfunc(ar) )
   allocate( filterwith(n) )
   j = 1
   do i = lbound(ar, dim=1), ubound(ar, dim=1)
      if ( testfunc(ar(i)) ) then
         filterwith(j) = ar(i)
         j = j + 1
      end if
   end do
 end function filterwith

end program Filter</lang>

Groovy

<lang groovy> def evens = [1, 2, 3, 4, 5].findAll{it % 2 == 0}</lang>

Haskell

In Haskell, a list is often more basic than an array:

<lang haskell>ary = [1..10] evens = [x | x <- ary, even x]</lang> or <lang haskell>evens = filter even ary</lang>

To do the same operation on an array, the simplest way it to convert it lazily into a list:

<lang haskell>import Data.Array

ary = listArray (1,10) [1..10] evens = listArray (1,n) l where

 n = length l
 l = [x | x <- elems ary, even x]</lang>

Note that the bounds must be known before creating the array, so the temporary list will be fully evaluated before the array is created.

IDL

The where() function can select elements on any logical expression. For example

<lang idl>result = array[where(NOT array AND 1)]</lang>

J

Solution:
With any verb (function) f that returns a boolean for each element of a vector v, the following is the generic solution: <lang j> (#~ f) v</lang>

Examples: <lang j> ] v=: 20 ?@$ 100 NB. vector of 20 random integers between 0 and 99 63 92 51 92 39 15 43 89 36 69 40 16 23 2 29 91 57 43 55 22

  v #~ -.2| v

92 92 36 40 16 2 22</lang>

Or using the generic form suggested above: <lang j> isEven=: 0 = 2&| NB. verb testing for even numbers

  (#~ isEven) v

92 92 36 40 16 2 22</lang>

We might decide that we use this pattern so often that it is worthwhile creating a new adverb select that filters an array using the verb to its left. <lang j> select=: adverb def '(#~ u)'

  isPrime=: 1&p:
  isEven select v

92 92 36 40 16 2 22

  isPrime select v

43 89 23 2 29 43

  (isEven *. isPrime) select v

2</lang>

Java

<lang java>int[] array = {1, 2, 3, 4, 5 }; List<Integer> evensList = new ArrayList<Integer>(); for (int i: array) {

   if (i % 2 == 0) evensList.add(i);

} int[] evens = evensList.toArray(new int[0]);</lang>

JavaScript

The standard way is to use the Array.prototype.filter function (

Works with: JavaScript version 1.6

):

<lang javascript>var arr = [1,2,3,4,5]; var evens = arr.filter(function(a) {return a % 2 == 0});</lang>

Other ways: <lang javascript>var arr = [1,2,3,4,5]; var evens = []; for (var i=0, ilen=arr.length; i<ilen; i++)

     if (arr[i] % 2 == 0)
             evens.push(arr[i]);</lang>
Works with: Firefox version 2.0

<lang javascript>var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; var evens = [i for (i in numbers) if (i % 2 == 0)];

function range(limit) {

 for(var i = 0; i < limit; i++) {
   yield i;
 }

}

var evens2 = [i for (i in range(100)) if (i % 2 == 0)];</lang>

Library: Functional

<lang javascript>Functional.select("+1&1", [1,2,3,4]) // [2, 4]</lang>

Liberty BASIC

<lang lb>' write random nos between 1 and 100 ' to array1 counting matches as we go dim array1(100) count=100 for i = 1 to 100

   array1(i) = int(rnd(0)*100)+1
   count=count-(array1(i) mod 2)

next

'dim the extract and fill it dim array2(count) for i = 1 to 100

   if not(array1(i) mod 2) then
       n=n+1
       array2(n)=array1(i)
   end if

next

for n=1 to count

   print array2(n)

next</lang>

Lisaac

<lang Lisaac>+ a, b : ARRAY[INTEGER]; a := ARRAY[INTEGER].create_with_capacity 10 lower 0; b := ARRAY[INTEGER].create_with_capacity 10 lower 0; 1.to 10 do { i : INTEGER;

 a.add_last i;

}; a.foreach { item : INTEGER;

 (item % 2 = 0).if {
   b.add_last item;
 };

};</lang>

<lang logo>to even? :n

 output equal? 0 modulo :n 2

end show filter "even? [1 2 3 4]  ; [2 4]

show filter [equal? 0 modulo ? 2] [1 2 3 4]</lang>

Lua

<lang lua>function filter(t, func)

 local ret = {}
 for i, v in ipairs(t) do ret[#ret+1] = func(v) and v or nil end
 return ret

end

function even(a) return a % 2 == 0 end

print(unpack(filter({1,2,3,4,5,6,7,8,9,10},even)))</lang>

Mathematica

Check for even integers: <lang Mathematica>Select[{4, 5, Pi, 2, 1.3, 7, 6, 8.0}, EvenQ]</lang> gives: <lang Mathematica>{4, 2, 6}</lang> To check also for approximate number (like 8.0 in the example above) a possible solution is: <lang Mathematica>Select[{4, 5, Pi, 2, 1.3, 7, 6, 8.0}, Mod[#, 2] == 0 &]</lang> gives: <lang Mathematica>{4, 2, 6, 8.}</lang> notice that the function returns 8. not 8 (the dot indicates that it is a float number, not an integer).


MAXScript

<lang maxscript>arr = #(1, 2, 3, 4, 5, 6, 7, 8, 9) newArr = for i in arr where (mod i 2 == 0) collect i</lang>

Nial

<lang nial>filter (= [0 first, mod [first, 2 first] ] ) 0 1 2 3 4 5 6 7 8 9 10 =0 2 4 6 8 10</lang>

Objective-C

Works with: Mac OS X version 10.5

<lang objc>NSArray *numbers = [NSArray arrayWithObjects:[NSNumber numberWithInt:1],

                                            [NSNumber numberWithInt:2],
                                            [NSNumber numberWithInt:3],
                                            [NSNumber numberWithInt:4],
                                            [NSNumber numberWithInt:5], nil];

NSPredicate *isEven = [NSPredicate predicateWithFormat:@"modulus:by:(SELF, 2) == 0"]; NSArray *evens = [numbers filteredArrayUsingPredicate:isEven];</lang>

Works with: GNUstep

<lang objc>#import <Foundation/Foundation.h>

@interface NSNumber ( ExtFunc ) -(int) modulo2; @end

@implementation NSNumber ( ExtFunc ) -(int) modulo2 {

 return [self intValue] % 2;

} @end

int main() {

 NSAutoreleasePool *pool = [[NSAutoreleasePool alloc] init];
 NSArray *numbers = [NSArray arrayWithObjects:[NSNumber numberWithInt:1],
                                              [NSNumber numberWithInt:2],
                                              [NSNumber numberWithInt:3],
                                              [NSNumber numberWithInt:4],
                                              [NSNumber numberWithInt:5], nil];
 NSPredicate *isEven = [NSPredicate predicateWithFormat:@"modulo2 == 0"];
 NSArray *evens = [numbers filteredArrayUsingPredicate:isEven];
 
 NSLog(@"%@", evens);


 [pool release];
 return 0;

}</lang>

OCaml

It is easier to do it with a list: <lang ocaml>let lst = [1;2;3;4;5;6] let even_lst = List.filter (fun x -> x mod 2 = 0) lst</lang>

Octave

<lang octave>arr = [1:100]; evennums = arr( mod(arr, 2) == 0 ); disp(evennums);</lang>

Oz

It is easier to do it with a list: <lang oz>declare Lst = [1 2 3 4 5] LstEven = {Filter Lst IsEven}</lang>

Pascal

Arrays are supported in all versions of pascal so this simple example will cover the entire gamut.

Works with: Delphi


Works with: Turbo Pascal

<lang pascal>const

numbers:array[0..9] of integer = (0,1,2,3,4,5,6,7,8,9);

for x = 1 to 10 do

    if odd(numbers[x]) then
        writeln( 'The number ',numbers[x],' is odd.');
    else
        writeln( 'The number ',numbers[x],' is even.');</lang>

The odd() function is a standard library function of pascal as is the function even().

Perl

<lang perl>my @a = (1, 2, 3, 4, 5, 6); my @even = grep { $_%2 == 0 } @a;</lang>

Perl 6

Works with: Rakudo version #24 "Seoul"

<lang perl6>my @a = 1, 2, 3, 4, 5, 6; my @even = grep * !% 2, @a;</lang>

Alternatively:

<lang perl6>my @even = @a.grep(* !% 2);</lang>

PHP

Using a standard loop <lang php>$arr = range(1,5); $evens = array(); foreach ($arr as $val){

     if ($val % 2 == 0) $evens[] = $val);

} print_r($evens);</lang>

Using a filter function <lang php>function is_even($var) { return(!($var & 1)); } $arr = range(1,5); $evens = array_filter($arr, "is_even"); print_r($evens);</lang>

PicoLisp

<lang PicoLisp>(filter '((N) (not (bit? 1 N)))

  (1 2 3 4 5 6 7 8 9) )</lang>

Output:

-> (2 4 6 8)

Pop11

Most natural solution in Pop11 would probably use list. Below we accumulate filtered elements on the stack and then allocate array for the result:

<lang pop11>;;; Generic filtering procedure which selects from ar elements

satisfying pred

define filter_array(ar, pred); lvars i, k;

   stacklength() -> k;
   for i from 1 to length(ar) do
      ;;; if element satisfies pred we leave it on the stack
      if pred(ar(i)) then ar(i) endif;
   endfor;
   ;;; Collect elements from the stack into a vector
   return (consvector(stacklength() - k));

enddefine;

Use it

filter_array({1, 2, 3, 4, 5},

            procedure(x); not(testbit(x, 0)); endprocedure) =></lang>

PowerShell

<lang powershell>$array = -15..37 $array | Where-Object { $_ % 2 -eq 0 }</lang>

Prolog

<lang prolog>evens(D, Es) :- findall(E, (member(E, D), E mod 2 =:= 0), Es).</lang>

Usage:

<lang prolog>?- evens([1,2,3,4,5,6,7,8,9,10],E). E = [2, 4, 6, 8, 10]</lang>

PureBasic

<lang PureBasic>Dim Tal.i(9) Dim Evens.i(0)

- Set up an array with random numbers

For i=0 To ArraySize(Tal())

 Tal(i)=Random(100)

Next

- Pick out all Even and save them

j=0 For i=0 To ArraySize(Tal())

 If Tal(i)%2=0
   ReDim Evens(j)    ; extend the Array as we find new Even's
   Evens(j)=tal(i)
   j+1
 EndIf

Next

- Display the result

PrintN("List of Randoms") For i=0 To ArraySize(Tal())

 Print(Str(Tal(i))+" ")

Next PrintN(#CRLF$+#CRLF$+"List of Even(s)") For i=0 To ArraySize(Evens())

 Print(Str(Evens(i))+" ")

Next</lang>

Output can look like

List of Randoms
32 35 89 91 11 33 12 22 42 43

List of Even(s)
32 12 22 42

Python

Works with: Python version 2.4

<lang python>values = range(10) evens = [x for x in values if not x & 1] ievens = (x for x in values if not x & 1) # lazy

  1. alternately but less idiomatic:

evens = filter(lambda x: not x & 1, values)</lang>

Alternative using the slice syntax with its optional "stride" expression:

<lang python>values = range(10) evens = values[::2]</lang>

This works for all versions of Python (at least as far back as 1.5). Lists (arrays) can be "sliced" by indexing them with a range (lower and upper bounds). Thus mylist[1:9] evaluates into a list from the second item (excluding the first item which is mylist[0], of course) up to but not including the ninth item. In Python the expression mylist[:] is synonymous with mylist[0:len(mylist)] ... returning a copy of the complete list. also mylist[:x] returns the first x items from the list and negative numbers can be used such that mylist[-x:] returns the last x items from the list. The relatively obscure and optional stride expression can skip items and/or force the evaluation from the end of the list downward towards it's lower elements. Thus mylist[::-1] returns a reversed copy of the list, mylist[::2] returns all even elements, mylist[1::2] returns all odd elements, and so on.

Since strings in Python can be treated as a sort of immutable list of characters then the slicing and extended slicing can also be used with them as well. Thus mystring[::-2] will return every other character from the reverse order of the string.

One can also assign to a slice (of a list or other mutable indexed object. Thus the following:

<lang python>values = range(10) values[::2] = [11,13,15,17,19] print values 11, 1, 13, 3, 15, 5, 17, 7, 19, 9</lang>

R

<lang R>a <- 1:100 evennums <- a[ a%%2 == 0 ] print(evennums)</lang>

Raven

<lang raven>[ 0 1 2 3 4 5 6 7 8 9 ] as nums group nums each

   dup 1 & if drop

list as evens</lang>

REBOL

<lang REBOL>a: [] repeat i 100 [append a i] ; Build and load array.

evens: [] repeat element a [if even? element [append evens element]]

print mold evens</lang>

Output:

[2 4 6 8 10 12 14 16 18 20 22 24 
26 28 30 32 34 36 38 40 42 44 46 48 50 
52 54 56 58 60 62 64 66 68 70 72 74 76 
78 80 82 84 86 88 90 92 94 96 98 100]

Ruby

<lang ruby>ary = [1,2,3,4,5,6] #or ary = (1..6).to_a even_ary = ary.select {|elem| elem.even?}

  1. => [2, 4, 6]</lang>

Sather

<lang sather>class MARRAY{T} < $ARR{T} is

 include ARRAY{T};
 filter_by(r:ROUT{T}:BOOL):SAME is
   o:MARRAY{T} := #;
   loop e ::= elt!;
     if r.call(e) then
       o := o.append(#MARRAY{T}(|e|));
     end;
   end;
   return o;
 end;

end;

class MAIN is

 main is
   a ::= #MARRAY{INT}(|5, 6, 7, 8, 9, 10, 11|);
   sel ::= a.filter_by( bind(_.is_even) );
   loop #OUT + sel.elt! + " "; end;
   #OUT + "\n";
 end;

end;</lang>

Seed7

<lang seed7>var array integer: arr is [] (1, 2, 3, 4, 5); var array integer: evens is 0 times 0; var integer: number is 0;

for number range arr do

 if not odd(number) then
   evens &:= [] (number);
 end if;

end for;</lang>

Scala

<lang scala>(1 to 100).filter(_ % 2 == 0)</lang>

Scheme

In the interactive prompt: <lang scheme>> (filter even? '(1 2 3 4 5 6 7 8 9 10)) (2 4 6 8 10)</lang> Or as a function: <lang scheme>(define (select-even lst)

 (filter even? lst))

(select-even '(1 2 3 4 5 6 7 8 9 10))</lang>


Slate

<lang slate>#(1 2 3 4 5) select: [| :number | number isEven].</lang>

Smalltalk

<lang smalltalk>#(1 2 3 4 5) select: [:number | number even]</lang>

SQL

Task: Select certain elements from an Array into a new Array in a generic way. To demonstrate, select all even numbers from an Array.

Works with: MS SQL

<lang sql>--Create the original array (table #nos) with numbers from 1 to 10 create table #nos (v int) declare @n int set @n=1 while @n<=10 begin insert into #nos values (@n) set @n=@n+1 end

--Select the subset that are even into the new array (table #evens) select v into #evens from #nos where v % 2 = 0

-- Show #evens select * from #evens

-- Clean up so you can edit and repeat: drop table #nos drop table #evens</lang>

'

Works with: MySQL

<lang sql>create temporary table nos (v int); insert into nos values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10); create temporary table evens (v int); insert into evens select v from nos where v%2=0; select * from evens order by v; /*2,4,6,8,10*/ drop table nos; drop table evens;</lang>

Or to be shorter, you could create the table evens directly from the query result : <lang sql>create temporary table evens select * from nos where v%2=0;</lang>

Standard ML

<lang sml>val ary = [1,2,3,4,5,6]; List.filter (fn x => x mod 2 = 0) ary</lang>

Tcl

Tcl doesn't really have a concept of a "number" per se - strictly speaking its only data type is the string (but a string can be interpreted as a number, of course). The generic way of getting certain elements from an array looks roughly like this:

<lang tcl>foreach key [array names arr] {if { <condition> } then {puts $arr($key)}}</lang>

In this case, we can do this particular challenge with: <lang tcl>foreach {key val} [array get srcArray] {

   if {[string is integer -strict $key] && !($key%2)} {
       set dstArray($key) $val
   }

}</lang>

If we were using Tcl's lists and interpreting the challenge to mean getting just the elements at index 0, 2, 4, ... <lang tcl>foreach {even odd} $srcList {

   lappend dstList $even

}</lang>

Toka

<lang toka>10 cells is-array table 10 cells is-array even {

 variable source
 [ swap source ! >r reset r> 0
   [ i source @ array.get
     dup 2 mod 0 <> [ drop ] ifTrue 
   ] countedLoop
   depth 0 swap [ i even array.put ] countedLoop
 ]

} is copy-even 10 0 [ i i table array.put ] countedLoop table 10 copy-even</lang>

UnixPipes

<lang bash>yes \ | cat -n | while read a; do ; expr $a % 2 >/dev/null && echo $a ; done</lang>

Ursala

Ursala doesn't have arrays, except when the run time system transparently converts a list to an array as needed for an external math library function call. However, selection can be done on lists.

Unary predicates

The most common way to select items from a list according to a unary predicate p is to write p*~, as shown below.

<lang Ursala>#import std

  1. import nat

x = <89,36,13,15,41,39,21,3,15,92,16,59,52,88,33,65,54,88,93,43>

  1. cast %nL

y = (not remainder\2)*~ x</lang>

The output will be

<36,92,16,52,88,54,88>

Binary predicates

Selection is so frequently useful that the language has a couple of other ways to do it. Selecting according to a binary predicate can be done like this. <lang Ursala>z = (not remainder)~| (36,<1,2,3,4,5,6,7,8,9,10,11,12>)</lang> The value of z will be the divisors of 36 appearing in the list.

<1,2,3,4,6,9,12>

This usage has the advantage over writing (not remainder/36)*~ with the operator above that it allows the 36 to be part of the argument rather than being hard coded into the function.

Operator suffixes

Many operators in Ursala allow suffixes that modify their semantics. For example, the suffix ihB on the identity function ~& makes it ~&ihB, a predicate to detect odd numbers by inspecting the binary representation. If an operator with this kind of suffix is further modified by appending an F, it becomes a selection filter. For example <lang Ursala>shortcut = ~&ihBF x</lang> using the x defined above will evaluate to

<89,13,15,41,39,21,3,15,59,33,65,93,43>

There are also suffixes corresponding to the ~| operator.

V

<lang v>[even? dup 2 / >int 2 * - zero?].

[1 2 3 4 5 6 7 8 9] [even?] filter =[2 4 6 8]</lang>

Wrapl

<lang wrapl>VAR l <- ALL 1:to(10);</lang> l will be the list [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] <lang wrapl>VAR m <- ALL l:values \ $ % 2 = 0;</lang> m will be the list [2, 4, 6, 8, 10]

XSLT

<lang xml><xsl:for-each select="nodes[@value mod 2 = 0]">

 <xsl:value-of select="@value" />

</xsl:for-each></lang>

XQuery

<lang xquery> (: Sequence of numbers from 1 to 10 :) let $array := (1 to 10)

(: Short version :) let $short := $array[. mod 2 = 0]

(: Long version with a FLWOR expression :) let $long := for $value in $array

            where $value mod 2 = 0
            return $value

(: Show the results :) return

 <result>
   <short>{$short}</short>
   <long>{$short}</long>
 </result>

</lang>

The output of this query is: <lang xml> <?xml version="1.0" encoding="UTF-8"?> <result>

  <short>2 4 6 8 10</short>
  <long>2 4 6 8 10</long>

</result> </lang>