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.
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;
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++
Compiler: Visual C++ 2005
Libraries: STL, 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() );
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.
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}
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)
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
Haskell
ary = [1..10] evens = [ x | x <- ary, even x ]
IDL
The where() function can select elements on any logical expression. For example
result = array[where(NOT array AND 1)]
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]);
Interpreter: Firefox 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)];
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
values = range(10) evens = [x for x in values if not(x & 1)] ievens = (x for x in values if not(x & 1)) # lazy
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
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.
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
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>
[ i source @ get-element
dup 2 mod 0 <> [ drop ] ifTrue
] iterate
depth [ i even put-element ] +iterate
]
} is copy-even
10 [ i i table put-element ] iterate table 10 copy-even