CODEWITHSUNDEEP
arraysangularjsmultidimensional-arrayunderscore.js
Gsk
Gsk

Reputation: 193

How to find repeated values in multidimensional array in angularjs

var arrayValues = [[2,3,5],[3,5]]
var commonArrayValues = _.intersection(arrayValues);

Currently it is working as, 

_.intersection([[2,3,5],[3,5]])
    Result: [2,3,5]

But it should work as, (i.e outer array should be removed)

_.intersection([2,3,5],[3,5])
    Expected Result: [3,5]

Anyone kindly give me a proper solutions. Thank you in advance.

Upvotes: 4

Views: 339

Answers (3)

Gruff Bunny
Gruff Bunny

Reputation: 27986

You can use apply with intersection to get what you want:

var result = _.intersection.apply(null, arrayValues);


var arrayValues = [[2,3,5],[3,5], [2,3,5,6]]

var result = _.intersection.apply(null, arrayValues);

document.getElementById('results').textContent = JSON.stringify(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.2/underscore.js"></script>

<pre id="results"></pre>

Upvotes: 2

Astaroth
Astaroth

Reputation: 783

The only way I can think of is using eval:

var arrayValues = [[2,3,5],[3,5]]
var evalString = '_.intersection(';

arrayValues.forEach(function (element, index){
  evalString += 'arrayValues['+index+'],';
});

evalString =evalString.slice(0, -1);
evalString += ');'
eval(evalString);

evalString would end being something like _.intersection(arrayValues[0],arrayValues[1],...,arrayValues[n]);

Upvotes: 0

Praveen Prasannan
Praveen Prasannan

Reputation: 7133

intersection *_.intersection(arrays)
Computes the list of values that are the intersection of all the arrays. Each value in the result is present in each of the arrays.

_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]); => [1, 2]

var arrayValues = [[2,3,5],[3,5]]

Here arrayValues is an array having 2 arrays. Where as _.intersection expects arrays as parameter and not an array having arrays.

_.intersection([2,3,5],[3,5])

Or

_.intersection(arrayValues[0],arrayValues[1])

will output as what you need.

Upvotes: 0

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