Find and remove object from array of objects

Question

I have 2 arrays of objects.
Array1:

[{ id: 2, ref: 1010101 },
 { id: 2, ref: 1010107 }]

Array2:

[{ id: 2, ref: 10010001 },
 { id: 2, ref: 10010002 },
 { id: 2, ref: 10010003 },
 { id: 2, ref: 10010006 },
 { id: 2, ref: 10010007 },
 { id: 2, ref: 10010008 },
 { id: 2, ref: 10010009 },
 { id: 2, ref: 10020005 },
 { id: 2, ref: 1010101 }]

I need to do 2 tasks:
1. Find what elements of array1 are in array2 . After find Items, Im going to do something with that elements.
2. Remove elements found from array2. After I do something with my found items, I need to remove each item found.

I was trying to do it with lodash but without success.

I need something like this:

if(elementFound) {
  //do something
  doSomething(item);
  //after it, remove from array2
  removeItemFromArray2(item);
}

Thanks, I think there is an easy way to do it with lodash.

Possible duplicate of Compare two Javascript Arrays and remove Duplicates — Mike Cluck
– Mike Cluck, Commented Mar 2, 2016 at 20:07
Since two objects with the same key/values aren't the same, you can only use indexOf if you stringify each object first. Depending on the size of the array, comparing strings and then slicing out the index from the original array might outperform using a nested filter which will do multiple loops. — Shilly
– Shilly, Commented Mar 2, 2016 at 20:17

Rajaprabhu Aravindasamy · Accepted Answer · 2016-03-02 20:09:10Z

2

Try to use Array.prototype.filter at this context to achieve what you want,

var filteredArray = Array2.filter(function(itm){
 return Array1.filter(function(itmIn){
   return itmIn.id === itm.id && itmIn.ref === itm.ref;
 }).length == 0;
});

answered Mar 2, 2016 at 20:09

Rajaprabhu Aravindasamy

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1 Comment

Baumannzone Over a year ago

what is length == 0 for? I dont get it

Shanoor · Accepted Answer · 2016-03-03 02:42:57Z

2

There is one specific lodash function for this:

var array1 = [
  { id: 2, ref: 1010101 },
  { id: 2, ref: 1010107 }
];

var array2 = [
  { id: 2, ref: 10010001 },
  { id: 2, ref: 10010002 },
  { id: 2, ref: 10010003 },
  { id: 2, ref: 10010006 },
  { id: 2, ref: 10010007 },
  { id: 2, ref: 10010008 },
  { id: 2, ref: 10010009 },
  { id: 2, ref: 10020005 },
  { id: 2, ref: 1010101 }
];

var res = _.differenceBy(array2, array1, 'ref');

document.write("<pre>" + JSON.stringify(res, null, 3) + "</pre>")

<script src="https://cdn.jsdelivr.net/lodash/4.6.1/lodash.min.js"></script>

answered Mar 3, 2016 at 2:42

Shanoor

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1 Comment

Baumannzone Over a year ago

I knew this lodash function, but I was looking for a find first (do something) and after it, remove it. But thanks anyway

Hamms · Accepted Answer · 2016-03-02 20:13:06Z

There certainly is! You can use _.remove to programmatically remove elements from array2 if they meet a certain predicate. In your case, that predicate would be "is the element in array1",

var array1 = [{ id: 2, ref: 1010101 },
 { id: 2, ref: 1010107 }]

var array2 = [{ id: 2, ref: 10010001 },
 { id: 2, ref: 10010002 },
 { id: 2, ref: 10010003 },
 { id: 2, ref: 10010006 },
 { id: 2, ref: 10010007 },
 { id: 2, ref: 10010008 },
 { id: 2, ref: 10010009 },
 { id: 2, ref: 10020005 },
 { id: 2, ref: 1010101 }]

_.remove(array2, function (item) {
  return _.find(array1, item);
});

document.write("<pre>" + JSON.stringify(array2, null, 2) + "</pre>")

<script src="https://cdn.jsdelivr.net/lodash/4.6.1/lodash.min.js"></script>

Nina Scholz · Accepted Answer · 2016-03-02 20:35:40Z

1

A solution with a linear O.

var array1 = [{ id: 2, ref: 1010101 }, { id: 2, ref: 1010107 }],
    array2 = [{ id: 2, ref: 10010001 }, { id: 2, ref: 10010002 }, { id: 2, ref: 10010003 }, { id: 2, ref: 10010006 }, { id: 2, ref: 10010007 }, { id: 2, ref: 10010008 }, { id: 2, ref: 10010009 }, { id: 2, ref: 10020005 }, { id: 2, ref: 1010101 }],
    result = function (a, b) {
        function k(x) { return x.id + '|' + x.ref; }
        var o = {};
        b.forEach(function (c) {
            o[k(c)] = true;
        });
        a.forEach(function (c) {
            o[k(c)] = false;
        });
        return b.filter(function (c) {
            return o[k(c)];
        });
    }(array1, array2);

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

edited Mar 2, 2016 at 20:35

answered Mar 2, 2016 at 20:19

Nina Scholz

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1 Comment

Baumannzone Over a year ago

This solutions it's very unclear for me, but thanks anyway.

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