Read an array with an array in Javascript?

Question

If a=[[1,[],"f",3],[3,[4,"x"]]] and b=[1,1].

I want to read a by b like a[1][1] to get [4,"x"]. Note that b is an array which should only consist of integers.

You could also do eval('a['+b.join('],[')+']') but requires the actual variable name as string and it's ugly.

Here are some of my functions:

Array.prototype.readByArray = function(a) {
    var c = this;
    for (var i = 0; i < a.length; i++) {
        c = c[a[i]];
    }
    return c;
};
Array.prototype.emptyByArray = function(a) {
    var c = this.readByArray(a);
    c.splice(0, c.length);
};
Array.prototype.concateByArray = function(a, e) {
    var c = this.readByArray(a);
    for (var i = 0; i < e.length; i++) {
        c.push(e[i]);
    }
};
Array.prototype.setByArray = function(a, e) {
    this.emptyByArray(a);
    this.readByArray(a).push(e);
};

This could be useful for reading a nested array in an imperative way in this example:

Array.prototype.readByArray=function(a){var c=this;for(var i=0;i<a.length;i++){c=c[a[i]];}return c;};
var a = [1,2,3,[1,2,3,[{x: 3},"test"],4],"foo","bar"]; //Your array
var b = [0]; //Reading stack
var s = '[\n'; //Output
while(b[0]<a.length){
    if(Array.isArray(a.readByArray(b))){
        s+=' '.repeat(b.length)+'[\n';
        b.push(-1);
    }else{
        s+=' '.repeat(b.length)+JSON.stringify(a.readByArray(b))+'\n';
    }
    b[b.length-1]++;
    while(b[b.length-1]>=a.readByArray(b.slice(0,-1)).length){
        b.pop();
        b[b.length-1]++;
        s+=' '.repeat(b.length)+']\n';
    }
}
console.log(s);

Is there any better way to do this? Are there native functions for this?

Protip: Don't change Array's prototype. It will give you more headaches than usefulness. — Derek 朕會功夫
– Derek 朕會功夫, Commented Aug 20, 2016 at 16:26
Why is extending native objects a bad practice? Many libraries in the last decade have tried to modify builtin object's prototype, and they all failed miserably. The general principle is that you only change what you own, and leave objects that you don't own untouched. — Derek 朕會功夫
– Derek 朕會功夫, Commented Aug 20, 2016 at 16:35

Nina Scholz · Accepted Answer · 2016-08-20 17:17:53Z

3

You could use Array#reduce for it.

You start with the whole array and return for every element of b a part of the array until all indices are used.

var a = [[1, [], "f", 3], [3, [4, "x"]]],
    b = [1, 1],
    result = b.reduce(function (v, i) {
        return v[i];
    }, a);
    
console.log(result);

ES6

var a = [[1, [], "f", 3], [3, [4, "x"]]],
    b = [1, 1],
    result = b.reduce((v, i) => v[i], a);
    
console.log(result);
result[0] = 42;
console.log(a);
result.splice(0, result.length, 'test');
console.log(a);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Aug 20, 2016 at 17:17

answered Aug 20, 2016 at 16:12

Nina Scholz

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5 Comments

bitten Over a year ago

With ES6, you are able to make it neater with const result = b.reduce((v, i) => v[i], a)

Anastasia Dunbar Over a year ago

Thank you, can I change the element inside of [4,"x"]?

Nina Scholz Over a year ago

yes, you can, because you get the reference to the array. please see second example.

Anastasia Dunbar Over a year ago

If I had to like set that array into an another element as result=["test"] how would I do that?

Nina Scholz Over a year ago

you could use Array#splice, for keeping the reference and delete all elements inside and move the new content to the reference. please see edit in the second example.

Redu · Accepted Answer · 2016-08-20 19:41:55Z

0

I had written a reusable generic code exactly for this purpose to get the nested object properties dynamically. Actually i was targeting objects but since in JS an array is a perfect object it also applies for arrays too. So lets see how it works in this particular case;

Object.prototype.getNestedValue = function(...a) {
  return a.length > 1 ? (this[a[0]] !== void 0 && this[a[0]].getNestedValue(...a.slice(1))) : this[a[0]];
};

var a = [[1,[],"f",3],[3,[4,"x"]]],
    b = [1,1],
    c = a.getNestedValue(...b);
console.log(c)

edited Aug 20, 2016 at 19:41

answered Aug 20, 2016 at 19:36

Redu

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