# Common list elements

Common list elements is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Given an integer array nums, find the common list elements.

Example

nums = [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]

output = [3,6,9]

## 11l

<lang 11l>F cle(nums)

```  V r = Set(nums[0])
L(num) nums[1..]
r = r.intersection(Set(num))
R r
```

print(cle([[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]]))</lang>

Output:
```Set([3, 6, 9])
```

procedure Common is

```  package Integer_Vectors is
Element_Type => Integer);
use Integer_Vectors;
```
```  function Common_Elements (Left, Right : Vector) return Vector is
Res : Vector;
begin
for E of Left loop
if Has_Element (Right.Find (E)) then
Res.Append (E);
end if;
end loop;
return Res;
end Common_Elements;
```
```  procedure Put (Vec : Vector) is
begin
Put ("[");
for E of Vec loop
Put (E'Image);  Put (" ");
end loop;
Put ("]");
New_Line;
end Put;
```
```  A : constant Vector := 2 & 5 & 1 & 3 & 8 & 9 & 4 & 6;
B : constant Vector := 3 & 5 & 6 & 2 & 9 & 8 & 4;
C : constant Vector := 1 & 3 & 7 & 6 & 9;
R : Vector;
```

begin

```  R := Common_Elements (A, B);
R := Common_Elements (R, C);
Put (R);
```

end Common;</lang>

Output:
`[ 3  9  6 ]`

## APL

APL has the built-in intersection function `∩`

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

```3 9 6 </lang>
```

## AppleScript

### AppleScriptObjC

<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>

on commonListElements(listOfLists)

```   set mutableSet to current application's class "NSMutableSet"'s setWithArray:(beginning of listOfLists)
repeat with i from 2 to (count listOfLists)
tell mutableSet to intersectSet:(current application's class "NSSet"'s setWithArray:(item i of listOfLists))
end repeat

return (mutableSet's allObjects()) as list
```

end commonListElements

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</lang>

Output:

<lang applescript>{3, 6, 9}</lang>

### 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. <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>

on commonListElements(listOfLIsts)

```   script o
property list1 : beginning of listOfLIsts
end script

set common to {}
set listCount to (count listOfLIsts)
repeat with i from 1 to (count o's list1)
set thisElement to {item i of o's list1}
if (thisElement is not in common) then
repeat with j from 2 to listCount
set OK to (item j of listOfLIsts contains thisElement)
if (not OK) then exit repeat
end repeat
if (OK) then set end of common to beginning of thisElement
end if
end repeat

return common
```

end commonListElements

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</lang>

## AutoHotkey

<lang AutoHotkey>Common_list_elements(nums){

```   counter := [], output := []
for i, num in nums
for j, d in num
if ((counter[d] := counter[d] ? counter[d]+1 : 1) = nums.count())
output.Push(d)
return output
```

} </lang> Examples:<lang 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</lang>

Output:
`[3, 6, 9]`

## AWK

<lang AWK>

1. syntax: GAWK -f COMMON_LIST_ELEMENTS.AWK

BEGIN {

```   PROCINFO["sorted_in"] = "@ind_num_asc"
nums = "[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]"
printf("%s : ",nums)
n = split(nums,arr1,"],?") - 1
for (i=1; i<=n; i++) {
gsub(/[\[\]]/,"",arr1[i])
split(arr1[i],arr2,",")
for (j in arr2) {
arr3[arr2[j]]++
}
}
for (j in arr3) {
if (arr3[j] == n) {
printf("%s ",j)
}
}
printf("\n")
exit(0)
```

} </lang>

Output:
```[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9] : 3 6 9
```

## Excel

### LAMBDA

Binding the names INTERSECT and INTERSECTCOLS to a pair of lambda expressions in the Excel WorkBook Name Manager:

<lang lisp>INTERSECT =LAMBDA(xs,

```   LAMBDA(ys,
FILTERP(
LAMBDA(x,
ELEM(x)(ys)
)
)(xs)
)
```

)

INTERSECTCOLS =LAMBDA(xs,

```   IF(1 < COLUMNS(xs),
INTERSECT(
FIRSTCOL(xs)
)(
INTERSECTCOLS(
TAILCOLS(xs)
)
),
xs
)
```

)</lang>

and also assuming the following generic bindings in Name Manager: <lang lisp>ELEM =LAMBDA(x,

```   LAMBDA(xs,
ISNUMBER(MATCH(x, xs, 0))
)
```

)

FILTERP =LAMBDA(p,

```   LAMBDA(xs,
FILTER(xs, p(xs))
)
```

)

FIRSTCOL =LAMBDA(xs,

```   INDEX(
xs,
SEQUENCE(ROWS(xs), 1, 1, 1),
1
)
```

)

TAILCOLS =LAMBDA(xs,

```   LET(
n, COLUMNS(xs) - 1,
```
```       IF(0 < n,
INDEX(
xs,
SEQUENCE(ROWS(xs), 1, 1, 1),
SEQUENCE(1, n, 2, 1)
),
NA()
)
)
```

)</lang>

Output:
 =INTERSECTCOLS(D2:F9) fx A B C D E F 1 Intersection of three columns 2 3 2 3 1 3 9 5 5 3 4 6 1 6 7 5 3 2 6 6 8 9 9 7 9 8 8 4 4 9 6

## F#

Of course it is possible to use sets but I thought the idea was not to? <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)));; </lang>

Output:
```[3; 9; 6]
```

## Factor

Note: in older versions of Factor, `intersect-all` was called `intersection`.

Works with: Factor version 0.99 2021-02-05

<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 .</lang>

Output:
```{ 3 6 9 }
```

## Go

Translation of: Wren

<lang go>package main

import "fmt"

func indexOf(l []int, n int) int {

```   for i := 0; i < len(l); i++ {
if l[i] == n {
return i
}
}
return -1
```

}

func common2(l1, l2 []int) []int {

```   // minimize number of lookups
c1, c2 := len(l1), len(l2)
shortest, longest := l1, l2
if c1 > c2 {
shortest, longest = l2, l1
}
longest2 := make([]int, len(longest))
copy(longest2, longest) // matching duplicates will be destructive
var res []int
for _, e := range shortest {
ix := indexOf(longest2, e)
if ix >= 0 {
res = append(res, e)
longest2 = append(longest2[:ix], longest2[ix+1:]...)
}
}
return res
```

}

func commonN(ll [][]int) []int {

```   n := len(ll)
if n == 0 {
return []int{}
}
if n == 1 {
return ll[0]
}
res := common2(ll[0], ll[1])
if n == 2 {
return res
}
for _, l := range ll[2:] {
res = common2(res, l)
}
return res
```

}

func main() {

```   lls := [][][]int{
{{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 := range lls {
fmt.Println("Intersection of", ll, "is:")
fmt.Println(commonN(ll))
fmt.Println()
}
```

}</lang>

Output:
```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]
```

<lang Haskell>import qualified Data.Set as Set

task :: Ord a => a -> [a] task [] = [] 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]]</lang>

Output:
`[3,6,9]`

## jq

Works with: jq

Works with gojq, the Go implementation of jq

The following definition of `intersection` does not place any restrictions on the arrays whose intersection is sought. The helper function, `ios`, might be independently useful and so is defined as a top-level filter. <lang jq># If a and b are sorted lists, and if all the elements respectively of a and b are distinct,

1. then [a,b] | ios will emit the stream of elements in the set-intersection of a and b.

def ios:

``` .[0] as \$a | .[1] as \$b
| if 0 == (\$a|length) or 0 == (\$b|length) then empty
elif \$a[0] == \$b[0] then \$a[0], ([\$a[1:], \$b[1:]] | ios)
elif \$a[0]  < \$b[0] then [\$a[1:], \$b] | ios
else [\$a, \$b[1:]] | ios
end ;
```
1. input: an array of arbitrary JSON arrays
2. output: their intersection as sets

def intersection:

``` def go:
if length == 1 then (.[0]|unique)
else [(.[0]|unique), (.[1:] | go)] | [ios]
end;
if length == 0 then []
elif any(.[]; length == 0) then []
else sort_by(length) | go
end;</lang>
```

<lang jq># The task: [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]</lang>

Output:
```[3,6,9]
```

## Julia

<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}:

```3
9
6
```

</lang>

## Mathematica /Wolfram Language

<lang Mathematica>Intersection[{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}]</lang>

Output:
`{3, 6, 9}`

## Nim

<lang Nim>import algorithm, sequtils

proc commonElements(list: openArray[seq[int]]): seq[int] =

``` var list = sortedByIt(list, it.len)   # Start with the shortest array.
for val in list[0].deduplicate():     # Check values only once.
block checkVal:
for i in 1..list.high:
if val notin list[i]:
break checkVal
```

echo commonElements([@[2,5,1,3,8,9,4,6], @[3,5,6,2,9,8,4], @[1,3,7,6,9]])</lang>

Output:
`@[3, 6, 9]`

## Perl

<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;</lang>

Output:
`3 6 9`

## Phix

```function intersection(sequence s)
sequence res = {}
if length(s) then
for i=1 to length(s[1]) do
integer candidate = s[1][i]
for j=2 to length(s) do
if not find(candidate,s[j]) then
exit
end if
end for
if bAdd then res &= candidate end if
end for
end if
return res
end function
?intersection({{2,5,1,3,8,9,4,6},{3,5,6,2,9,8,4},{1,3,7,6,9}})
?intersection({{2,2,1,3,8,9,4,6},{3,5,6,2,2,2,4},{2,3,7,6,2}})
```

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.

Output:
```{3,9,6}
{2,3,6}
```

## Python

### Without Duplicates

<lang python>"""Find distinct common list elements using set.intersection."""

def common_list_elements(*lists):

```   return list(set.intersection(*(set(list_) for list_ in lists)))
```

if __name__ == "__main__":

```   test_cases = [
([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 case in test_cases:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
```

</lang>

Output:
```intersection of ([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]) is [9, 3, 6]
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]
```

### With Duplicates

<lang python>"""Find common list elements using collections.Counter (multiset)."""

from collections import Counter from functools import reduce from itertools import chain

def common_list_elements(*lists):

```   counts = (Counter(list_) for list_ in lists)
intersection = reduce(lambda x, y: x & y, counts)
return list(chain.from_iterable([elem] * cnt for elem, cnt in intersection.items()))
```

if __name__ == "__main__":

```   test_cases = [
([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 case in test_cases:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
```

</lang>

Output:
```intersection of ([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]) is [9, 3, 6]
intersection of ([2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]) is [2, 2, 3, 6]
```

## Quackery

<lang Quackery> [ behead sort swap witheach

```   [ sort [] temp put
[ over [] !=
over [] != and while
over 0 peek
over 0 peek = iff
swap join temp put
over 0 peek
over 0 peek < if swap
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</lang>
```
Output:
`[ 3 6 9 ]`

## Raku

<lang perl6>put [∩] [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9,3];</lang>

Output:
`6 9 3`

## REXX

This REXX version properly handles the case of duplicate entries in a list   (which shouldn't happen in a true list). <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.*/

1. = words(a) /*the number of sets that are specified*/
```           do j=1  for #                        /*process each set  in  a list of sets.*/
@.j= translate( word(a, j), ,'],[')  /*extract   a   "  from "   "   "   "  */
end   /*j*/
```

\$= /*the list of common elements (so far).*/

```  do k=1  for #-1                               /*use the last set as the base compare.*/
do c=1  for words(@.#);  x= word(@.#, c)   /*obtain an element from a set.        */
do f=1  for #-1                         /*verify that the element is in the set*/
if wordpos(x, @.f)==0  then iterate c   /*Is in the set?  No, then skip element*/
end   /*f*/
if wordpos(x, \$)==0  then \$= \$ x           /*Not already in the set?  Add common. */
end      /*c*/
end         /*k*/
/*stick a fork in it,  we're all done. */
```

say 'the list of common elements in all sets: ' "["translate(space(\$), ',', " ")']'</lang>

output   when using the default inputs:
```the list of common elements in all sets:  [3,6,9]
```

## Ring

<lang ring> nums = [[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]] sumNums = [] result = []

for n = 1 to len(nums)

```   for m = 1 to len(nums[n])
next
```

next

sumNums = sort(sumNums) for n = len(sumNums) to 2 step -1

```   if sumNums[n] = sumNums[n-1]
del(sumNums,n)
ok
```

next

for n = 1 to len(sumNums)

```   flag = list(len(nums))
for m = 1 to len(nums)
flag[m] = 1
ind = find(nums[m],sumNums[n])
if ind < 1
flag[m] = 0
ok
next
flagn = 1
for p = 1 to len(nums)
if flag[p] = 1
flagn = 1
else
flagn = 0
exit
ok
next
if flagn = 1
ok
```

next

see "common list elements are: " showArray(result)

func showArray(array)

```    txt = ""
see "["
for n = 1 to len(array)
txt = txt + array[n] + ","
next
txt = left(txt,len(txt)-1)
txt = txt + "]"
see txt
```

</lang>

Output:
```common list elements are: [3,6,9]
```

## Ruby

<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(:&) </lang>

Output:
```[3, 9, 6]
```

## Wren

Library: 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. <lang ecmascript>import "/seq" for Lst

var common2 = Fn.new { |l1, l2|

```   // minimize number of lookups
var c1 = l1.count
var c2 = l2.count
var shortest = (c1 < c2) ? l1 : l2
var longest = (l1 == shortest) ? l2 : l1
longest = longest.toList // matching duplicates will be destructive
var res = []
for (e in shortest) {
var ix = Lst.indexOf(longest, e)
if (ix >= 0) {
longest.removeAt(ix)
}
}
return res
```

}

var commonN = Fn.new { |ll|

```   var n = ll.count
if (n == 0) return ll
if (n == 1) return ll[0]
var first2 = common2.call(ll[0], ll[1])
if (n == 2) return first2
return ll.skip(2).reduce(first2) { |acc, l| common2.call(acc, l) }
```

}

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("Intersection of %(ll) is:")
System.print(commonN.call(ll))
System.print()
```

}</lang>

Output:
```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, 2]
```