Filter

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

AppleScript

set array to {1, 2, 3, 4, 5}
set evens to {}
repeat with i in array
    -- very important -- list index starts at 1 not 0
    if (i mod 2 = 0) then set evens to evens & i
end repeat

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 = (int*)malloc(memoryReqd*sizeof(int));
       for(i=0,j=0; i<length; i++)
       {
               if(0 == arr[i]%2)
               {
                       result[j++]=arr[i];
               }
       }

Test our resultant array:

       for(i=0; i<memoryReqd; i++)
       {
               printf("%d\n",result[i]);
       }

C++

Works with: Visual C++ version 2005

Library: STL

Library: Boost.Lambda

 using namespace std;
 using namespace boost::lambda;
 vector<int> ary(10);
 int i = 0;
 for_each(ary.begin(), ary.end(), _1 = ++var(i)); // init array
 vector<int> evens;
 remove_copy_if(ary.begin(), ary.end(), back_inserter(evens), _1 % 2); // filter copy

C#

       // .NET 1.x solution
       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() );

       // .NET 2.0 solution
       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() );

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

;; 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))))

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:

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

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:

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

D

Library: Tango

<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

}</d> Output:

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

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

def evens = [1, 2, 3, 4, 5].findAll{it % 2 == 0}

Haskell

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

 ary = [1..10]
 evens = [x | x <- ary, even x]

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

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]

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

   ] v=: 20 ?@$ 100  NB. random vector
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

With any verb f that returns a boolean for each element of a vector v, the following is the generic solution:

   (#~f) v

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]);

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]);

Works with: Firefox version 2.0

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)];

Library: Functional

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

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]

MAXScript

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

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

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

 my @list = (1, 2, 3, 4, 5, 6);
 my @even = grep { 0 == $_%2; } @list;

PHP

Using a standard loop

 $arr = range(1,5);
 $evens = array();
 foreach ($arr as $val){
       if ($val % 2 == 0) $evens[] = $val);
 }
 print_r($evens);

Using a filter function

 function is_even($var) { return(!($var & 1)); }
 $arr = range(1,5);
 $evens = array_filter($arr, "is_even");
 print_r($evens);

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) =>

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

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

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

values = range(10)
evens = values[::2]

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.

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

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

Raven

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

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]

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

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:

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

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

V

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

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