JavaScript indexOf didn't return good results

Question

I'am trying to find the index of my array by the function indexOf, but I can't get a right result.

var points =[
    ["2.408","38.8"],
    ["2.410","38.8"],
    ["2.410","38.76"]
];
var position = points.indexOf(["2.408","38.8"]);

I think it should return 0 rather than -1, so I campared the two array like blow.

console.log(points[0]===["2.408","38.8"])

Then I got false.

I don't understand why it's not the true.

I'll appreciate for your suggestion...

["indexOf() compares searchElement to elements of the Array using strict equality (the same method used by the === or triple-equals operator)."](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf) so with that you try `console.log(["2.408","38.8"]===["2.408","38.8"])` and now you see why it fails. — epascarello, Jan 23 '18 at 12:23
See https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript — Barmar, Jan 23 '18 at 12:41

Nenad Vracar · Accepted Answer · 2018-01-23T12:37:29.537

2

You can do this with findIndex and every methods.

var points = [["2.408", "38.8"], ["2.410", "38.8"],["2.410", "38.76"]];
const arr = ["2.408","38.8"];

const i = points.findIndex(a => {
  return a.length == arr.length && arr.every((e, i) => a[i] == e)
})

console.log(i)

edited Jan 23 '18 at 12:37

answered Jan 23 '18 at 12:25

Nenad Vracar

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if there duplicates elements in the array with the same values? – zabusa Jan 23 '18 at 12:27
your code doesn't take into account indexes and duplicated values like `["2.408", "2.408"]` – pwolaq Jan 23 '18 at 12:28
@charlietfl no, it won't. it will output that `["38.8", "2.408"]` is in array and it isn't – pwolaq Jan 23 '18 at 12:30
@pwolaq ok...that case makes sense...first one you mentioned doesn't – charlietfl Jan 23 '18 at 12:31
@charlietfl it also does but if you make that change in `points`, not the searching value – pwolaq Jan 23 '18 at 12:32
Updated the answer, I think it covers all the cases now. – Nenad Vracar Jan 23 '18 at 12:35
Thanks, I take your answer – Tengda Su Jan 23 '18 at 12:48

score 1 · Answer 2 · answered Jan 23 '18 at 12:43

You are not searching for the same Array.

When you create an Array, you are creating an Object and storing a reference to it in a variable. When you search the Array of Arrays, the find logic will be comparing by reference, not by deep value. That's why the indexOf is giving you -1; you aren't actually searching for anything that is really in the Array of Arrays.

const a = ["foo", "bar"],
  b = ["foo", "bar"];

const arr = [a];

console.log(arr.indexOf(["foo", "bar"])); // -1, not the right reference
console.log(arr.indexOf(b)); // -1, again, not the right reference
console.log(arr.indexOf(a)); // 0, the right reference

Pac0 · Answer 3 · 2018-01-23T12:49:32.170

You can use findIndex and a helper function to create a comparison to any array.

    var allElementsEquals = function(refArray) {
        return function(array) {
          if (array.length !== refArray.length) return false;
       for (var i = 0; i < array.length; i++) { if (array[i] !== refArray[i]) return false; }
          return true
        }
    }

    // test 1 
    var points =[
        ["2.408","38.8"],
        ["2.410","38.8"],
        ["2.410","38.76"]
    ];
    var position = points.findIndex(allElementsEquals(["2.408","38.8"]));
    console.log(position);


    // test 2
    var points =[
        ["2.410","38.8"],
        ["2.408","38.8"],
        ["2.410","38.76"]
    ];
    var position = points.findIndex(allElementsEquals(["2.408","38.8"]));
    console.log(position);
    
    // test 3
        var points =[
        ["2.410","38.8"],
        ["2.409","38.8"],
        ["2.410","38.76"]
    ];
    var position = points.findIndex(allElementsEquals(["2.408","38.8"]));
    console.log(position);

    // test 4
        var points =[
        ["2.410","38.8"],
        ["38.8","2.408"],
        ["2.410","38.76"]
    ];
    var position = points.findIndex(allElementsEquals(["2.408","38.8"]));
    console.log(position);

JavaScript indexOf didn't return good results

3 Answers3