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I'm trying to create a map or dictionary style data structure in JavaScript. Each string key points to an array of int, and I'd like to do a few simple things with the arrays that the keys point to.

I have a bunch of animals, each with an ID (1,2,3,...). I want to put these in a map so I know that 1,4,5 are cats and that 2,3,6 are dogs.

Here's what I have for code to explain it better.

var myMap = {};
myMap["cats"] = new Array(1,4,5);   //animals 1,4,5 are cats
myMap["dogs"] = new Array(2,3,6);   //animals 2,3,6 are dogs

1) How would I add something to an array? For example, if animal #7 is a cat, would the following code be correct?

myMap["cats"].push(7);   //so that myMap["cats"] is now an array with [1,4,5,7]

2) How would I sort the map so that the keys are ordered by the number of items in their arrays? In this case, myMap["cats"] would be in front of myMap["dogs"] because the array for "cats" has more items than the array for "dogs". Would the following code be correct?

myMap.sort(function(a,b){return myMap[b].length - myMap[a].length});

If there's a much more efficient way to do this in JavaScript, please let me know. Thank you so much!

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  • 1
    Be careful with new Array();: if you pass a single number to it, it will initialize an array with that length, not an array with that number inside. Using array literals such as [1,4,5] is preferred. Commented May 10, 2013 at 1:50

3 Answers 3

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8

Seems you've answered your own first question.

As for the second question, you'll need another Array to maintain a sort order of the map keys.

var mapKeys = Object.keys(myMap);

mapKeys.sort(function(a,b){return myMap[b].length - myMap[a].length});

Then you can iterate the mapKeys to obtain the collections in your sorted order.

mapKeys.forEach(function(key) {
    console.log("Processing ", key);
    var indices = myMap[key];
});
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Comments

3

1) Yes, push would work (as would unshift or splice).

2) Object keys are inherently unordered and as such they do not have a sort function :-) You could add a function to your map to return the values in sorted order however:

myMap.sort = function sortMap(sortFunc) {
    var results = [];
    for (key in this) {
        if (this.hasOwnProperty(key) && this[key] !== sortMap) {
            results.push(this[key]);
        }
    }
    return results.sort(sortFunc);
};

myMap.sort(function(a, b) { return a.length - b.length; });

Some notes:

  • Don't use Array or new Array unless you need to (e. g. Array(12).join("-") to create a string of 11 dashes). Instead use the array literal syntax []. It's clearer (and in most browsers actually faster).
  • When in doubt, open up MDN and the console in your browser and try it out.
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1

How would I sort the map so that the keys are ordered by the number of items in their arrays?

Maps are not ordered — there is no difference between { 'cats': [1,4,5], 'dogs': [2,3,6] } and { 'dogs': [2,3,6], 'cats': [1,4,5] }.

As for the rest of your question — yes, that's all correct. But instead of writing this:

var myMap = {};
myMap["cats"] = new Array(1,4,5);   //animals 1,4,5 are cats
myMap["dogs"] = new Array(2,3,6);   //animals 2,3,6 are dogs

I'd recommend writing this:

var myMap = {
    'cats': [1,4,5], //animals 1,4,5 are cats
    'dogs': [2,3,6]  //animals 2,3,6 are dogs  
};

or perhaps this:

var animals = [ 'cats', 'dogs', 'dogs', 'cats', 'cats', 'dogs' ];
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12 Comments

there is a difference between the pair of objects you first mentioned. jsfiddle.net/martco/dvdHq/1
@MartinCortez - yes, but that's because they are different objects not because their keys are in different source order ;-) jsfiddle.net/dvdHq/2
@SeanVieira i'm not questioning why they're different. i'm just pointing out that it's wrong to say that there is no difference.
@MartinCortez: That makes no sense. You must realize that {} === {} also evaluates to false (because each occurrence of {} instantiates a new object); would you therefore insist that there is a difference between {} and {}?
i'm not insisting anything besides that it's wrong to say there is no difference, as you did above
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