Include element in array if is not present in other array

Question

I have this array:

let idsClass = [1,2]

and i have this other array:

let idsQuiz = [{id: 1}]

I need to know the id numbers that have in idsClass that is missing in idsQuiz.

So, i need an new array containing this elements, for this example, this is the correct output:

[2]

I tryed something like:

 idsClass.forEach(item => { if (!idsQuizz.includes({id:item})) { console.log('dont find this: ' + item) } })

But my console is printing the two elements

Does this answer your question? How to merge two arrays in JavaScript and de-duplicate items — fingeron
– fingeron, Commented Jan 21, 2020 at 18:56
When I copy-paste your code and run it, the value of idsQuiz is [1,2] — Simon Crane
– Simon Crane, Commented Jan 21, 2020 at 18:58
sorry i forgot to specify that the second array is an array of object — bla
– bla, Commented Jan 21, 2020 at 19:09
this problem is very poorly defined. The title says to include elements in the array if it's not in the other array, yet the question says to put in the array all elements from the other array. The expected output is integers, yet the starting array is objects. This is certainly being overcomplicated. — Klaycon
– Klaycon, Commented Jan 21, 2020 at 19:13

Jing Yu · Accepted Answer · 2020-01-21 19:29:40Z

3

Try to use filter built-in function to check existence.

let idsClass = [1,2];


let idsQuiz = [{id: 1}];

idsClass.forEach(item => {
  if (idsQuiz.filter(e => e.id === item).length === 0) {
    idsQuiz.push({id: item});
  }
});

console.log(idsQuiz); // output is [{"id": 1}, {"id": 2}]

edited Jan 21, 2020 at 19:29

answered Jan 21, 2020 at 18:57

Jing Yu

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4 Comments

bla Over a year ago

There's a way to make this work if the second array is an array of object? My second element is an array of object and i try: idsClass.forEach(item => { if (!idsQuizz.includes({id:item})) { console.log('dont find this: ' + item) } }) but is returning the two elements

Jing Yu Over a year ago

Yes, there is. Check it's key with includes function. if (! idsQuiz.includes(item.id)) {

bla Over a year ago

if i put this way, item.id is undefined, i need to compare the key of the idsQuiz.id

Jing Yu Over a year ago

You could use filter function in case.

Mike Reddington · Accepted Answer · 2020-01-21 19:40:41Z

basic solution

let idsClass = [1,2,2,4,5,6];
let idsQuiz = [1,4];
for(let i=0;i<idsClass.length;i++){
    let j=0;
    for(;j<idsQuiz.length;j++){
        if(idsClass[i] === idsQuiz[j]){
            break;
        }
    }
    if(j === idsQuiz.length) {
        idsQuiz.push(idsClass[i]);
    }
}
console.log(idsQuiz);

solution using inbuilt methods

let idsClass = [1,2,2,4,5,6];
let idsQuiz = [1,4];

// Adds all elements of idsClass into idsQuiz. may contain duplicates
idsQuiz.push(...idsClass);

//Removes duplicates from idsQuiz
idsQuiz = idsQuiz.filter((val,index,self) => self.indexOf(val) === index);
console.log(idsQuiz);

your code adds element into idsQuiz whenever an element in idsClass doesn't match with an element in idsQuiz. Instead, iterate through whole idsQuiz list and then add if it doesn't exist for each idsClass element.

idsClass = [1,2,3]
idsQuiz = [1]

i = 0, j=0 (no element is added in idsQuiz)
i=1, j=0 (2 is added in idsQuiz)
i=1, j=1 (no element is added in idsQuiz)
i=2, j=0 (3 is added in idsQuiz)
i=2, j=1 (3 is added in idsQuiz)
i=2, j=2 (no element is added in idsQuiz)
i=2, j=3 (no element is added in idsQuiz)

Klaycon · Accepted Answer · 2020-01-21 19:29:09Z

1

There's probably a more efficient method, but I think this way is easy to understand:

let idsClass = [1,2,3,4,5,6];
let idsQuiz = [{id: 1},{id: 4}];

let result = idsClass.filter(n => !idsQuiz.some(o => o.id == n));

console.log(result);

Simply use a filter() on the idsClass. filter will return a new array with only the elements that pass the inner function. In this case, it only returns elements that aren't in the idsQuiz.

answered Jan 21, 2020 at 19:29

Klaycon

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