Javascript find index of missing elements of two arrays

Question

I have the following JavaScript where I have some lists:

var deleteLinks = $(\".remove-button .remove-from-cart\");
deleteLinks.on(\'click\', function(ev){
            


        

Answer 1

As I can see from your code, you need to compare innerText of the two objects and find the indices of missing elemnts.

var missing_indices = [];
for(var i=0;i<currentText.length;i++){
    var found = false;
    for(var j=0;j<newHTML.length;j++){
        if(currentText[i].innerText == newHTML[j].innerText){
            found = true;
            break;
        }
    }
    if(!found){
        missing_indices.push(i);
    }
}

Then use missing_indices to remove missing elements from currentHTML

Answer 2

To compare two array's value's there is many different approaches to simply see if a value is in another array. You might use array.indexOf(value) to return a position of an element in another array, if the result is greater than expected -1, it is present or the opposite missing. You might also use array.includes(value), and another way is to simply see if a value is not present by using !array.includes(value).

So knowing we can use !array.includes(value) to see if an array does not include a value. We must now loop through one array to grab entries and compare against another array for items that are not in the other array. We will use array.forEach() for this. We could possibly design more recursive functions using array.some(), but I am just trying to give you some simple examples.

Example 1.

// Data
let a = [1, 2, 3, 4, 5]; // Array A
let b = [1, 0, 9, 3, 5]; // Array B

// Basic principals
function check(a, b) {
    // Loop through A using array.some() => value
    a.forEach(value => {
        // B did not include value
        if (!b.includes(value)) {
            // Output
            console.log("B doesn't have", value, "at position", a.indexOf(value), "in A")
        }
    });
    // Loop through B using array.some() => value
    b.forEach(value => {
        // A did not include value
        if (!a.includes(value)) {
            // Output
            console.log("A doesn't have", value, "at position", b.indexOf(value), "in B")
        }
    });
}

// We are checking both Arrays A and B
check([1, 2, 3, 4, 5], [1, 0, 9, 3, 5]);

Example 2

Instead of just outputting to the console for demonstration purposes let's make a prototype for an array element, and then call a comparative function against array B. We will return an array with the [value, position] if that value is not in array B but indeed in array A.

// Data
let a = [1, 2, 3, 4, 5]; // Array A
let b = [1, 0, 9, 3, 5]; // Array B

/*
    Check an array for non-duplicates against another array.
    If the value of item in array A is present in array B,
    return [false, -1]. If value of item in array A is
    not present in B, return value and position of value
    in array A.
 */

Array.prototype.check = function(b) {
    /*
    * return if true
    *   [value, position]
    * else
    *   [false, -1]
    * */
    return a.map(v => {
       return (!b.includes(v) ? [v, a.indexOf(v)] : [false, -1])
    })
};

// Array A is checking against array B
console.log(a.check(b));

/*  Output:
    (A checking against B)
    [
      [ false, -1 ], // Array element 1 found in B
      [ 2, 1 ],  // Array element 2 not found in B, and has position 1 in A
      [ false, -1 ],  // Array element 3 found in B
      [ 4, 3 ], // Array element 4 not found in B, and has position 3 in A
      [ false, -1 ] // Array element 5 found in B
    ]
*/

Answer 3

If you have access to lodash, you can use _.difference to solve this in one line.

var a = ['a', 'b', 'c'],
  b = ['b'],
  result = [];
_.difference(a, b).forEach(function(t) {result.push(a.indexOf(t))});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>

Answer 4

Assuming that you have unique elements in an array A and only one occurrence in the array B. We can just use set to add all the elements of array A and remove the elements from the set when you iterate over Array B. Use findIndex to look up for elements remaining in the set.

const a = [1,2,3,4,5];
const b = [2,1,5,4];
let set = new Set();
a.forEach(ele => set.add(ele));
b.forEach(ele => {
  if(set.has(ele)){
    set.remove(ele);
  }
});
let res = [];
for(let s of set) {
  if(a.findIndex(s) !== -1) {
    res.push({
     arr:  a,
     pos: a.findeIndex(s)
    });
  }else {
    res.push({
     arr:  b,
     pos: b.findeIndex(s)
    });
  }
}

The res array contains the index and the array in which the element is present.