JavaScript function that takes a multidimensional array and a single array, and finds matches of the single array in the multi-d array

Question

I tried looking at a similar question in stack overflow but it didn't cater to exactly what I'm trying to do and I keep trying to make it work but continuously get the wrong results. At this point I'm simply at a loss. Example of what I mean

var masterArray = [[1,2,5,6],[5,13,7,8],[9,11,13,15],[13,14,15,16],[1,9,11,12]]
var compareArray = [9,11,13,15,1,2,5,6]

function intersect(a, b) {
//code to compare both arrays and find a match
}

console.log(Intersect(compareArray, masterArray))

And the output would be

[9,11,13,15]
[1,2,5,6]
[5,13]
[13,15]

Show (add to your question) your best effort so far and explain where and why you are stuck. — PM 77-1
– PM 77-1, Commented Feb 11, 2017 at 19:49
Sorry, I missed typing that one, but yes. [1,9,11] would be one of the outputs as well. — Jovanny
– Jovanny, Commented Feb 12, 2017 at 16:00

ibrahim mahrir · Accepted Answer · 2017-02-11 20:02:36Z

1

Use Array.prototype.reduce to get an array of all the intersections like this:

var masterArray = [[1,2,5,6],[5,13,7,8],[9,11,13,15],[13,14,15,16],[1,9,11,12]];
var compareArray = [9,11,13,15,1,2,5,6];


function intersect(multi, simple) {
  return multi.reduce(function(res, b) {
    var intersection = simple.reduce(function(r, e) { // get the intersection of the current array and compareArray (simple)
      if(b.indexOf(e) != -1) // if the current element of the current array is also in the compareArray then push it into the intersection array
        r.push(e);
      return r;
    }, []);
    
    res.push(intersection); // push the intersection array into the result array
    return res;
  }, []);
}

console.log(intersect(masterArray, compareArray));

edited Feb 11, 2017 at 20:02

answered Feb 11, 2017 at 19:57

ibrahim mahrir

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RomanPerekhrest · Accepted Answer · 2017-02-11 21:31:41Z

1

The solution using RegExp object with specific regex pattern(converting compareArray numbers into regex alternation group items).
Additional functions used: String.prototype.match(), Array.prototype.join(), Array.prototype.map()

var masterArray = [[1,2,5,6],[5,13,7,8],[9,11,13,15],[13,14,15,16],[1,9,11,12]],
    compareArray = [9,11,13,15,1,2,5,6];

function intersect(compareArray, masterArray) {
    var pattern = new RegExp('\\b(' + compareArray.join('|') + ')\\b', 'g'),
        result = [];

    masterArray.forEach(function(v) {
        var matches = v.join(' ').match(pattern);
        if (matches.length) result.push(matches.map(Number));
    });

    return result;
}

console.log(intersect(compareArray, masterArray));

edited Feb 11, 2017 at 21:31

answered Feb 11, 2017 at 20:22

RomanPerekhrest

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Rahul · Accepted Answer · 2017-02-11 19:56:26Z

0

As per your logic, you have wrong output in your question :

var masterArray = [[1,2,5,6],[5,13,7,8],[9,11,13,15],[13,14,15,16],[1,9,11,12]]
var compareArray = [9,11,13,15,1,2,5,6]


function intersect(a, b) {
  var temp = [];
   k = 0;
   for(i = 0; i < b.length; i++){
      temp1 = []
      for(j=0;j<b[i].length;j++){
         if($.inArray(b[i][j], a) > -1){
            temp1.push(b[i][j]);

         }
      }
      temp.push(temp1);
   }
   return temp;
}

console.log(intersect(compareArray, masterArray));

Check this jsfiddle.

Give it a try, this should work.

answered Feb 11, 2017 at 19:56

Rahul

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Comments

trincot · Accepted Answer · 2017-02-11 20:50:58Z

0

You could use this ES6 function:

function intersect(a, b) {
    a = new Set(a); // for faster lookup
    return b.map(c => c.filter(a.has.bind(a))).filter(Boolean);
}
// Demo
const masterArray = [[1,2,5,6],[5,13,7,8],[9,11,13,15],[13,14,15,16],[1,9,11,12]];
const compareArray = [9,11,13,15,1,2,5,6];
console.log(intersect(compareArray, masterArray).join('\n'));

The .filter(Boolean) part is optional, and only needed to exclude arrays that have no matching numbers at all, which otherwise would be represented by empty arrays in the result.

Using Set will improve the performance, as lookup in a Set (with has) can be done in near-constant time.

answered Feb 11, 2017 at 20:50

trincot

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Nina Scholz · Accepted Answer · 2017-02-11 20:52:48Z

You could map the filtered result of the masterArray.

var masterArray = [[1, 2, 5, 6], [5, 13, 7, 8], [9, 11, 13, 15], [13, 14, 15, 16], [1, 9, 11, 12]],
    compareArray = [9, 11, 13, 15, 1, 2, 5, 6],
    result = masterArray.map(function (a) {
        return a.filter(function (b) {
            return compareArray.indexOf(b) !== -1;
        });
    });

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

ES6

var masterArray = [[1, 2, 5, 6], [5, 13, 7, 8], [9, 11, 13, 15], [13, 14, 15, 16], [1, 9, 11, 12]],
    compareArray = [9, 11, 13, 15, 1, 2, 5, 6],
    result = masterArray.map(a => a.filter(b => compareArray.includes(b)));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

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