Comparing between 2 different nested arrays in JavaScript

Question

I've been on this for a while but haven't been able to arrive at my desired result. I'd like to compare items in two arrays based on their indexes. Something like this:

const first = 0;
const second = 1;
const result = []; // should equal (false, true, false)

const final = [[[2,3],[1,1]],[[4,2],[3,0]],[[0,3],[1,1]]];

for (let i = 0; i < final.length; i++) {
    for (let j = 0; j < final[i].length; j++){
//Comparison between ((2 and 3) && (1 and 1)) with a single result.
//Should also compare between ((4 and 2) && (3 and 0)) and many other nested arrays
//This logic is not the right one as it can't do the comparison as intended.
      if (final[i][j][first] > final[i][j][second]) {
result.push(true);
    } else result.push(false);
  }

I hope this is understood, enough. There is a very similar problem like this one. I don't know if it's right to post it here or open another ask a question, but they're both similar.

Thanks very much.

`const result = []; // should equal (true, true, false)` !? with your condition : `final[i][j][first] > final[i][j][second])` I think it's like : `(false, true, false)` — Xupitan, Jun 04 '22 at 17:40

Xupitan · Accepted Answer · 2022-06-05T16:07:44.557

3

You can try it :

const first = 0;
const second = 1;

const final = [[[2,3],[1,1]],[[4,2],[3,0]],[[0,3],[1,1]]];

const result = final.map(([e1, e2]) => (e1[first] > e1[second] && e2[first] > e2[second]))
console.log(result)

// with non arrow function
function compareArr(arr, first, second) {
  return arr.map(function (ele) {
    return ele[0][first] > ele[0][second] && ele[1][first] > ele[1][second]
  })
}
console.log(compareArr(final, first, second))

// with non map function :
function compareArr1(arr, first, second) {
  let result1 = []
  for(let i = 0; i < arr.length; i++) {
    result1[i] = true
    for(let j = 0; j < arr[i].length; j++) {
      if(!(arr[i][j][first] > arr[i][j][second]))
        result1[i] = false
      }
  }
  return result1
}
console.log(compareArr1(final, first, second))

UPDATE : edit with suggest from @yes sir

Update : about ([e1,e2]) :

normal : Array.map((ele) => .....) with structure of ele is [a, b]

It can use same as : Array.map(([a, b]) => .....)

Document of it : Destructuring assignment

edited Jun 05 '22 at 16:07

answered Jun 04 '22 at 17:48

Xupitan

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`e1[first] > e1[second] && e2[first] > e2[second]` instead of `e1[first] > e1[second] && e1[first] > e1[second]`, i can't edit it. – yes sir Jun 04 '22 at 18:00
1

oh sorry ! i'm lazy ^^! – Xupitan Jun 04 '22 at 18:01
Great! Thanks very much. But I'm trying to understand the arrow function, it seems hard, can you help to elaborate? – Khalifah Jun 04 '22 at 18:03
with arrow function, you have context is larger than function norma – Xupitan Jun 04 '22 at 18:06
you can read more in here : https://stackoverflow.com/questions/34361379/are-arrow-functions-and-functions-equivalent-interchangeable – Xupitan Jun 04 '22 at 18:09
@Xupitan Sorry to disturb you, again. Can you, PLEASE, help to restructure the callback from arrow function to a normal function expression? Just the callback. – Khalifah Jun 05 '22 at 05:35
I don't just wanna copy without fully understanding. – Khalifah Jun 05 '22 at 05:35
@Xupitan can you give me a brief explanation about this function parameters: ```([e1,e2])``` – Khalifah Jun 05 '22 at 06:41
i 've edited answer, you can read more documents of javascript . have a good day ! – Xupitan Jun 05 '22 at 09:35

Tayeb HAMDAOUI · Answer 2 · 2022-06-04T18:21:16.563

Xupitan used functional programming, which is the better approach in my opinion, but I tried to extend your code.

what you previously lacked was that you weren't keeping tracking of the first array, you were comparing every nested array, this approach is also valid but you need to then compare each results e.g., result[i] && result[i+1].

const first = 0;
const second = 1;
const result = []; // should equal (true, true, false)

const final = [[[2,3],[1,1]],[[4,2],[3,0]],[[0,3],[1,1]]];

let k=0

for (let i = 0; i < final.length; i++) {
    for (let j = 0; j < final[i].length; j++){
        //Comparison between ((2 and 3) && (1 and 1)) with a single result.
        //Should also compare between ((4 and 2) && (3 and 0)) and many other nested arrays
        //This logic is not the right one as it can't do the comparison as intended.
        if (final[i][j][first] > final[i][j][second]) {
            if(k > 0){
                result[i] = true
            }
            result[i] = true
        }else{
            //first false exit
            result[i] = false;
            break;
        } 
        k++
  }
  k=0
}

console.log(result)

Comparing between 2 different nested arrays in JavaScript

2 Answers2