multidimensional index by array of indices in JavaScript

Question

Is there a way in JavaScript to select a element of a multidimential array. Where the depth/rank/dimensionality is variable and the keys are given by a array of indices. Such that i don't have handle every possible dimentional depth separately. concretely speaking i want do get rid of switch cases like here:

/**
 * set tensor value by index
 * @type {array} indices [ index1, index2, index3 ] -> length == rank.
 * @type {string} value.
 */
tensor.prototype.setValueByIndex = function( indices, value ) {
    var rank = indices.length;

    switch(rank) {
        case 0:
            this.values[0] = value;
        break;
        case 1:
            this.values[indices[0]] = value;
        break;
        case 2:
            this.values[indices[0]][indices[1]] = value;
        break;
        case 3:
            this.values[indices[0]][indices[1]][indices[2]] = value;
        break;
    }
}

Where this.values is a multidimensional array.

such that i get something that looks more like this:

/**
 * set tensor value by index
 * @type {array} indices, [ index1, index2, index3 ] -> length == rank
 * @type {string} value
 */
tensor.prototype.setValueByIndex = function( indices, value ) {
    var rank = indices.length;

    this.values[ indices ] = value;
}

Thank you in advance!

possible duplicate of array.contains(obj) in JavaScript

Peter O.
– Peter O.

2012-12-08 09:22:33 +00:00
Commented Dec 8, 2012 at 9:22 — Peter O.
– Peter O., Commented Dec 8, 2012 at 9:22

Bergi · Accepted Answer · 2012-06-13 16:57:00Z

2

tensor.prototype.setValueByIndex = function( indices, value ) {
    var array = this.values;
    for (var i = 0; i < indices.length - 1; ++i) {
        array = array[indices[i]];
    }
    array[indices[i]] = value;
}

This uses array to point to the nested array we are currently at and reads through the indicies for find the next array value from the current array. Once we reach the last index in the indices list, we have found the array where we want to deposit the value. The final index is the slot in that final array where we deposit the value.

edited Jun 13, 2012 at 16:57

Bergi

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answered Jun 13, 2012 at 16:51

apsillers

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Comments

Nayjest · Accepted Answer · 2012-06-13 16:43:21Z

1

Something like this:

tensor.prototype.setValueByIndex = function( indexes, value ) {
    var ref = this.values;  
    if (!indexes.length) indexes = [0];  
    for (var i = 0; i<indexes.length;i++) {
       if (typeof ref[i] === 'undefined') ref[i] = [];
       if (ref[i] instanceof Array) {  
           ref = ref[i];
       } else {
           throw Error('There is already value stored') 
       }
    } 
    ref = value;
}

answered Jun 13, 2012 at 16:43

Nayjest

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Comments

Bergi · Accepted Answer · 2012-06-13 16:56:13Z

1

Why would you want to do that? I'd say writing

tensor.values[1][5][8][2] = value;

is much more perspicuous than

tensor.setValues([1, 5, 8, 2], value);

If you really need to do that, it would be a simple loop over the array:

tensor.prototype.setValueByIndex = function(indices, value) {
    var arr = this.values;
    for (var i=0; i<indices.length-1 && arr; i++)
        arr = arr[indices[i]];
    if (arr)
        arr[indices[i]] = value;
    else
        throw new Error("Tensor.setValueByIndex: Some index pointed to a nonexisting array");
};

answered Jun 13, 2012 at 16:56

Bergi

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2 Comments

Kaj Dijkstra Over a year ago

Yes it looks trivial but. I want to use a variable "number of indices" to create a abstract tensors such that the rank is equal to the "number of indices". such that all the operations can be defined in a abstract way instead of defining all the operations separately for the different ranks and dimensions

Kaj Dijkstra Over a year ago

Thanks for the solution. It turns out to be pretty simple :P, Thanks!.

Guffa · Accepted Answer · 2012-06-13 16:59:13Z

1

Like this?

tensor.prototype.setValueByIndex = function( indices, value ) {
  var t = this, i;
  for (i = 0; i < indices.length - 1; i++) t = t[indices[i]];
  t[indices[i]] = value;
}

edited Jun 13, 2012 at 16:59

answered Jun 13, 2012 at 16:30

Guffa

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1 Comment

Bergi Over a year ago

No, he wants a set instead of a get

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