How to find objects that are in several arrays using jQuery

Question

I have five arrays that can contain the same value across them.

var arr1 = ['1','2','3','4'];
var arr2 = ['1','3']
var arr3 = ['1','2','3']
var arr4 = ['1','2','3','4','5'];
var arr5 = ['1','3','4']

How can I find all of the numbers that are in all of the arrays using jQuery?

The numbers 1 and 3 are in all of the arrays?

You mean like [array intersection](http://stackoverflow.com/questions/1885557/simplest-code-for-array-intersection-in-javascript)? — Stryner, Sep 23 '15 at 18:11

Andi AR · Accepted Answer · 2015-09-23T18:26:17.103

3

Check this out

Filtering array by duplicate elements

JsFiddle http://jsfiddle.net/nWjcp/87/

var arr1 = ['1','2','3','4'];
var arr2 = ['1','3']
var arr3 = ['1','2','3']
var arr4 = ['1','2','3','4','5'];
var arr5 = ['1','3','4']

var arrays = [
   arr1,arr2,arr3,arr4,arr5
    ];


var result = arrays.shift().filter(function(v) { // Filtering
    return arrays.every(function(a) {       // Seek duplicate
        return a.indexOf(v) !== -1;    
    });
});




   alert( JSON.stringify(result,null,4) ); // ['1','3']

edited Sep 23 '15 at 18:26

answered Sep 23 '15 at 18:14

Andi AR

2,678
2
23
28

This is the better way to go. – Etheryte Sep 23 '15 at 19:17
Why the null and 4 ? To make things harder to understand ? does it contribute something to the answer ? not to mention that every+filter+indexOf are all need polyfills – Royi Namir Sep 23 '15 at 19:50
1

@RoyiNamir Only if you need to support IE8, in which case you need a polyfill for pretty much everything. – Etheryte Sep 29 '15 at 15:36

score 2 · Answer 2 · answered Sep 23 '15 at 18:12

First off, let's make a more manageable dataset:

var arrays = [
    ['1', '2', '3', '4'],
    ['1', '3'],
    ['1', '2', '3'],
    ['1', '2', '3', '4', '5'],
    ['1', '3', '4']
];

Then we'll want to get all the unique elements:

var elements = [].concat.apply([], arrays).filter(function(value, index, self) {
    return self.indexOf(value) === index;
}); //["1", "2", "3", "4", "5"]

And then finally, check this set against each of our arrays, keeping only the elements that are present in all of them:

var out = elements.filter(function(item) {
    return arrays.reduce(function(present, array) {
        present = present && (array.indexOf(item) !== -1);
        return present;
    }, true);
}); //["1", "3"]

Royi Namir · Answer 3 · 2015-09-23T20:20:36.757

I don't like the usage of methods without crossbrowser support :

Here is a simple solution without "extra" non-jquery-included functions (every , inArray,filter(array)):

var arr1 = ['1','2','3','4'];
var arr2 = ['1','3']
var arr3 = ['1','2','3']
var arr4 = ['1','2','3','4','5'];
var arr5 = ['1','3','4']

var arrs = [arr1,arr2,arr3,arr4,arr5];

var obj={};

$.each(arrs,function (i,arr){
     $.each(arr,function (j,n){
      obj[n]=(+obj[n] || 0) + 1;
     });
});


for (item in obj) 
{  
     if(obj.hasOwnProperty(item ) && obj[item]==arrs.length) console.log(item) //1,3
}

score 0 · Answer 4 · answered Sep 23 '15 at 18:30

Find a long solution here. At first, let us find the array with least items so that search can be optimized:

var arr1 = ['1','2','3','4'];
var arr2 = ['1','3', '8'];
var arr3 = ['1','2','3'];
var arr4 = ['1','2','3','4','5'];
var arr5 = ['1','3','4'];

var arrayOfArray = [arr1, arr2, arr3, arr4, arr5];
var arrayWithMin = null;
var outputArray = [];
$.each(arrayOfArray, function(index, arrayItem) {
    arrayWithMin = (arrayWithMin == null) ? arrayItem : (arrayWithMin.length > arrayItem.length ? arrayItem : arrayWithMin);
});

$.each(arrayWithMin, function(i, searchItem) {
    if($.inArray(searchItem, arr1) > -1 
       && $.inArray(searchItem, arr2) > -1 
       && $.inArray(searchItem, arr3) > -1 
       && $.inArray(searchItem, arr4) > -1
       && $.inArray(searchItem, arr5) > -1) {
        outputArray.push(searchItem);
    }
});

document.write(outputArray.join(', '));

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

How to find objects that are in several arrays using jQuery

4 Answers4

Filtering array by duplicate elements