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Common list elements

From Rosetta Code
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]


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences



Ada[edit]

with Ada.Text_Io;
with Ada.Containers.Vectors;
 
procedure Common is
 
package Integer_Vectors is
new Ada.Containers.Vectors (Index_Type => Positive,
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
use Ada.Text_Io;
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;
Output:
[ 3  9  6 ]

APL[edit]

APL has the built-in intersection function

      ∩/ (2 5 1 3 8 9 4 6) (3 5 6 2 9 8 4) (1 3 7 9 6)
3 9 6

AppleScript[edit]

AppleScriptObjC[edit]

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
Output:
{3, 6, 9}

Core language only[edit]

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.

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

AutoHotkey[edit]

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
}
 
Examples:
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
Output:
[3, 6, 9]

AWK[edit]

 
# 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)
}
 
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[edit]

LAMBDA[edit]

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

(See LAMBDA: The ultimate Excel worksheet function)

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
)
)

and also assuming the following generic bindings in Name Manager:

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()
)
)
)
Output:
fx =INTERSECTCOLS(D2:F9)
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#[edit]

Of course it is possible to use sets but I thought the idea was not to?

 
// 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->[email protected])|>List.distinct|>List.filter(fun n->nums|>Array.forall(fun g->List.contains n g)));;
 
Output:
[3; 9; 6]

Factor[edit]

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

Works with: Factor version 0.99 2021-02-05
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 .
Output:
{ 3 6 9 }

Go[edit]

Translation of: Wren
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()
}
}
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]

Haskell[edit]

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]]
Output:
[3,6,9]

Julia[edit]

 
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
 

Nim[edit]

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

Perl[edit]

@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;
Output:
3 6 9

Phix[edit]

function intersection(sequence s)
    sequence res = {}
    if length(s) then
        for i=1 to length(s[1]) do
            integer candidate = s[1][i]
            bool bAdd = not find(candidate,res)
            for j=2 to length(s) do
                if not find(candidate,s[j]) then
                    bAdd = false
                    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[edit]

Without Duplicates[edit]

"""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}")
 
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[edit]

"""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}")
 
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[edit]

  [ behead sort swap witheach
[ sort [] temp put
[ over [] !=
over [] != and while
over 0 peek
over 0 peek = iff
[ behead temp take
swap join temp put
swap behead drop ] again
over 0 peek
over 0 peek < if swap
behead drop again ]
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
Output:
[ 3 6 9 ]

Raku[edit]

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

REXX[edit]

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

/*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.*/
#= 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($), ',', " ")']'
output   when using the default inputs:
the list of common elements in all sets:  [3,6,9]

Ring[edit]

 
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])
add(sumNums,nums[n][m])
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
add(result,sumNums[n])
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
 
Output:
common list elements are: [3,6,9]

Wren[edit]

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.

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) {
res.add(e)
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()
}
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]