8

I have 2 arrays that they are identical at first but the user may remove (not able to add but just remove) items from the second array. I want to find items that are in the first array but not in the second one.

I can think of several ways to do this, but as these arrays can be very large, I'm curious to find out if anyone can offer a more efficient way:

$.grep( firstArray, function( n, i ){
  return $.inArray(n, secondArray) == -1;
});
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2
  • Please share those two arrays Commented Mar 4, 2018 at 10:15
  • If it works I think your way is quite efficient Commented Mar 4, 2018 at 10:17

5 Answers 5

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22

You could try to make use of filter and indexOf array methods as below:

var firstArray = [1,2,3,4,5,6];
var secondArray = [3,4,6];

var result = firstArray.filter(item => secondArray.indexOf(item) == -1);

console.log(result);

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2 Comments

I don't get why people downvote correct answers. If you have no clue, simply shut up and learn. This is the correct answer.
As efficiency was explicitly mentioned by the author: Assuming sorted arrays one might want to do a binary search
7

Assuming the arrays have the same order, then you could filter with an index as closure.

var array1 = [1, 2, 3, 4, 5, 6, 7, 1, 2, 3],
    array2 = [2, 4, 6, 7, 2],
    missing = array1.filter((i => a => a !== array2[i] || !++i)(0));
    
console.log(missing);

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13 Comments

This is imposing additional restrictions and also far less readable than Christos' answer.
@connexo, the mentiond answer relies on unique values, which may not be given.
@NinaScholz great answer, Performance Comparison
@NinaScholz , could you please explain this part , a => a !== array2[i] || !++i)(0) ,thanks
@BrandonMcConnell, i need i for array2. if i would be the index of array1, it does not work to filter without an additional offset.
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2

Do a .filter on the array and check if the array 2 does not have the element using .includes

var a1 = [1,2,3,4,5,6];
var a2 = [1,3,5];

var absent = a1.filter(e=>!a2.includes(e));

console.log(absent);

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2 Comments

Thanks. let me check the performance time of these algorithms. because actually, I know a way to do it but looking for better performance.
@AshkanMobayenKhiabani yea but according this is a small and fast enough code to make this task done. .includes is better than .indexOf performance wise if you just need to check array contains an element or not.
1 
function arr_diff (a1, a2) {
var a = [], diff = [];
for (var i = 0; i < a1.length; i++) {
    a[a1[i]] = true;
}
for (var i = 0; i < a2.length; i++) {
    if (a[a2[i]]) {
        delete a[a2[i]];
    } else {
        a[a2[i]] = true;
    }
}
for (var k in a) {
    diff.push(k);
}
return diff;

}

Use this function to get difference between two sets.
A:Set 1
B:Set 2
A-B:Elements that are present in A but, not in B
Basic set theory

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1 Comment

your answer has great performance, but as I can only mark one answer, I chose Nina's as its shorter. +1
0

As user can delete items. So you can add those deleted items in a different array.

var deltedItems = [];
var position = 0;

function onDeleteItem(value){
   deltedItems[position] = value;
   position++;
}

Here, you can find all your deleted items in deltedItems variable. By using this logic you can eliminate search cost of your program.

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