Compress Json Objects by Field JQuery/JS

Question

I have the following JSON:

[{"state":"MN","value":10},{"state":"MN","value":10},{"state":"MN","value":10},{"state":"NY","value":8},{"state":"NY","value":8},{"state":"NY","value":8}]

is there a command in JQuery/JS or any other HTML compatible language that will allow me to convert the above JSON into the following:

[{"state":"MN","value":[10,10,10]},{"state":"NY","value":[8,8,8]}]

Thanks!

Simply iterate the original object and create the desired output — Selvakumar Arumugam
– Selvakumar Arumugam, Commented Apr 19, 2012 at 21:25

Selvakumar Arumugam · Accepted Answer · 2012-04-19 21:37:17Z

3

Below looping will give you the desired output,

DEMO

var data = [{"state":"MN","value":10},{"state":"MN","value":10},{"state":"MN","value":10},{"state":"NY","value":8},{"state":"NY","value":8},{"state":"NY","value":8}];

var tmp = {};
var t;
$.each (data, function (i, val) {
   if (tmp.hasOwnProperty(val.state)) {
      t = tmp[val.state];
      t.push(val.value);
   } else {
      tmp[val.state] = [val.value];
   }   
});

var output = [];
for (i in tmp) {
    output.push({'state': i, 'value': tmp[i]});   
}

Proof:

enter image description here

answered Apr 19, 2012 at 21:37

Selvakumar Arumugam

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1 Comment

Florian Margaine Over a year ago

Dude, you're +1'ing non-jQuery answers are uses $.each in an answer that completely doesn't need it at all? Other than this, nice code :)

Tronix117 · Accepted Answer · 2012-04-19 22:11:46Z

3

you can use:

var input=[{"state":"MN","value":10},{"state":"MN","value":10},{"state":"MN","value":10},{"state":"NY","value":8},{"state":"NY","value":8},{"state":"NY","value":8}]

var tmp={}, output=[];
while(o=input.pop()) (tmp[o.state] || (tmp[o.state]=[])).push(o.value); //you could even stop here if it is enough for you
for(k in tmp) output.push({state:k,value:tmp[k]});

edited Apr 19, 2012 at 22:11

answered Apr 19, 2012 at 21:38

Tronix117

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3 Comments

Tronix117 Over a year ago

I like simplifying and optimizing everything ^^

Selvakumar Arumugam Over a year ago

I wouldn't call optimized as you are calling array .shift() which is an overkill. Instead you can use .pop which doesn't shift the elements.. but again a for loop iterate is generally faster in this case as we gonna just read.

Tronix117 Over a year ago

With V8, it shows better results with shift, jsperf.com/pop-vs-shift-on-a-array but it seems in not the case everywhere and pop seems to be faster. I change it to pop right now ;)

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