Jaccard index

From Rosetta Code
Jaccard index 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.
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The Jaccard index, also known as the Jaccard similarity coefficient, is a statistic used for gauging the similarity and diversity of sample sets. It was developed by Paul Jaccard, originally giving the French name coefficient de communauté, and independently formulated again by T. Tanimoto. Thus, the Tanimoto index or Tanimoto coefficient are also used in some fields. However, they are identical in generally taking the ratio of Intersection over Union. The Jaccard coefficient measures similarity between finite sample sets, and is defined as the size of the intersection divided by the size of the union of the sample sets:

J(A, B) = |A ∩ B|/|A ∪ B|

Define sets as follows, using any linear data structure:

A = {}
B = {1, 2, 3, 4, 5}
C = {1, 3, 5, 7, 9}
D = {2, 4, 6, 8, 10}
E = {2, 3, 5, 7}
F = {8}

Write a program that computes the Jaccard index for every ordered pairing (to show that J(A, B) and J(B, A) are the same) of these sets, including self-pairings.

APL

<lang apl>task←{

   jaccard ← (≢∩)÷(≢∪)
   A ← ⍬
   B ← 1 2 3 4 5
   C ← 1 3 5 7 9
   D ← 2 4 6 8 10
   E ← 2 3 5 7
   F ← ,8
   '.ABCDEF' ⍪ 'ABCDEF' , ∘.jaccard⍨ A B C D E F

}</lang>

Output:
. A            B            C     D     E   F
A 1 0            0            0     0     0  
B 0 1            0.4285714286 0.25  0.5   0  
C 0 0.4285714286 1            0     0.5   0  
D 0 0.25         0            1     0.125 0.2
E 0 0.5          0.5          0.125 1     0  
F 0 0            0            0.2   0     1  

BQN

<lang bqn>Jaccard ← ≡◶⟨∊ ÷○(+´) ∊∘∾, 1⟩

a ← ⟨⟩ b ← ⟨1,2,3,4,5⟩ c ← ⟨1,3,5,7,9⟩ d ← ⟨2,4,6,8,10⟩ e ← ⟨2,3,5,7⟩ f ← ⟨8⟩

Jaccard⌜˜ ⟨a,b,c,d,e,f⟩</lang>

Output:
┌─                                                           
╵ 1                   0                   0     0     0   0  
  0                   1 0.42857142857142855  0.25   0.5   0  
  0 0.42857142857142855                   1     0   0.5   0  
  0                0.25                   0     1 0.125 0.2  
  0                 0.5                 0.5 0.125     1   0  
  0                   0                   0   0.2     0   1  
                                                            ┘

Factor

Works with: Factor version 0.99 2021-06-02

<lang factor>USING: assocs formatting grouping kernel math math.combinatorics prettyprint sequences sequences.repeating sets ;

jaccard ( seq1 seq2 -- x )
   2dup [ empty? ] both? [ 2drop 1 ]
   [ [ intersect ] [ union ] 2bi [ length ] bi@ / ] if ;

{ { } { 1 2 3 4 5 } { 1 3 5 7 9 } { 2 4 6 8 10 } { 2 3 5 7 } { 8 } } [ 2 <combinations> ] [ 2 repeat 2 group append ] bi [ 2dup jaccard "%u %u -> %u\n" printf ] assoc-each</lang>

Output:
{ } { 1 2 3 4 5 } -> 0
{ } { 1 3 5 7 9 } -> 0
{ } { 2 4 6 8 10 } -> 0
{ } { 2 3 5 7 } -> 0
{ } { 8 } -> 0
{ 1 2 3 4 5 } { 1 3 5 7 9 } -> 3/7
{ 1 2 3 4 5 } { 2 4 6 8 10 } -> 1/4
{ 1 2 3 4 5 } { 2 3 5 7 } -> 1/2
{ 1 2 3 4 5 } { 8 } -> 0
{ 1 3 5 7 9 } { 2 4 6 8 10 } -> 0
{ 1 3 5 7 9 } { 2 3 5 7 } -> 1/2
{ 1 3 5 7 9 } { 8 } -> 0
{ 2 4 6 8 10 } { 2 3 5 7 } -> 1/8
{ 2 4 6 8 10 } { 8 } -> 1/5
{ 2 3 5 7 } { 8 } -> 0
{ } { } -> 1
{ 1 2 3 4 5 } { 1 2 3 4 5 } -> 1
{ 1 3 5 7 9 } { 1 3 5 7 9 } -> 1
{ 2 4 6 8 10 } { 2 4 6 8 10 } -> 1
{ 2 3 5 7 } { 2 3 5 7 } -> 1
{ 8 } { 8 } -> 1

Julia

<lang julia>J(A, B) = begin i, u = length(A ∩ B), length(A ∪ B); u == 0 ? 1//1 : i // u end

A = Int[] B = [1, 2, 3, 4, 5] C = [1, 3, 5, 7, 9] D = [2, 4, 6, 8, 10] E = [2, 3, 5, 7] F = [8] testsets = [A, B, C, D, E, F]

println("Set A Set B J(A, B)\n", "-"^44) for a in testsets, b in testsets

   println(rpad(isempty(a) ? "[]" : a, 18), rpad(isempty(b) ? "[]" : b, 18),
       replace(string(J(a, b)), "//" => "/"))

end

</lang>

Output:
Set A             Set B             J(A, B)
--------------------------------------------
[]                []                1/1
[]                [1, 2, 3, 4, 5]   0/1
[]                [1, 3, 5, 7, 9]   0/1
[]                [2, 4, 6, 8, 10]  0/1
[]                [2, 3, 5, 7]      0/1
[]                [8]               0/1
[1, 2, 3, 4, 5]   []                0/1
[1, 2, 3, 4, 5]   [1, 2, 3, 4, 5]   1/1
[1, 2, 3, 4, 5]   [1, 3, 5, 7, 9]   3/7
[1, 2, 3, 4, 5]   [2, 4, 6, 8, 10]  1/4
[1, 2, 3, 4, 5]   [2, 3, 5, 7]      1/2
[1, 2, 3, 4, 5]   [8]               0/1
[1, 3, 5, 7, 9]   []                0/1
[1, 3, 5, 7, 9]   [1, 2, 3, 4, 5]   3/7
[1, 3, 5, 7, 9]   [1, 3, 5, 7, 9]   1/1
[1, 3, 5, 7, 9]   [2, 4, 6, 8, 10]  0/1
[1, 3, 5, 7, 9]   [2, 3, 5, 7]      1/2
[1, 3, 5, 7, 9]   [8]               0/1
[2, 4, 6, 8, 10]  []                0/1
[2, 4, 6, 8, 10]  [1, 2, 3, 4, 5]   1/4
[2, 4, 6, 8, 10]  [1, 3, 5, 7, 9]   0/1
[2, 4, 6, 8, 10]  [2, 4, 6, 8, 10]  1/1
[2, 4, 6, 8, 10]  [2, 3, 5, 7]      1/8
[2, 4, 6, 8, 10]  [8]               1/5
[2, 3, 5, 7]      []                0/1
[2, 3, 5, 7]      [1, 2, 3, 4, 5]   1/2
[2, 3, 5, 7]      [1, 3, 5, 7, 9]   1/2
[2, 3, 5, 7]      [2, 4, 6, 8, 10]  1/8
[2, 3, 5, 7]      [2, 3, 5, 7]      1/1
[2, 3, 5, 7]      [8]               0/1
[8]               []                0/1
[8]               [1, 2, 3, 4, 5]   0/1
[8]               [1, 3, 5, 7, 9]   0/1
[8]               [2, 4, 6, 8, 10]  1/5
[8]               [2, 3, 5, 7]      0/1
[8]               [8]               1/1

Phix

with javascript_semantics
include sets.e

function jaccard(sequence a, b)
    integer i = length(intersection(a,b)),
            u = length(union(a,b))
    return iff(u=0?1:i/u)
end function

constant tests = {{},               -- A
                  {1, 2, 3, 4, 5},  -- B
                  {1, 3, 5, 7, 9},  -- C
                  {2, 4, 6, 8, 10}, -- D
                  {2, 3, 5, 7},     -- E
                  {8}}              -- F

for i=1 to length(tests) do
    for j=i to length(tests) do
        string s = sprintf("J(%c,%c)",{'A'+i-1,'A'+j-1})
        atom jij = jacard(tests[i],tests[j])
        if i!=j then
            atom jji = jacard(tests[j],tests[i])
            assert(jji==jij)
            s &= sprintf(" = J(%c,%c)",{'A'+j-1,'A'+i-1})
        end if
        printf(1,"%s = %g\n",{s,jij})
    end for
end for
Output:
J(A,A) = 1
J(A,B) = J(B,A) = 0
J(A,C) = J(C,A) = 0
J(A,D) = J(D,A) = 0
J(A,E) = J(E,A) = 0
J(A,F) = J(F,A) = 0
J(B,B) = 1
J(B,C) = J(C,B) = 0.428571
J(B,D) = J(D,B) = 0.25
J(B,E) = J(E,B) = 0.5
J(B,F) = J(F,B) = 0
J(C,C) = 1
J(C,D) = J(D,C) = 0
J(C,E) = J(E,C) = 0.5
J(C,F) = J(F,C) = 0
J(D,D) = 1
J(D,E) = J(E,D) = 0.125
J(D,F) = J(F,D) = 0.2
J(E,E) = 1
J(E,F) = J(F,E) = 0
J(F,F) = 1

Perl

<lang perl>#!/usr/bin/perl

use strict; use warnings;

my %sets = (

 A => [],
 B => [1, 2, 3, 4, 5],
 C => [1, 3, 5, 7, 9],
 D => [2, 4, 6, 8, 10],
 E => [2, 3, 5, 7],
 F => [8],
 );

use Data::Dump 'dd'; dd \%sets;

for my $left (sort keys %sets )

 {
 for my $right (sort keys %sets )
   {
   my %union;
   $union{ $_ }++ for @{ $sets{$left} }, @{ $sets{$right} };
   print "J($left,$right) = ",
     %union ? (grep $_ == 2, values %union) / (keys %union) : 1, "\n";
   }
 }</lang>
Output:
{
  A => [],
  B => [1 .. 5],
  C => [1, 3, 5, 7, 9],
  D => [2, 4, 6, 8, 10],
  E => [2, 3, 5, 7],
  F => [8],
}
J(A,A) = 1
J(A,B) = 0
J(A,C) = 0
J(A,D) = 0
J(A,E) = 0
J(A,F) = 0
J(B,A) = 0
J(B,B) = 1
J(B,C) = 0.428571428571429
J(B,D) = 0.25
J(B,E) = 0.5
J(B,F) = 0
J(C,A) = 0
J(C,B) = 0.428571428571429
J(C,C) = 1
J(C,D) = 0
J(C,E) = 0.5
J(C,F) = 0
J(D,A) = 0
J(D,B) = 0.25
J(D,C) = 0
J(D,D) = 1
J(D,E) = 0.125
J(D,F) = 0.2
J(E,A) = 0
J(E,B) = 0.5
J(E,C) = 0.5
J(E,D) = 0.125
J(E,E) = 1
J(E,F) = 0
J(F,A) = 0
J(F,B) = 0
J(F,C) = 0
J(F,D) = 0.2
J(F,E) = 0
J(F,F) = 1

Prolog

<lang prolog> show([]). show([X|Xs]):- write(X), show(Xs).

j(N,M,X):- M > 0 -> X is N/M; X is 1.

task:- L = [[], [1,2,3,4,5], [1,3,5,7,9], [2,4,6,8,10], [2,3,5,7], [8]],

   forall((member(A,L), member(B,L)), (
       findall(X, (member(X,A), member(X,B)), I), length(I,N),
       findall(X, (member(X,B), not(member(X,A))), T), append(A,T,U), length(U,M),
       j(N,M,J), show(["A = ",A,", B = ",B,", J = ",J]), nl)).

</lang>

Output:
?- task. 
A = [], B = [], J = 1
A = [], B = [1,2,3,4,5], J = 0
A = [], B = [1,3,5,7,9], J = 0
A = [], B = [2,4,6,8,10], J = 0
A = [], B = [2,3,5,7], J = 0
A = [], B = [8], J = 0
A = [1,2,3,4,5], B = [], J = 0
A = [1,2,3,4,5], B = [1,2,3,4,5], J = 1
A = [1,2,3,4,5], B = [1,3,5,7,9], J = 0.42857142857142855
A = [1,2,3,4,5], B = [2,4,6,8,10], J = 0.25
A = [1,2,3,4,5], B = [2,3,5,7], J = 0.5
A = [1,2,3,4,5], B = [8], J = 0
A = [1,3,5,7,9], B = [], J = 0
A = [1,3,5,7,9], B = [1,2,3,4,5], J = 0.42857142857142855
A = [1,3,5,7,9], B = [1,3,5,7,9], J = 1
A = [1,3,5,7,9], B = [2,4,6,8,10], J = 0
A = [1,3,5,7,9], B = [2,3,5,7], J = 0.5
A = [1,3,5,7,9], B = [8], J = 0
A = [2,4,6,8,10], B = [], J = 0
A = [2,4,6,8,10], B = [1,2,3,4,5], J = 0.25
A = [2,4,6,8,10], B = [1,3,5,7,9], J = 0
A = [2,4,6,8,10], B = [2,4,6,8,10], J = 1
A = [2,4,6,8,10], B = [2,3,5,7], J = 0.125
A = [2,4,6,8,10], B = [8], J = 0.2
A = [2,3,5,7], B = [], J = 0
A = [2,3,5,7], B = [1,2,3,4,5], J = 0.5
A = [2,3,5,7], B = [1,3,5,7,9], J = 0.5
A = [2,3,5,7], B = [2,4,6,8,10], J = 0.125
A = [2,3,5,7], B = [2,3,5,7], J = 1
A = [2,3,5,7], B = [8], J = 0
A = [8], B = [], J = 0
A = [8], B = [1,2,3,4,5], J = 0
A = [8], B = [1,3,5,7,9], J = 0
A = [8], B = [2,4,6,8,10], J = 0.2
A = [8], B = [2,3,5,7], J = 0
A = [8], B = [8], J = 1
true.

Raku

<lang perl6>sub J(\A, \B) { A ∪ B ?? (A ∩ B) / (A ∪ B) !! A ∪ B == A ∩ B ?? 1 !! 0 }

my %p =

 A => < >,
 B => <1 2 3 4 5>,
 C => <1 3 5 7 9>,
 D => <2 4 6 8 10>,
 E => <2 3 5 7>,
 F => <8>,

.say for %p.sort; say ; say "J({.join: ','}) = ", J |%p{$_} for [X] <A B C D E F> xx 2;</lang>

Output:
A => ()
B => (1 2 3 4 5)
C => (1 3 5 7 9)
D => (2 4 6 8 10)
E => (2 3 5 7)
F => 8

J(A,A) = 1
J(A,B) = 0
J(A,C) = 0
J(A,D) = 0
J(A,E) = 0
J(A,F) = 0
J(B,A) = 0
J(B,B) = 1
J(B,C) = 0.428571
J(B,D) = 0.25
J(B,E) = 0.5
J(B,F) = 0
J(C,A) = 0
J(C,B) = 0.428571
J(C,C) = 1
J(C,D) = 0
J(C,E) = 0.5
J(C,F) = 0
J(D,A) = 0
J(D,B) = 0.25
J(D,C) = 0
J(D,D) = 1
J(D,E) = 0.125
J(D,F) = 0.2
J(E,A) = 0
J(E,B) = 0.5
J(E,C) = 0.5
J(E,D) = 0.125
J(E,E) = 1
J(E,F) = 0
J(F,A) = 0
J(F,B) = 0
J(F,C) = 0
J(F,D) = 0.2
J(F,E) = 0
J(F,F) = 1

Wren

Library: Wren-set
Library: Wren-trait
Library: Wren-fmt

Note that the Set object in the above module is implemented as a Map and consequently the iteration order (and the order in which elements are printed) is undefined. <lang ecmascript>import "./set" for Set import "./trait" for Indexed import "./fmt" for Fmt

var jaccardIndex = Fn.new { |a, b|

   if (a.count == 0 && b.count == 0) return 1
   return a.intersect(b).count / a.union(b).count

}

var a = Set.new([]) var b = Set.new([1, 2, 3, 4, 5]) var c = Set.new([1, 3, 5, 7, 9]) var d = Set.new([2, 4, 6, 8, 10]) var e = Set.new([2, 3, 5, 7]) var f = Set.new([8]) var isets = Indexed.new([a, b, c, d, e, f]) for (se in isets) {

   var i = String.fromByte(se.index + 65)
   var v = se.value
   v = v.toList.sort() // force original sorted order
   Fmt.print("$s = $n", i, v)

} System.print() for (se1 in isets) {

   var i1 = String.fromByte(se1.index + 65)
   var v1 = se1.value
   for (se2 in isets) {
       var i2 = String.fromByte(se2.index + 65)
       var v2 = se2.value
       Fmt.print("J($s, $s) = $h", i1, i2, jaccardIndex.call(v1, v2))
   }

}</lang>

Output:
A = []
B = [1, 2, 3, 4, 5]
C = [1, 3, 5, 7, 9]
D = [2, 4, 6, 8, 10]
E = [2, 3, 5, 7]
F = [8]

J(A, A) = 1       
J(A, B) = 0       
J(A, C) = 0       
J(A, D) = 0       
J(A, E) = 0       
J(A, F) = 0       
J(B, A) = 0       
J(B, B) = 1       
J(B, C) = 0.428571
J(B, D) = 0.25    
J(B, E) = 0.5     
J(B, F) = 0       
J(C, A) = 0       
J(C, B) = 0.428571
J(C, C) = 1       
J(C, D) = 0       
J(C, E) = 0.5     
J(C, F) = 0       
J(D, A) = 0       
J(D, B) = 0.25    
J(D, C) = 0       
J(D, D) = 1       
J(D, E) = 0.125   
J(D, F) = 0.2     
J(E, A) = 0       
J(E, B) = 0.5     
J(E, C) = 0.5     
J(E, D) = 0.125   
J(E, E) = 1       
J(E, F) = 0       
J(F, A) = 0       
J(F, B) = 0       
J(F, C) = 0       
J(F, D) = 0.2     
J(F, E) = 0       
J(F, F) = 1