javascript find value from array based on associative array key value

Question

I have an associative array:

var array1 = {
    a1: {
        id: "box1",
        title: "box 1"
    },
    a2: {
        id: "box2",
        title: "box 2"
    },
    a3: {
        id: "box3",
        title: "box 3"
    }
};

I then have another array which has references to the first array:

var array_order = ["a3:positionA", "a2:postitionB", "a1:positionC"];

I want to loop through the first list and then use the second list to find the position text

I am using jQuery so I have

$.each(array1, function(i,o) {
  something in here where I can use i to find out what position. e.g. if a1 I would get positionC
}

Possibly irrelevant side note: your array1 variable is not an array. JavaScript only has one kind of arrays: good old zero-based numeric arrays such as array_order. — Álvaro González
– Álvaro González, Commented May 18, 2011 at 10:30

Lance Roberts · Accepted Answer · 2013-01-03 17:10:22Z

0

Easier and faster iterate over array_order instead of array1:

for( var i = 0, len = array_order.length; i < len; i++ ) {
    var ref = array_order[i].split(':');

    if( ref.length === 2 && ({}).hasOwnProperty.call( array1, ref[0] ) ) {
        var array1Property = array1[ ref[0] ];
        var array1PropertyPosition = ref[1];
    }
}

Iteration over array1 properties can be realized also, but it's significantly slower:

for( var prop in array1 )
    if( ({}).hasOwnProperty.call( array1, prop ) )
        for( var i = 0, len = array_order.length; i < len; i++ ) {
            var ref = array_order[i].split(':');
            if( ref.length === 2 && prop === ref[1] ) {
                var array1Property = array1[ prop ];
                var array1PropertyPosition = ref[1];
            }
        }

And there is no need in jQuery.

edited Jan 3, 2013 at 17:10

Lance Roberts

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answered May 18, 2011 at 10:39

Phillip Kovalev

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2 Comments

John Over a year ago

I ended up iterating over array_order like you said

hornetbzz Over a year ago

Is there any need to double "=" in ref.length = 2 ?

Archan Mishra · Accepted Answer · 2011-05-18 11:04:04Z

0

Ok first things first JS calls associative arrays as objects only.

next to iterate over each key in object

for(var prop in array1) {
    if(array1.hasOwnProperty(prop))
        for(key in array_order){
           if( key === prop){
           doSomethingwith(array_order[key]);

           }

        }
}

answered May 18, 2011 at 11:04

Archan Mishra

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2 Comments

Phillip Kovalev Over a year ago

Iteration over Array using for( in ) are too slow. Try to launch following example in the firebug: gist.github.com/978743

Phillip Kovalev Over a year ago

Correction: for( in ) slower versus for( var .. ++ ), except v8, there for( in ) faster due optimizations.

Raynos · Accepted Answer · 2011-05-18 11:21:46Z

0

var temp_array_order = {};
$.each(array_order, function(key, val) { 
    temp_array_order[val.substr(0,2)] = val.substr(3); 
};
$.each(array1, function(key, val) {
    var category = temp_array_order[key];
    ...
});

answered May 18, 2011 at 11:21

Raynos

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8 Comments

Phillip Kovalev Over a year ago

jQuery .each too slow versus simple for iteration over array, and your solution are suppose, that array1 properties names length equals 2 chars always.

Phillip Kovalev Over a year ago

var category = temp_array_order[key]; there is must be check, that it's temp_array_order own property, not one, which coming from prototypes chain.

Raynos Over a year ago

@PhillipKovalev meh. The code is smaller. Prototype checking isnt really needed, $.each uses a for loop internally.

Phillip Kovalev Over a year ago

I'm sorry, I was wrong about jQuery .each perfomance -- it's best of all. But I disagree with needless of property ownership checking.

Raynos Over a year ago

@Phillip but it's not needed because $.each uses for (var i = 0; ... i++) It does not use for ... in loop.

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