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

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 algol>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));

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

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.

$ 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

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>

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.

module SelectFromArray

import StdEnv

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

array :: {Int}
array = {x \\ x <- [1 .. 10]}

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

Start :: {!Int}
Start = {x \\ x <-: array | isEven x}

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

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

Or using a list comprehension:

EvenNumbers = [X || X <- Numbers, X rem 2 == 0].

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

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

 result = array[where(NOT array AND 1)]

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 = new int[] {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

<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>

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>

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]

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

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

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

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:

 let lst = [1;2;3;4;5;6]
 let even_lst = List.filter (fun x -> x mod 2 = 0) lst

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:

 Lst = [1 2 3 4 5]
 LstEven = {Filter Lst fun{$ X} X mod 2 == 0 end}

Perl

Perl 6

Works with: Rakudo version #21 "Seattle"

Alternatively:

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>

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:

;;; 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) =>

PowerShell

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

Prolog

 evens(D, Es) :- findall(E, (member(E, D), E mod 2 =:= 0), Es).

Usage:

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

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

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

Raven

[ 0 1 2 3 4 5 6 7 8 9 ] as nums
group nums each
    dup 1 & if drop
list as evens

Ruby

<lang ruby>ary = [1,2,3,4,5,6] #or ary = (1..6).to_a even_ary = ary.select{|el|el%2==0}

=> [2, 4, 6]</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;

Scala

val a = Array(1, 2, 3, 4, 5)
val even = a.filter{x => x % 2 == 0}

Or

val even = for(val x <- a; x % 2 == 0) yield x

Slate

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

Smalltalk

#(1 2 3 4 5) select: [:number | number even]

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

--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

'

Works with: MySQL

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;

Or to be shorter, you could create the table evens directly from the query result :

create temporary table evens select * from nos where v%2=0;

Standard ML

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

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

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

UnixPipes

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

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.

import std
import nat

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

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 differs from writing (not remainder/36)*~ with the operator above because 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

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

[1 2 3 4 5 6 7 8 9] [even?] filter
=[2 4 6 8]

Wrapl

 VAR l <- ALL 1:to(10);

l will be the list [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

 VAR m <- ALL l:values \ $ % 2 = 0;

m will be the list [2, 4, 6, 8, 10]

XSLT

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

</lang>