Common list elements: Difference between revisions

added RPL
(added RPL)
 
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=={{header|11l}}==
<langsyntaxhighlight lang="11l">F cle(nums)
V r = Set(nums[0])
L(num) nums[1..]
Line 19:
R r
 
print(cle([[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]]))</langsyntaxhighlight>
 
{{out}}
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=={{header|Action!}}==
{{libheader|Action! Tool Kit}}
<langsyntaxhighlight Actionlang="action!">INCLUDE "D2:SORT.ACT" ;from the Action! Tool Kit
 
DEFINE PTR="CARD"
Line 135:
arrays(2)=a7
Test(arrays,lengths,3)
RETURN</langsyntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Common_list_elements.png Screenshot from Atari 8-bit computer]
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=={{header|Ada}}==
<langsyntaxhighlight Adalang="ada">with Ada.Text_Io;
with Ada.Containers.Vectors;
 
Line 202:
R := Common_Elements (R, C);
Put (R);
end Common;</langsyntaxhighlight>
{{out}}
<pre>[ 3 9 6 ]</pre>
 
=={{header|ALGOL 68}}==
<syntaxhighlight lang="algol68">
BEGIN # find common elements of lists #
 
PRIO COMMON = 1;
# returns the common elements of a and b #
OP COMMON = ( []INT a, b )[]INT:
IF INT a len = ( UPB a - LWB a ) + 1;
INT b len = ( UPB b - LWB b ) + 1;
a len < 1 OR b len < 1
THEN # one or both lists is/are empty #
[]INT()
ELIF a len < b len
THEN # both lists are non-empty, b is shorter #
b COMMON a
ELSE # both lists are non-empty, a is at most as long as b #
[ 1 : b len ]INT result;
[ LWB a : UPB a ]BOOL used;
FOR i FROM LWB a TO UPB a DO used[ i ] := FALSE OD;
INT r pos := 0;
FOR b pos FROM LWB b TO UPB b DO
BOOL found := FALSE;
FOR a pos FROM LWB a TO UPB a WHILE NOT found DO
IF NOT used[ a pos ] THEN
IF ( found := a[ a pos ] = b[ b pos ] ) THEN
result[ r pos +:= 1 ] := b[ b pos ];
used[ a pos ] := TRUE
FI
FI
OD
OD;
result[ : r pos ]
FI # COMMON # ;
# returns the common elements in the lists in nums #
OP COMMON = ( [][]INT nums )[]INT:
IF 1 UPB nums < 1 LWB nums THEN # no lists #
[]INT()
ELIF 1 UPB nums = 1 LWB nums THEN # only one list #
nums[ LWB nums ]
ELSE # two or more lists #
FLEX[ 1 : 0 ]INT result;
result := nums[ LWB nums ] COMMON nums[ LWB nums + 1 ];
FOR i FROM LWB nums + 2 TO UPB nums DO
result := result COMMON nums[ i ]
OD;
result
FI # COMMON # ;
 
print( ( COMMON [][]INT( ( 2, 5, 1, 3, 8, 9, 4, 6 )
, ( 3, 5, 6, 2, 9, 8, 4 )
, ( 1, 3, 7, 6, 9 )
)
)
)
END
</syntaxhighlight>
{{out}}
<pre>
+3 +6 +9
</pre>
 
=={{header|APL}}==
APL has the built-in intersection function <code>∩</code>
 
<langsyntaxhighlight APLlang="apl"> ∩/ (2 5 1 3 8 9 4 6) (3 5 6 2 9 8 4) (1 3 7 9 6)
3 9 6 </langsyntaxhighlight>
 
=={{header|AppleScript}}==
===AppleScriptObjC===
<langsyntaxhighlight lang="applescript">use AppleScript version "2.4" -- OS X 10.10 (Yosemite) or later
use framework "Foundation"
use sorter : script "Insertion Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Insertion_sort#AppleScript>
Line 229 ⟶ 290:
set commonElements to commonListElements({{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}})
tell sorter to sort(commonElements, 1, -1)
return commonElements</langsyntaxhighlight>
 
{{output}}
<syntaxhighlight lang ="applescript">{3, 6, 9}</langsyntaxhighlight>
 
===Core language only===
The requirement for AppleScript 2.3.1 is only for the 'use' command which loads the "Insertion Sort" script. If the sort's instead loaded with the older 'load script' command or copied into the code, this will work on systems as far back as Mac OS X 10.5 (Leopard) or earlier. Same output as above.
<langsyntaxhighlight lang="applescript">use AppleScript version "2.3.1" -- Mac OS X 10.9 (Mavericks) or later.
use sorter : script "Insertion Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Insertion_sort#AppleScript>
 
Line 262 ⟶ 323:
set commonElements to commonListElements({{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}})
tell sorter to sort(commonElements, 1, -1)
return commonElements</langsyntaxhighlight>
 
=={{header|Arturo}}==
 
<syntaxhighlight lang="rebol">commonElements: function [subsets][
if zero? size subsets -> return []
if 1 = size subsets -> return first subsets
 
result: first subsets
 
loop slice subsets 1 dec size subsets 'subset [
result: intersection result subset
]
return result
]
 
print commonElements [
[2 5 1 3 8 9 4 6]
[3 5 6 2 9 8 4]
[1 3 7 6 9]
]</syntaxhighlight>
 
{{out}}
 
<pre>3 6 9</pre>
 
=={{header|AutoHotkey}}==
<langsyntaxhighlight AutoHotkeylang="autohotkey">Common_list_elements(nums){
counter := [], output := []
for i, num in nums
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return output
}
</syntaxhighlight>
</lang>
Examples:<langsyntaxhighlight AutoHotkeylang="autohotkey">nums := [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]
output := Common_list_elements(nums)
return</langsyntaxhighlight>
{{out}}
<pre>[3, 6, 9]</pre>
 
=={{header|AWK}}==
<syntaxhighlight lang="awk">
<lang AWK>
# syntax: GAWK -f COMMON_LIST_ELEMENTS.AWK
BEGIN {
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exit(0)
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9] : 3 6 9
</pre>
 
=={{header|BQN}}==
<syntaxhighlight lang="bqn">(∊/⊣)´ ⟨2,5,1,3,8,9,4,6⟩‿⟨3,5,6,2,9,8,4⟩‿⟨1,3,7,6,9⟩</syntaxhighlight>
{{out}}
<pre>⟨ 3 9 6 ⟩</pre>
 
=={{header|CLU}}==
<syntaxhighlight lang="clu">contains = proc [T: type] (a: array[T], v: T) returns (bool)
where T has equal: proctype (T,T) returns (bool)
for i: T in array[T]$elements(a) do
if i=v then return(true) end
end
return(false)
end contains
 
common = proc [T: type] (lists: ssT) returns (sT)
where T has equal: proctype (T,T) returns (bool)
sT = sequence[T]
aT = array[T]
ssT = sequence[sequence[T]]
cur: aT := sT$s2a(ssT$bottom(lists))
for i: int in int$from_to(2, ssT$size(lists)) do
next: aT := aT$[]
for e: T in sT$elements(lists[i]) do
if contains[T](cur, e) then
aT$addh(next,e)
end
end
cur := next
end
return(sT$a2s(cur))
end common
 
start_up = proc ()
si = sequence[int]
ssi = sequence[sequence[int]]
nums: ssi := ssi$[
si$[2,5,1,3,8,9,4,6],
si$[3,5,6,2,9,8,4],
si$[1,3,7,6,9]
]
po: stream := stream$primary_output()
for i: int in si$elements(common[int](nums)) do
stream$puts(po, int$unparse(i) || " ")
end
end start_up</syntaxhighlight>
{{out}}
<pre>3 6 9</pre>
 
=={{header|Excel}}==
===LAMBDA===
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{{Works with|Office 365 Betas 2021}}
<langsyntaxhighlight lang="lisp">INTERSECT
=LAMBDA(xs,
LAMBDA(ys,
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xs
)
)</langsyntaxhighlight>
 
and also assuming the following generic bindings in Name Manager:
<langsyntaxhighlight lang="lisp">ELEM
=LAMBDA(x,
LAMBDA(xs,
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)
)
)</langsyntaxhighlight>
 
{{Out}}
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|
|}
 
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
 
 
<syntaxhighlight lang="Delphi">
const Set1: set of byte = [2,5,1,3,8,9,4,6];
const Set2: set of byte = [3,5,6,2,9,8,4];
const Set3: set of byte = [1,3,7,6,9];
 
 
procedure CommonListElements(Memo: TMemo);
{Using Delphi "sets" to find common elements}
var I,Start,Stop: integer;
var Common: set of byte;
var S: string;
begin
{Uses "*" intersection set operator to}
{ find items common to all three sets}
Common:=Set1 * Set2 * Set3;
Memo.Lines.Add('Common Elements in');
Memo.Lines.Add(' [2,5,1,3,8,9,4,6]');
Memo.Lines.Add(' [3,5,6,2,9,8,4]');
Memo.Lines.Add(' [1,3,7,6,9]: ');
S:='';
{Display the common items}
for I:=0 to 9 do
if I in Common then S:=S+IntToStr(I)+',';
Memo.Lines.Add(S);
end;
 
</syntaxhighlight>
{{out}}
<pre>
Common Elements in
[2,5,1,3,8,9,4,6]
[3,5,6,2,9,8,4]
[1,3,7,6,9]:
3,6,9,
Elapsed Time: 5.260 ms.
</pre>
 
 
=={{header|EasyLang}}==
<syntaxhighlight>
nums[][] = [ [ 2 5 1 3 8 9 4 6 ] [ 3 5 6 2 9 8 4 ] [ 1 3 7 6 9 ] ]
#
found = 1
for e in nums[1][]
for l = 2 to len nums[][]
found = 0
for x in nums[l][]
if e = x
found = 1
break 1
.
.
if found = 0
break 1
.
.
if found = 1
r[] &= e
.
.
print r[]
</syntaxhighlight>
 
=={{header|F_Sharp|F#}}==
Of course it is possible to use sets but I thought the idea was not to?
<langsyntaxhighlight lang="fsharp">
// Common list elements. Nigel Galloway: February 25th., 2021
let nums=[|[2;5;1;3;8;9;4;6];[3;5;6;2;9;8;4];[1;3;7;6;9]|]
printfn "%A" (nums|>Array.reduce(fun n g->n@g)|>List.distinct|>List.filter(fun n->nums|>Array.forall(fun g->List.contains n g)));;
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 487 ⟶ 692:
Note: in older versions of Factor, <code>intersect-all</code> was called <code>intersection</code>.
{{works with|Factor|0.99 2021-02-05}}
<langsyntaxhighlight lang="factor">USING: prettyprint sets ;
 
{ { 2 5 1 3 8 9 4 6 } { 3 5 6 2 9 8 4 } { 1 3 7 6 9 } } intersect-all .</langsyntaxhighlight>
{{out}}
<pre>
{ 3 6 9 }
</pre>
 
=={{header|FreeBASIC}}==
<syntaxhighlight lang="freebasic">dim as integer nums(1 to 3, 1 to 8) = {{2,5,1,3,8,9,4,6}, {3,5,6,2,9,8,4}, {1,3,7,6,9} }
redim as integer outp(0)
dim as integer i, j
dim as boolean found
 
function is_in( s() as integer, n as integer, r as integer ) as boolean
for i as uinteger = 1 to ubound(s,2)
if s(r,i)=n then return true
next i
return false
end function
 
for i = 1 to 8
found = true
for j = 2 to 3
if not is_in( nums(), nums(1,i), j ) then found = false
next j
if found then
redim preserve as integer outp(1 to 1+ubound(outp))
outp(ubound(outp)) = nums(1,i)
end if
next i
 
for i = 1 to ubound(outp)
print outp(i);" ";
next i</syntaxhighlight>
{{out}}<pre>3 9 6</pre>
 
=={{header|Go}}==
{{trans|Wren}}
<langsyntaxhighlight lang="go">package main
 
import "fmt"
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fmt.Println()
}
}</langsyntaxhighlight>
 
{{out}}
Line 568 ⟶ 802:
[3 6 2 2]
</pre>
 
=={{header|Haskell}}==
<langsyntaxhighlight Haskelllang="haskell">import qualified Data.Set as Set
 
task :: Ord a => [[a]] -> [a]
Line 575 ⟶ 810:
task xs = Set.toAscList . foldl1 Set.intersection . map Set.fromList $ xs
 
main = print $ task [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]</langsyntaxhighlight>
{{out}}
<pre>[3,6,9]</pre>
 
=={{header|J}}==
 
<syntaxhighlight lang="j"> 2 5 1 3 8 9 4 6([-.-.)3 5 6 2 9 8 4([-.-.)1 3 7 6 9
3 9 6</syntaxhighlight>
 
Or,
 
<syntaxhighlight lang="j"> ;([-.-.)&.>/2 5 1 3 8 9 4 6;3 5 6 2 9 8 4;1 3 7 6 9
3 9 6</syntaxhighlight>
 
=={{header|jq}}==
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on the arrays whose intersection is sought. The helper function, `ios`,
might be independently useful and so is defined as a top-level filter.
<langsyntaxhighlight lang="jq"># If a and b are sorted lists, and if all the elements respectively of a and b are distinct,
# then [a,b] | ios will emit the stream of elements in the set-intersection of a and b.
def ios:
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elif any(.[]; length == 0) then []
else sort_by(length) | go
end;</langsyntaxhighlight>
<langsyntaxhighlight lang="jq"># The task:
[[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]</langsyntaxhighlight>
{{out}}
<pre>
[3,6,9]
</pre>
 
=={{header|K}}==
{{works with|ngn/k}}<syntaxhighlight lang=K>{x^x^y}/(2 5 1 3 8 9 4 6;1 3 7 6 9;3 5 6 2 9 8 4)
3 9 6</syntaxhighlight>
 
=={{header|Ksh}}==
<langsyntaxhighlight lang="ksh">
#!/bin/ksh
 
Line 658 ⟶ 907:
done
 
print "( ${output[@]} )"</langsyntaxhighlight>
{{out}}
<pre>( 3 9 6 )</pre>
 
=={{header|Julia}}==
<langsyntaxhighlight lang="julia">
julia> intersect([2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9])
3-element Array{Int64,1}:
Line 669 ⟶ 918:
9
6
</syntaxhighlight>
</lang>
 
=={{header|Lambdatalk}}==
<syntaxhighlight lang="scheme">
{def intersection
{def intersection.r
{lambda {:a :b :c :d}
{if {A.empty? :a}
then :d
else {intersection.r {A.rest :a} :b :c
{if {and {> {A.in? {A.first :a} :b} -1}
{> {A.in? {A.first :a} :c} -1}}
then {A.addlast! {A.first :a} :d}
else :d} }}}}
 
{lambda {:a :b :c}
{A.sort! < {intersection.r :a :b :c {A.new}}} }}
-> intersection
 
{intersection
{A.new 2 5 1 3 8 9 4 6}
{A.new 3 5 6 2 9 8 4}
{A.new 1 3 7 6 9}
}
-> [3,6,9]
</syntaxhighlight>
 
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<langsyntaxhighlight Mathematicalang="mathematica">Intersection[{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}]</langsyntaxhighlight>
{{out}}
<pre>{3, 6, 9}</pre>
 
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">
common_elems(lst):=block(map(setify,lst),apply(intersection,%%),listify(%%))$
nums:[[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]]$
common_elems(nums);
</syntaxhighlight>
{{out}}
<pre>
[3,6,9]
</pre>
 
=={{header|Nim}}==
<langsyntaxhighlight Nimlang="nim">import algorithm, sequtils
 
proc commonElements(list: openArray[seq[int]]): seq[int] =
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result.add val
 
echo commonElements([@[2,5,1,3,8,9,4,6], @[3,5,6,2,9,8,4], @[1,3,7,6,9]])</langsyntaxhighlight>
 
{{out}}
Line 694 ⟶ 980:
 
=={{header|Perl}}==
<langsyntaxhighlight lang="perl">@nums = ([2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]);
map { print "$_ " if @nums == ++$c{$_} } @$_ for @nums;</langsyntaxhighlight>
{{out}}
<pre>3 6 9</pre>
 
=={{header|Phix}}==
<!--<langsyntaxhighlight Phixlang="phix">(phixonline)-->
<span style="color: #008080;">function</span> <span style="color: #000000;">intersection</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>
Line 720 ⟶ 1,006:
<span style="color: #0000FF;">?</span><span style="color: #000000;">intersection</span><span style="color: #0000FF;">({{</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">8</span><span style="color: #0000FF;">,</span><span style="color: #000000;">9</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">},{</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">9</span><span style="color: #0000FF;">,</span><span style="color: #000000;">8</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">},{</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">7</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">,</span><span style="color: #000000;">9</span><span style="color: #0000FF;">}})</span>
<span style="color: #0000FF;">?</span><span style="color: #000000;">intersection</span><span style="color: #0000FF;">({{</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">8</span><span style="color: #0000FF;">,</span><span style="color: #000000;">9</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">},{</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">},{</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">7</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">}})</span>
<!--</langsyntaxhighlight>-->
Note that a (slightly more flexible) intersection() function is also defined in sets.e, so you could just include that instead, and use it the same way.
{{out}}
Line 731 ⟶ 1,017:
 
===Without Duplicates===
<langsyntaxhighlight lang="python">"""Find distinct common list elements using set.intersection."""
 
def common_list_elements(*lists):
Line 746 ⟶ 1,032:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
</syntaxhighlight>
</lang>
 
{{out}}
Line 755 ⟶ 1,041:
 
===With Duplicates===
<langsyntaxhighlight lang="python">"""Find common list elements using collections.Counter (multiset)."""
 
from collections import Counter
Line 777 ⟶ 1,063:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
</syntaxhighlight>
</lang>
 
{{out}}
Line 787 ⟶ 1,073:
=={{header|Quackery}}==
 
<langsyntaxhighlight Quackerylang="quackery"> [ behead sort swap witheach
[ sort [] temp put
[ over [] !=
Line 801 ⟶ 1,087:
2drop temp take ] ] is common ( [ [ --> [ )
 
' [ [ 2 5 1 3 8 9 4 6 ] [ 3 5 6 2 9 8 4 ] [ 1 3 7 6 9 ] ] common echo</langsyntaxhighlight>
 
{{out}}
Line 808 ⟶ 1,094:
 
=={{header|Raku}}==
<syntaxhighlight lang="raku" perl6line>put [∩] [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9,3];</langsyntaxhighlight>
{{out}}
<pre>6 9 3</pre>
Line 814 ⟶ 1,100:
=={{header|REXX}}==
This REXX version properly handles the case of duplicate entries in a list &nbsp; (which shouldn't happen in a true list).
<langsyntaxhighlight lang="rexx">/*REXX program finds and displays the common list elements from a collection of sets. */
parse arg a /*obtain optional arguments from the CL*/
if a='' | a="," then a= '[2,5,1,3,8,9,4,6] [3,5,6,2,9,8,4] [1,3,7,6,9]' /*defaults.*/
Line 831 ⟶ 1,117:
end /*k*/
/*stick a fork in it, we're all done. */
say 'the list of common elements in all sets: ' "["translate(space($), ',', " ")']'</langsyntaxhighlight>
{{out|output|text=&nbsp; when using the default inputs:}}
<pre>
Line 838 ⟶ 1,124:
 
=={{header|Ring}}==
<langsyntaxhighlight lang="ring">
nums = [[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]]
sumNums = []
Line 891 ⟶ 1,177:
txt = txt + "]"
see txt
</syntaxhighlight>
</lang>
{{out}}
<pre>
common list elements are: [3,6,9]
</pre>
 
=={{header|RPL}}==
Can handle duplicates.
{{works with|HP|48}}
« SWAP LIST→ → n
« 'n' DECR DUP 3 + ROLL - 2 +
ROLL DROP n →LIST
» » '<span style="color:blue">POPL</span>' STO <span style="color:grey">@ ''( {list} idx_item_to_remove → {list} )''</span>
« '''IF''' OVER SIZE OVER SIZE < '''THEN''' SWAP '''END'''
0 → a j
« { } SWAP
'''WHILE''' 'j' INCR a SIZE ≤ '''REPEAT'''
a j GET
'''IF''' DUP2 POS '''THEN'''
LASTARG ROT SWAP <span style="color:blue">POPL</span>
ROT ROT + SWAP
'''ELSE''' DROP '''END'''
'''END''' DROP
» » '<span style="color:blue">INTER</span>' STO
 
{ 2 5 1 3 8 9 4 6 } { 3 5 6 2 9 8 4 } { 1 3 7 6 9 } <span style="color:blue">INTER</span> <span style="color:blue">INTER</span>
{ 2 2 1 3 8 9 4 6 } { 3 5 6 2 2 2 4 } { 2 3 7 6 2 } <span style="color:blue">INTER</span> <span style="color:blue">INTER</span>
{{out}}
<pre>
2: { 3 6 9 }
1: { 3 6 2 2 }
</pre>
 
=={{header|Ruby}}==
<langsyntaxhighlight lang="ruby">nums = [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]
p nums.inject(&:intersection) # or nums.inject(:&)
</syntaxhighlight>
</lang>
{{out}}
<pre>[3, 9, 6]
</pre>
 
=={{header|V (Vlang)}}==
{{trans|go}}
<syntaxhighlight lang="v (vlang)">fn index_of(l []int, n int) int {
for i in 0..l.len {
if l[i] == n {
return i
}
}
return -1
}
fn common2(l1 []int, l2 []int) []int {
// minimize number of lookups
c1, c2 := l1.len, l2.len
mut shortest, mut longest := l1.clone(), l2.clone()
if c1 > c2 {
shortest, longest = l2.clone(), l1.clone()
}
mut longest2 := longest.clone()
mut res := []int{}
for e in shortest {
ix := index_of(longest2, e)
if ix >= 0 {
res << e
longest2 << longest2[ix+1..]
}
}
return res
}
fn common_n(ll [][]int) []int {
n := ll.len
if n == 0 {
return []int{}
}
if n == 1 {
return ll[0]
}
mut res := common2(ll[0], ll[1])
if n == 2 {
return res
}
for l in ll[2..] {
res = common2(res, l)
}
return res
}
fn main() {
lls := [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]],
]
for ll in lls {
println("Intersection of $ll is:")
println(common_n(ll))
println('')
}
}</syntaxhighlight>
{{out}}
<pre>Intersection of [[2, 5, 1, 3, 8, 9, 4, 6] [3, 5, 6, 2, 9, 8, 4] [1, 3, 7, 6, 9]] is:
[3, 6, 9]
 
Intersection of [[2 2 1 3 8 9 4 6] [3 5 6 2 2 2 4] [2, 3, 7, 6, 2]] is:
[3, 6, 2, 2]</pre>
 
'''Alternative:'''
<syntaxhighlight lang="vlang">fn main()
{
lls := [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]],
]
for ll in lls {
println("Intersection of $ll is:")
println(common_list_elements(ll))
println('')
}
 
}
 
fn common_list_elements(md_arr [][]int) []int {
mut counter := map[int]int{}
mut output := []int{}
 
for sd_arr in md_arr {
for value in sd_arr {
if counter[value] == counter[value] {counter[value] = counter[value] + 1} else { 1 } {
if counter[value] >= md_arr.len && output.any(it == value) == false {
output << value
}
}
}
}
return output
}</syntaxhighlight>
{{out}}
<pre>Intersection of [[2, 5, 1, 3, 8, 9, 4, 6] [3, 5, 6, 2, 9, 8, 4] [1, 3, 7, 6, 9]] is:
[3, 6, 9]
 
Intersection of [[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]] is:
[2, 3, 6]</pre>
 
=={{header|Wren}}==
{{libheader|Wren-seq}}
As we're dealing here with lists rather than sets, some guidance is needed on how to deal with duplicates in each list in the general case. A drastic solution would be to remove all duplicates from the result. Instead, the following matches duplicates - so if List A contains 2 'a's and List B contains 3 'a's, there would be 2 'a's in the result.
<langsyntaxhighlight ecmascriptlang="wren">import "./seq" for Lst
 
var common2 = Fn.new { |l1, l2|
Line 945 ⟶ 1,363:
System.print(commonN.call(ll))
System.print()
}</langsyntaxhighlight>
 
{{out}}
Line 954 ⟶ 1,372:
Intersection of [[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]] is:
[2, 3, 6, 2]
</pre>
<br>
Since the above was written, we can also now offer a library based solution.
<syntaxhighlight lang="wren">import "./seq" for Lst
 
var lls = [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]]
]
 
for (ll in lls) {
System.print(Lst.intersect(ll[0], Lst.intersect(ll[1], ll[2])))
}</syntaxhighlight>
 
{{out}}
<pre>
[3, 9, 6]
[2, 2, 3, 6]
</pre>
 
=={{header|XPL0}}==
A 32-bit integer is used to specify a set of values 0..31.
The [-1] terminator helps determine the number of lists.
<syntaxhighlight lang="xpl0">int IntSize, Nums, Sets, Ans, N, ListSize, Set, I;
[IntSize:= @Nums - @IntSize; \number of bytes in an integer
Nums:= [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9], [-1]];
Sets:= 0; \find the number of lists = number of Sets
while Nums(Sets, 0) > 0 do Sets:= Sets+1;
Ans:= -1;
for N:= 0 to Sets-1 do
[ListSize:= (Nums(N+1) - Nums(N)) / IntSize;
Set:= 0;
for I:= 0 to ListSize-1 do \Set = union (or) of list elements
Set:= Set or 1<<Nums(N, I);
Ans:= Ans & Set; \Answer is intersection (&) of Sets
];
I:= 0;
while Ans do \show common list elements
[if Ans & 1 then
[IntOut(0, I); ChOut(0, ^ )];
Ans:= Ans>>1;
I:= I+1;
];
]</syntaxhighlight>
 
{{out}}
<pre>
3 6 9
</pre>
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