Common sorted list: Difference between revisions

Content added Content deleted

Inline

Latest revision as of 16:40, 3 April 2024

Given an integer array nums, the goal is create common sorted list with unique elements.

Example

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

output = [1,3,4,5,7,8,9]

11l

F csl(nums)
   Set[Int] r
   L(num) nums
      r.update(num)
   R sorted(Array(r))

print(csl([[5, 1, 3, 8, 9, 4, 8, 7], [3, 5, 9, 8, 4], [1, 3, 7, 9]]))

Output:

[1, 3, 4, 5, 7, 8, 9]

Action!

Library: Action! Tool Kit

INCLUDE "D2:SORT.ACT" ;from the Action! Tool Kit

DEFINE PTR="CARD"

PROC PrintArray(BYTE ARRAY a BYTE len)
  BYTE i

  Print("  [")
  IF len>0 THEN
    FOR i=0 TO len-1
    DO
      PrintB(a(i))
      IF i<len-1 THEN Put(' ) FI
    OD
  FI
  PrintE("]")
RETURN

BYTE FUNC Contains(BYTE ARRAY a BYTE len,value)
  BYTE i

  IF len>0 THEN
    FOR i=0 TO len-1
    DO
      IF a(i)=value THEN RETURN(1) FI
    OD
  FI
RETURN (0)

PROC CommonListElements(PTR ARRAY arrays
  BYTE ARRAY lengths BYTE count
  BYTE ARRAY res BYTE POINTER resLen)
  
  BYTE ARRAY a
  BYTE i,j,len,value,cnt,maxcnt

  resLen^=0
  IF count=0 THEN
    RETURN
  ELSEIF count=1 THEN
    MoveBlock(res,a,len) RETURN
  FI

  FOR i=0 TO count-1
  DO
    a=arrays(i) len=lengths(i)
    IF len>0 THEN
      FOR j=0 TO len-1
      DO
        value=a(j)
        IF Contains(res,resLen^,value)=0 THEN
          res(resLen^)=value resLen^==+1
        FI
      OD
    FI
  OD
  SortB(res,resLen^,0)
RETURN

PROC Test(PTR ARRAY arrays BYTE ARRAY lengths BYTE count)
  BYTE ARRAY res(100)
  BYTE len,i

  CommonListElements(arrays,lengths,count,res,@len)
  PrintE("Input:")
  FOR i=0 TO count-1
  DO
    PrintArray(arrays(i),lengths(i))
  OD
  PrintE("Output:")
  PrintArray(res,len) PutE()
RETURN

PROC Main()
  PTR ARRAY arrays(3)
  BYTE ARRAY
    lengths(3)=[8 5 4],
    a1(8)=[5 1 3 8 9 4 8 7],
    a2(5)=[3 5 9 8 4],
    a3(4)=[1 3 7 9],
    a4(8)=[5 1 1 1 9 9 8 7],
    a5(5)=[5 5 9 9 9],
    a6(4)=[1 7 7 9]
  BYTE len

  Put(125) PutE() ;clear the screen
  
  arrays(0)=a1 arrays(1)=a2 arrays(2)=a3
  Test(arrays,lengths,3)

  arrays(0)=a4 arrays(1)=a5 arrays(2)=a6
  Test(arrays,lengths,3)

  lengths(0)=0
  Test(arrays,lengths,3)
RETURN

Output:

Screenshot from Atari 8-bit computer

Input:
  [5 1 3 8 9 4 8 7]
  [3 5 9 8 4]
  [1 3 7 9]
Output:
  [1 3 4 5 7 8 9]

Input:
  [5 1 1 1 9 9 8 7]
  [5 5 9 9 9]
  [1 7 7 9]
Output:
  [1 5 7 8 9]

Input:
  []
  [5 5 9 9 9]
  [1 7 7 9]
Output:
  [1 5 7 9]

Ada

with Ada.Text_Io;
with Ada.Containers.Vectors;

procedure Sorted is

   package Integer_Vectors is
     new Ada.Containers.Vectors (Index_Type   => Positive,
                                 Element_Type => Integer);
   use Integer_Vectors;

   package Vector_Sorting is
     new Integer_Vectors.Generic_Sorting;
   use Vector_Sorting;

   procedure Unique (Vec : in out Vector) is
      Res : Vector;
   begin
      for E of Vec loop
         if Res.Is_Empty or else Res.Last_Element /= E then
            Res.Append (E);
         end if;
      end loop;
      Vec := Res;
   end Unique;

   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 := 5 & 1 & 3 & 8 & 9 & 4 & 8 & 7;
   B : constant Vector := 3 & 5 & 9 & 8 & 4;
   C : constant Vector := 1 & 3 & 7 & 9;
   R : Vector          := A & B & C;
begin
   Sort (R);
   Unique (R);
   Put (R);
end Sorted;

Output:

[ 1  3  4  5  7  8  9 ]

APL

Works with: Dyalog APL

csl ← (⊂∘⍋⌷⊢)∪∘∊

Output:

      csl (5 1 3 8 9 4 8 7)(3 5 9 8 4)(1 3 7 9)
1 3 4 5 7 8 9

AppleScript

use AppleScript version "2.4" -- OS X 10.10 (Yosemite) or later
use framework "Foundation"

local nums, output

set nums to current application's class "NSArray"'s arrayWithArray:({{5, 1, 3, 8, 9, 4, 8, 7}, {3, 5, 9, 8, 4}, {1, 3, 7, 9}})
set output to (nums's valueForKeyPath:("@distinctUnionOfArrays.self"))'s sortedArrayUsingSelector:("compare:")
return output as list

Output:

{1, 3, 4, 5, 7, 8, 9}

Or, as a composition of slightly more commonly-used generic functions (given that distinctUnionOfArrays is a relatively specialised function, while concat/flatten and nub/distinct are more atomic, and more frequently reached for):

use AppleScript version "2.4"
use framework "Foundation"


------------------- COMMON SORTED ARRAY ------------------
on run
    sort(nub(concat({¬
        {5, 1, 3, 8, 9, 4, 8, 7}, ¬
        {3, 5, 9, 8, 4}, ¬
        {1, 3, 7, 9}})))
end run


-------------------- GENERIC FUNCTIONS -------------------

-- concat :: [[a]] -> [a]
on concat(xs)
    ((current application's NSArray's arrayWithArray:xs)'s ¬
        valueForKeyPath:"@unionOfArrays.self") as list
end concat


-- nub :: [a] -> [a]
on nub(xs)
    ((current application's NSArray's arrayWithArray:xs)'s ¬
        valueForKeyPath:"@distinctUnionOfObjects.self") as list
end nub


-- sort :: Ord a => [a] -> [a]
on sort(xs)
    ((current application's NSArray's arrayWithArray:xs)'s ¬
        sortedArrayUsingSelector:"compare:") as list
end sort

Output:

{1, 3, 4, 5, 7, 8, 9}

Arturo

print sort unique flatten [[5 1 3 8 9 4 8 7]  [3 5 9 8 4]  [1 3 7 9]]

Output:

1 3 4 5 7 8 9

AutoHotkey

Common_sorted_list(nums){
    elements := [], output := []
    for i, num in nums
        for j, d in num
            elements[d] := true
    for val, bool in elements
        output.push(val)
    return output
}

Examples:

nums := [[5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9]]
output := Common_sorted_list(nums)
return

Output:

[1, 3, 4, 5, 6, 7, 9]

AWK

# syntax: GAWK -f COMMON_SORTED_LIST.AWK
BEGIN {
    PROCINFO["sorted_in"] = "@ind_num_asc"
    nums = "[5,1,3,8,9,4,8,7],[3,5,9,8,4],[1,3,7,9]"
    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) {
      printf("%s ",j)
    }
    printf("\n")
    exit(0)
}

Output:

[5,1,3,8,9,4,8,7],[3,5,9,8,4],[1,3,7,9] : 1 3 4 5 7 8 9

BQN

Joins all the arrays, deduplicate, sort.

  arr ← ⟨⟨5,1,3,8,9,4,8,7⟩, ⟨3,5,9,8,4⟩, ⟨1,3,7,9⟩⟩
⟨ ⟨ 5 1 3 8 9 4 8 7 ⟩ ⟨ 3 5 9 8 4 ⟩ ⟨ 1 3 7 9 ⟩ ⟩
  CSL ← ∧·⍷∾
∧(⍷∾)
  CSL arr
⟨ 1 3 4 5 7 8 9 ⟩

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define COUNTOF(a) (sizeof(a)/sizeof(a[0]))

void fatal(const char* message) {
    fprintf(stderr, "%s\n", message);
    exit(1);
}

void* xmalloc(size_t n) {
    void* ptr = malloc(n);
    if (ptr == NULL)
        fatal("Out of memory");
    return ptr;
}

int icompare(const void* p1, const void* p2) {
    const int* ip1 = p1;
    const int* ip2 = p2;
    return (*ip1 < *ip2) ? -1 : ((*ip1 > *ip2) ? 1 : 0);
}

size_t unique(int* array, size_t len) {
    size_t out_index = 0;
    int prev;
    for (size_t i = 0; i < len; ++i) {
        if (i == 0 || prev != array[i])
            array[out_index++] = array[i];
        prev = array[i];
    }
    return out_index;
}

int* common_sorted_list(const int** arrays, const size_t* lengths, size_t count, size_t* size) {
    size_t len = 0;
    for (size_t i = 0; i < count; ++i)
        len += lengths[i];
    int* array = xmalloc(len * sizeof(int));
    for (size_t i = 0, offset = 0; i < count; ++i) {
        memcpy(array + offset, arrays[i], lengths[i] * sizeof(int));
        offset += lengths[i];
    }
    qsort(array, len, sizeof(int), icompare);
    *size = unique(array, len);
    return array;
}

void print(const int* array, size_t len) {
    printf("[");
    for (size_t i = 0; i < len; ++i) {
        if (i > 0)
            printf(", ");
        printf("%d", array[i]);
    }
    printf("]\n");
}

int main() {
    const int a[] = {5, 1, 3, 8, 9, 4, 8, 7};
    const int b[] = {3, 5, 9, 8, 4};
    const int c[] = {1, 3, 7, 9};
    size_t len = 0;
    const int* arrays[] = {a, b, c};
    size_t lengths[] = {COUNTOF(a), COUNTOF(b), COUNTOF(c)};
    int* sorted = common_sorted_list(arrays, lengths, COUNTOF(arrays), &len);
    print(sorted, len);
    free(sorted);
    return 0;
}

Output:

[1, 3, 4, 5, 7, 8, 9]

C++

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>

template<typename T>
std::vector<T> common_sorted_list(const std::vector<std::vector<T>>& ll) {
    std::set<T> resultset;
    std::vector<T> result;
    for (auto& list : ll)
        for (auto& item : list)
            resultset.insert(item);
    for (auto& item : resultset)
        result.push_back(item);
    
    std::sort(result.begin(), result.end());
    return result;
}

int main() {
    std::vector<int> a = {5,1,3,8,9,4,8,7};
    std::vector<int> b = {3,5,9,8,4};
    std::vector<int> c = {1,3,7,9};
    std::vector<std::vector<int>> nums = {a, b, c};
    
    auto csl = common_sorted_list(nums);
    for (auto n : csl) std::cout << n << " ";
    std::cout << std::endl;
    
    return 0;
}

Output:

1 3 4 5 7 8 9

Clojure

Efficient transducer & sorted-set version

(defn common-sorted [& colls]
  (into (sorted-set) cat colls))

(common-sorted [5 1 3 8 9 4 8 7] [3 5 9 8 4] [1 3 7 9])
;; => #{1 3 4 5 7 8 9}

Point-free, list-based version

(def common-sorted (comp sort distinct concat))

(common-sorted [5 1 3 8 9 4 8 7] [3 5 9 8 4] [1 3 7 9])
;; => (1 3 4 5 7 8 9)

EasyLang

nums[][] = [ [ 2 5 1 3 8 9 4 6 ] [ 3 5 6 2 9 8 4 ] [ 1 3 7 6 9 ] ]
# 
proc sort . d[] .
   for i = 1 to len d[] - 1
      for j = i + 1 to len d[]
         if d[j] < d[i]
            swap d[j] d[i]
         .
      .
   .
.
# 
for l to len nums[][]
   for e in nums[l][]
      h[] &= e
   .
.
sort h[]
for e in h[]
   if e <> last or len r[] = 0
      r[] &= e
      last = e
   .
.
print r[]

Excel

LAMBDA

Binding the name COMMONSORTED to the following lambda expression in the Name Manager of the Excel WorkBook:

(See LAMBDA: The ultimate Excel worksheet function)

Works with: Office 365 betas 2021

COMMONSORTED
=LAMBDA(grid,
    SORT(
        UNIQUE(
            CONCATCOLS(grid)
        )
    )
)

and also assuming the following generic bindings in the Name Manager for the WorkBook:

CONCATCOLS
=LAMBDA(cols,
    LET(
        nRows, ROWS(cols),
        ixs, SEQUENCE(
            COLUMNS(cols) * nRows, 1,
            1, 1
        ),

        FILTERP(
            LAMBDA(x, 0 < LEN("" & x))
        )(
            INDEX(cols,
                LET(
                    r, MOD(ixs, nRows),

                    IF(0 = r, nRows, r)
                ),
                1 + QUOTIENT(ixs - 1, nRows)
            )
        )
    )
)


FILTERP
=LAMBDA(p,
    LAMBDA(xs,
        FILTER(xs, p(xs))
    )
)

The formula in cell B2 defines a dynamic array of values, additionally populating further cells in column B.

Output:

	fx	=COMMONSORTED(C2:E9)
	A	B	C	D	E
1		Common sorted
2		1	5	3	1
3		3	1	5	3
4		4	3	9	7
5		5	8	8	9
6		7	9	4
7		8	4
8		9	8
9			7

F#

// Common sorted list. Nigel Galloway: February 25th., 2021
let nums=[|[5;1;3;8;9;4;8;7];[3;5;9;8;4];[1;3;7;9]|]
printfn "%A" (nums|>Array.reduce(fun n g->n@g)|>List.distinct|>List.sort)

Output:

[1; 3; 4; 5; 7; 8; 9]

Factor

Note: in older versions of Factor, union-all is called combine.

Works with: Factor version 0.99 2021-02-05

USING: formatting kernel sets sorting ;

{ { 5 1 3 8 9 4 8 7 } { 3 5 9 8 4 } { 1 3 7 9 } }
dup union-all natural-sort
"Sorted union of %u is:\n%u\n" printf

Output:

Sorted union of { { 5 1 3 8 9 4 8 7 } { 3 5 9 8 4 } { 1 3 7 9 } } is:
{ 1 3 4 5 7 8 9 }

FreeBASIC

Dim As Integer nums(2, 7) = {{5,1,3,8,9,4,8,7},{3,5,9,8,4},{1,3,7,9}}
Dim Shared As Integer sumNums()

Sub Sum(arr() As Integer, vals As Integer)
    Redim Preserve arr(Ubound(arr) + 1)
    arr(Ubound(arr)) = vals
End Sub

Sub Del(arr() As Integer, index As Integer)
    For i As Integer = index To Ubound(arr) - 1
        arr(i) = arr(i + 1)
    Next i
    Redim Preserve arr(Ubound(arr) - 1)
End Sub

Sub showArray(arr() As Integer)
    Dim As String txt = ""
    Print "[";
    For n As Integer = 1 To Ubound(arr)
        txt &= Str(arr(n)) & ","
    Next n
    txt = Left(txt, Len(txt) - 1)
    txt &= "]"
    Print txt
End Sub

Sub Sort(arr() As Integer)
    Dim As Integer i, j
    For i = 0 To Ubound(arr) - 1
        For j = i + 1 To Ubound(arr)
            If arr(i) > arr(j) Then Swap arr(i), arr(j)
        Next j
    Next i
End Sub

For n As Integer = 0 To Ubound(nums, 1)
    For m As Integer = 0 To Ubound(nums, 2)
        Sum(sumNums(), nums(n, m))
    Next m
Next n

Sort(sumNums())
For n As Integer = Ubound(sumNums) To 1 Step -1
    If sumNums(n) = sumNums(n - 1) Then
        Del(sumNums(), n)
    End If
Next n

Sort(sumNums())

Print "common sorted list elements are: "; 
showArray(sumNums())

Sleep

Output:

common sorted list elements are: [1,3,4,5,7,8,9]

Go

Translation of: Wren

package main

import (
    "fmt"
    "sort"
)

func distinctSortedUnion(ll [][]int) []int {
    var res []int
    for _, l := range ll {
        res = append(res, l...)
    }
    set := make(map[int]bool)
    for _, e := range res {
        set[e] = true
    }
    res = res[:0]
    for key := range set {
        res = append(res, key)
    }
    sort.Ints(res)
    return res
}

func main() {
    ll := [][]int{{5, 1, 3, 8, 9, 4, 8, 7}, {3, 5, 9, 8, 4}, {1, 3, 7, 9}}
    fmt.Println("Distinct sorted union of", ll, "is:")
    fmt.Println(distinctSortedUnion(ll))
}

Output:

Distinct sorted union of [[5 1 3 8 9 4 8 7] [3 5 9 8 4] [1 3 7 9]] is:
[1 3 4 5 7 8 9]

Haskell

import Data.List (nub, sort)

-------------------- COMMON SORTED LIST ------------------

commonSorted :: Ord a => [[a]] -> [a]
commonSorted = sort . nub . concat

--------------------------- TEST -------------------------
main :: IO ()
main =
  print $
    commonSorted
      [ [5, 1, 3, 8, 9, 4, 8, 7],
        [3, 5, 9, 8, 4],
        [1, 3, 7, 9]
      ]

Output:

[1,3,4,5,7,8,9]

J

csl =: /:~@~.@;

Output:

   nums =: 5 1 3 8 9 4 8 7;3 5 9 8 4;1 3 7 9
   csl nums
1 3 4 5 7 8 9

JavaScript

(() => {
    "use strict";

    // --------------- COMMON SORTED LIST ----------------

    // commonSorted :: Ord a => [[a]] -> [a]
    const commonSorted = xs =>
        sort(nub(concat(xs)));


    // ---------------------- TEST -----------------------
    const main = () =>
        commonSorted([
            [5, 1, 3, 8, 9, 4, 8, 7],
            [3, 5, 9, 8, 4],
            [1, 3, 7, 9]
        ]);


    // --------------------- GENERIC ---------------------

    // concat :: [[a]] -> [a]
    const concat = xs =>
        xs.flat(1);


    // nub :: Eq a => [a] -> [a]
    const nub = xs => [...new Set(xs)];


    // sort :: Ord a => [a] -> [a]
    const sort = xs =>
        // An (ascending) sorted copy of xs.
        xs.slice().sort();

    return main();
})();

Output:

[1, 3, 4, 5, 7, 8, 9]

jq

Works with: jq

Works with gojq, the Go implementation of jq

Assuming the input is a single array of arrays, we can simply chain the `add` and `unique` filters:

jq -nc '[[5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9]] | add | unique'

Output:

[1,3,4,5,7,8,9]

If the arrays are presented individually in an external stream, we could use the `-s` command-line option.

Alternatively, here's a stream-oriented version of `add` that could be used:

def add(s): reduce s as $x (null; . + $x);

For example:

jq -nc '
    def add(s): reduce s as $x (null; . + $x);
    add(inputs) | unique' <<< '[5,1,3,8,9,4,8,7] [3,5,9,8,4] [1,3,7,9]'

Julia

julia> sort(union([5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9]))
7-element Array{Int64,1}:
 1
 3
 4
 5
 7
 8
 9

julia> sort(union([2, 3, 4], split("3.14 is not an integer", r"\s+")), lt=(x, y) -> "$x" < "$y")
8-element Array{Any,1}:
 2
 3
  "3.14"
 4
  "an"
  "integer"
  "is"
  "not"

Mathematica/Wolfram Language

Union[{5, 1, 3, 8, 9, 4, 8, 7}, {3, 5, 9, 8, 4}, {1, 3, 7, 9}]

Output:

{1, 3, 4, 5, 7, 8, 9}

Maxima

common_sorted(lst):=unique(flatten(lst))$
nums:[[5,1,3,8,9,4,8,7],[3,5,9,8,4],[1,3,7,9]]$
common_sorted(nums);

Output:

[1,3,4,5,7,8,9]

Nim

We could use a HashSet or an IntSet to deduplicate. We have rather chosen to use the procedure deduplicate from module sequtils applied to the sorted list.

import algorithm, sequtils

proc commonSortedList(list: openArray[seq[int]]): seq[int] =
  sorted(concat(list)).deduplicate(true)

echo commonSortedList([@[5,1,3,8,9,4,8,7], @[3,5,9,8,4], @[1,3,7,9]])

Output:

@[1, 3, 4, 5, 7, 8, 9]

Pascal

Free Pascal

Program CommonSortedList;
{$mode ObjFPC}{$H+}

Uses sysutils,fgl;

Type 
  tarr = array Of array Of integer;

Const List1: tarr = ((5,1,3,8,9,4,8,7), (3,5,9,8,4), (1,3,7,9));

Var list : specialize TFPGList<integer>;

Procedure addtolist(arr : tarr);
Var i : integer;
  arr2 : array Of integer;
Begin
  For arr2 In arr Do
    For i In arr2 Do
      If (list.indexof(i) = -1) {make sure number isn't in list already}
        Then list.add(i);
End;

Function CompareInt(Const Item1,Item2: Integer): Integer;
Begin
  result := item1 - item2;
End;

Var i : integer;
Begin
  list := specialize TFPGList<integer>.create;
  addtolist(list1);
  List.Sort(@CompareInt); {quick sort the list}
  For i In list Do
    write(i:4);
  list.destroy;
End.

Output:

   1   3   4   5   7   8   9

Perl

@c{@$_}++ for [5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9];
print join ' ', sort keys %c;
@c{@$_}++ for [qw<not all is integer ? is not ! 4.2>];
print join ' ', sort keys %c;

Output:

1 3 4 5 7 8 9
! 1 3 4 4.2 5 7 8 9 ? all integer is not

Phix

Library: Phix/basics

?unique(join({{5,1,3,8,9,4,8,7}, {3,5,9,8,4}, {1,3,7,9}, split("not everything is an integer")},{}))

Note the join(x,{}): the 2nd param is needed to prevent it putting 32 (ie the acsii code for a space) in the output.

Output:

(Unexpectedly rather Yoda-esque)

{1,3,4,5,7,8,9,"an","everything","integer","is","not"}

Python

'''Common sorted list'''

from itertools import chain


# ------------------------- TEST -------------------------
# main :: IO ()
def main():
    '''Sorted union of lists'''

    print(
        sorted(nub(concat([
            [5, 1, 3, 8, 9, 4, 8, 7],
            [3, 5, 9, 8, 4],
            [1, 3, 7, 9]
        ])))
    )


# ----------------------- GENERIC ------------------------

# concat :: [[a]] -> [a]
# concat :: [String] -> String
def concat(xs):
    '''The concatenation of all the elements in a list.
    '''
    return list(chain(*xs))


# nub :: [a] -> [a]
def nub(xs):
    '''A list containing the same elements as xs,
       without duplicates, in the order of their
       first occurrence.
    '''
    return list(dict.fromkeys(xs))


# MAIN ---
if __name__ == '__main__':
    main()

Output:

[1, 3, 4, 5, 7, 8, 9]

Quackery

uniquewith is defined at Remove duplicate elements#Quackery.

  ' [ [ 5 1 3 8 9 4 8 7 ]
      [ 3 5 9 8 4 ]
      [ 1 3 7 9 ] ]
 
  [] swap witheach [ witheach join ]
  uniquewith > echo

Output:

[ 1 3 4 5 7 8 9 ]

Raku

put sort keys [∪] [5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9];
put sort keys [∪] [5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9], [<not everything is an integer so why not avoid special cases # ~ 23 4.2>];

Output:

1 3 4 5 7 8 9
# 1 3 4 4.2 5 7 8 9 23 an avoid cases everything integer is not so special why ~

REXX

/*REXX pgm creates and displays a  common sorted list  of a specified collection of sets*/
parse arg a                                      /*obtain optional arguments from the CL*/
if a='' | a=","  then a= '[5,1,3,8,9,4,8,7]  [3,5,9,8,4]  [1,3,7,9]'     /*default sets.*/
x= translate(a, ,'],[')                          /*extract elements from collection sets*/
se= words(x)
#= 0;            $=                              /*#: number of unique elements; $: list*/
$=                                               /*the list of common elements (so far).*/
    do j=1  for se;  _= word(x, j)               /*traipse through all elements in sets.*/
    if wordpos(_, $)>0  then iterate             /*Is element in the new list? Yes, skip*/
    $= $ _;   #= # + 1;    @.#= _                /*add to list; bump counter; assign──►@*/
    end   /*j*/
$=
call eSort #                                     /*use any short (small)  exchange sort.*/
                 do k=1  for #;   $= $  @.k      /*rebuild the $ list, it's been sorted.*/
                 end   /*k*/

say 'the list of sorted common elements in all sets: ' "["translate(space($), ',', " ")']'
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
eSort: procedure expose @.; arg h 1 z; do while h>1; h= h%2; do i=1  for z-h; j= i; k= h+i
        do while @.k<@.j; t=@.j; @.j=@.k; @.k=t; if h>=j then leave; j=j-h; k=k-h; end;end
        end;     return                          /*this sort was used 'cause of brevity.*/

output when using the default inputs:

the list of sorted common elements in all sets:  [1,3,4,5,7,8,9]

Ring

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

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

sumNums = sort(sumNums)

see "common sorted list elements are: "
showArray(sumNums)

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 sorted list elements are: [1,3,4,5,7,8,9]

RPL

Works with: HP version 48G

≪ ∑LIST SORT → nums
  ≪ { }
     1 nums SIZE FOR j
        nums j GET
        IF DUP2 POS THEN DROP ELSE + END 
     NEXT
≫ ≫ 'COMMON' STO

{ { 5 1 3 8 9 4 8 7 } { 3 5 9 8 4 } { 1 3 7 9 } } COMMON

Output:

1: { 1 3 4 5 7 8 9 }

Ruby

nums = [5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9]
p nums.inject(:+).sort.uniq

Output:

[1, 3, 4, 5, 7, 8, 9]

Uiua

F ← ⊏⍏.◴/◇⊂

Output:

F {[5 1 3 8 9 4 8 7] [3 5 9 8 4] [1 3 7 9]}
[1 3 4 5 7 8 9]

V (Vlang)

fn main() {
    nums := [[5,1,3,8,9,4,8,7], [3,5,9,8,4], [1,3,7,9]]
    println("Distinct sorted union of $nums is:")
    println(common_sorted_list(nums))
}

fn common_sorted_list(nums [][]int) []int {
    mut elements, mut output := map[int]bool{}, []int{}
	
    for num in nums {
        for value in num {
            elements[value] = true
        }
    }

    for key, _ in elements {
        output << key
    }
    output.sort()
    return output
}

Output:

Distinct sorted union of [[5 1 3 8 9 4 8 7] [3 5 9 8 4] [1 3 7 9]] is:
[1 3 4 5 7 8 9]

Wren

Library: Wren-seq

Library: Wren-sort

import "./seq" for Lst
import "./sort" for Sort

var distinctSortedUnion = Fn.new { |ll| 
    var res = ll.reduce([]) { |acc, l| acc + l }
    res = Lst.distinct(res)
    Sort.insertion(res)
    return res
}

var ll = [[5, 1, 3, 8, 9, 4, 8, 7], [3, 5, 9, 8, 4], [1, 3, 7, 9]]
System.print("Distinct sorted union of %(ll) is:")
System.print(distinctSortedUnion.call(ll))

Output:

Distinct sorted union of [[5, 1, 3, 8, 9, 4, 8, 7], [3, 5, 9, 8, 4], [1, 3, 7, 9]] is:
[1, 3, 4, 5, 7, 8, 9]