如何在数组数组中找到元素的索引？

Question

Here is a simple example of an array that contains at least one other array. I want a way to find the index of an array, within an array. However, the code I have below does not work:

var arr = [1,2,[1,2]];
console.log(arr.indexOf([1,2])); //-> -1

for (var i = 0; i < arr.length; i++) {
  if (arr[i] == [1,2])
    return 'true' // does not return true
}

Intuitively, this should work but does not:

if ([1,2] == [1,2]) return 'true' // does not return true

Can someone explain why it does not work, and offer an alternative solution? Thanks!

Answer 1

No, but you can check it yourself:

var a = [1,2], b = [1,2];

a.length === b.length && a.every(function(x,y) { return x === b[y]; });

Answer 2

Arrays in JavaScript are compared by reference not by value. That is why

console.log([1,2] == [1,2])

returns false .

You need a custom function that could compare arrays. If you want to check only the first level of nesting you can use this code:

var compareArrays = function(a, b) {
    if (a.length !== b.length) {
        return false;
    }

    for (var i = 0, len = a.length; i < len; i++) {
        if (a[i] !== b[i]) {
            return false;
        }
    }

    return true;
}

Answer 3

You are confusing the definitions of similar objects vs. the same object. Array.prototype.indexOf() compares the objects using the strict equality comparison algorithm . For (many) objects, this means an object reference comparison (ie the object's unique identifier, almost like its memory address).

In your example above, you are trying to treat two similar arrays as though they were the same array, and that's why it doesn't work. To do what you are trying to do, you need to use the same array. Like this:

var testArr = [1,2]; //a specific object with a specific reference
var arr = [1,2,testArr]; //the same array, not a different but similar array
console.log(arr.indexOf(testArr)); //returns 2

If you just want to know where arrays occur in the parent array, use Array.isArray() :

...
if(Array.isArray(arr[i])) {
    //do something
}
...

Hopefully that helps!