Common list elements: Difference between revisions

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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]

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Tokenize

Sequences

Ada

<lang Ada>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;</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>

<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:

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

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:

(See LAMBDA: The ultimate Excel worksheet function)

Works with: Office 365 Betas 2021

<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:

	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#

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]

Haskell

<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,

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 ;

input: an array of arbitrary JSON arrays
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>

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>

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
     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]])</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]
            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

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
         [ 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</lang>

Output:

[ 3 6 9 ]

Raku

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.*/

= 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])
       add(sumNums,nums[n][m])
   next

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

   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

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]

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) {
           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()

}</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]

Revision as of 18:32, 8 August 2021

AppleScriptObjC

Core language only

LAMBDA

Without Duplicates

With Duplicates