Is This the Best Way to Grab Common Elements from a Hash of Arrays

Is this the best way to grab common elements from a Hash of arrays?

Behold the power of inject! ;)

[[1,2,3],[1,3,5],[1,5,6]].inject(&:&)
=> [1]

As Jordan mentioned, if your version of Ruby lacks support for &-notation, just use

inject{|acc,elem| acc & elem}

How do I breakdown common elements in hash of arrays in perl?

This can be done as two simple hash conversions:

• Build a hash that lists all of the lots each item is in

• Convert that to a hash that lists all items for each lot combination

Then just dump the last hash in a convenient form

This is the code.

use strict;
use warnings 'all';
use feature 'say';

my %test = (
    Lot1 => [ "A", "B", "C" ],
    Lot2 => [ "A", "B", "C" ],
    Lot3 => ["C"],
    Lot4 => [ "E", "F" ],
);

my %items;

for my $lot ( keys %test ) {
    for my $item ( @{ $test{$lot} } ) {
        push @{ $items{$item} }, $lot;
    }
}

my %lots;

for my $item ( keys %items ) {
    my $lots = join '!', sort @{ $items{$item} };
    push @{ $lots{$lots} }, $item;
}

for my $lots ( sort keys %lots ) {

    my @lots = split /!/, $lots;
    my $items = join '', @{ $lots{$lots} };

    $lots = join ', ', @lots;
    $lots =~ s/.*\K,/ and/;

    printf "%s %s %s\n", $lots, @lots > 1 ? 'have' : 'has', $items;
}

output

Lot1 and Lot2 have AB
Lot1, Lot2 and Lot3 have C
Lot4 has EF

It generates an %items hash that looks like this

{
  A => ["Lot2", "Lot1"],
  B => ["Lot2", "Lot1"],
  C => ["Lot2", "Lot3", "Lot1"],
  E => ["Lot4"],
  F => ["Lot4"],
}

and from that a %lots hash that looks like this

{
  "Lot1!Lot2" => ["A", "B"],
  "Lot1!Lot2!Lot3" => ["C"],
  "Lot4" => ["E", "F"],
}

How to use HashSet to find common elements in two Comparable arrays?

EDIT I forgot to mention that I imported the Apache Commons Array Utils. Very useful.

I figured it out. Thanks for all your help. I have a main method that calls an instance of the class 3 times, and 3 test methods, but those are irrelevant. Here is what was giving me trouble, and now it works. :-)

public int getComparisons() {
        return comparisons;
    } //method to return number of comparisons
    public static Comparable[] findCommonElements(Comparable[][] collections) {
        /*
        I LEARNED THAT WE HAD TO USE MORE THAN TWO ARRAYS, SO IT WAS BACK
        TO THE DRAWING BOARD FOR ME. I FIGURED IT OUT, THOUGH.
        */
        Comparable[] arr1 = collections[0]; //set initial values to 1 Dimensional arrays so the test methods can read their respective values
        Comparable[] arr2 = collections[1];
        Comparable[] arr3 = collections[2];

        /*
        THE FOLLOWING BLOCK OF CODE TAKES ALL THE PERMUTATIONS OF THE 3 ARRAYS (i.e. 1,2,3; 1,3,2; 2,1,3, etc),
        DETERMINES WHICH ARRAY IS THE SHORTEST, AND ADDS THE LONGER TWO ARRAYS TO A QUERY ARRAY.
         */
        if(arr1.length < arr2.length && arr1.length < arr3.length || arr2.length <= arr3.length) { //shortest array will become hash array. the other two will become a combined query array.
            hashArray = arr1;  //these will be utilized below to put into Sets
            queryArray = ArrayUtils.addAll(arr2, arr3);
        }
        else if(arr2.length < arr1.length && arr2.length < arr3.length || arr1.length <= arr3.length) {
            hashArray = arr2;
            queryArray = ArrayUtils.addAll(arr1, arr3);
        }
        else if(arr3.length < arr1.length && arr3.length < arr2.length || arr1.length <= arr2.length) {
            hashArray = arr3;
            queryArray = ArrayUtils.addAll(arr1, arr2);
        }
        HashSet<Comparable> intersectionSet = new HashSet<>(); //initialize Sets
        HashSet<Comparable> arrayToHash = new HashSet<>();
        for(Comparable element : hashArray) { //add shorter array to hashedArray Set
            arrayToHash.add(element);
        }
        //NOTE FROM THE JAVADOC ON THE IMPLEMENTATION OF .contains() USING HASHSET COMPARISONS
        /**
         * <p>This class offers constant time performance for the basic operations
         * (<tt>add</tt>, <tt>remove</tt>, <tt>contains</tt> and <tt>size</tt>),
         * assuming the hash function disperses the elements properly among the
         * buckets.
         */
        for(Comparable element : queryArray) {
            if(element != null) {
                comparisons++; // increment comparisons with each search
            }
            if(arrayToHash.contains(element)) { //search for matches and add to intersectionSet (.contains uses the equals method to determine if an object is within array)
                intersectionSet.add(element);
            }
        }
        return intersectionSet.toArray(new Comparable[0]); //return Set as Array defined in method signature
    }

How do I find common elements from n arrays

I advocate for hashes in such cases: you'll have time proportional to common size of both arrays.

Since most major languages offer hashtable in their standard libraries, I hardly need to show your how to implement such solution.

Compare arrays in Ruby and print common elements

Hmmm... Not that I'm a judge, by why not use the Array#& method?

Set Intersection — Returns a new array containing elements common to the two arrays, excluding any duplicates. The order is preserved from the original array.

arr1 = [1, 2, 3, 4, 5]
arr2 = [1, 3, 5, 7, 9]
arr1 & arr2 # => [1, 3, 5]

How do I find common elements from n arrays

I advocate for hashes in such cases: you'll have time proportional to common size of both arrays.

Since most major languages offer hashtable in their standard libraries, I hardly need to show your how to implement such solution.

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