Remove elements from JSON with duplicate array members in JavaScript

Question

Is there a simple way to remove JSON elements with duplicate "array members"?

This is my example JSON object:

{
"temperature": [
  {"datetime":"2011-01-27T11:40:50.000Z", "value":15},
  {"datetime":"2011-01-27T11:40:50.000Z", "value":16}, <-- this one should be removed
  {"datetime":"2011-01-27T11:41:00.000Z", "value":14},
  {"datetime":"2011-01-27T11:41:10.000Z", "value":15},
  {"datetime":"2011-01-27T11:41:10.000Z", "value":15}, <-- this one should be removed
  {"datetime":"2011-01-27T11:41:10.000Z", "value":14}, <-- this one should be removed
  {"datetime":"2011-01-27T11:41:20.000Z", "value":16}
  ]
}

As for my example I want to remove the whole JSON array if "datetime" is a duplicate of any other "datetime" inside other JSON arrays.

I tried doing it this way but it gave me the seemingly opposite result:

var datetimes = [];
for(var i = 0; i < obj.temperature.length; i++) {
  if($.inArray(obj.temperature[i].datetime, datetimes)) {
    obj.temperature.splice(i,1);
  }else {
    datetimes.push(obj.temperature[i].datetime);
  }
}

Try it here

Siderite Zackwehdex · Accepted Answer · 2016-03-31 18:16:53Z

1

You can cheat it a little and add the items as properties to an object, with the date as the property key. This makes it unique and then all you have to do is recreate the array from the property values. Something like this:

var dict={}
obj.temperature.forEach(function(t) {
  dict[t.datetime]=t;
});
var arr=[];
for(var datetime in dict) {
  arr.push(dict[datetime]);
}

Fiddle here: https://jsfiddle.net/cmfanjox/

edited Mar 31, 2016 at 18:16

answered Mar 31, 2016 at 18:13

Siderite Zackwehdex

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5 Comments

Linus Over a year ago

Can you provide a working jsfiddle? I can't get it to work.

Siderite Zackwehdex Over a year ago

sorry, your object was also named obj. Now it should work.

Linus Over a year ago

It works almost perfectly now, however it picks the last item not the first: jsfiddle.net/ra48f46c/2

Linus Over a year ago

Thanks alot, it works now, although it seems my error was that I should've set i to 1 and not to 0. I'll have to compare performance speed with your solution and Lea Fox's. I'll mark the fastest one as best answer I guess.

Linus Over a year ago

Okay, I've tested this further with much more data and it the end your solution was the only one that worked consistently. Thanks for the help!

Lea Fox · Accepted Answer · 2016-03-31 18:21:44Z

1

var i = 1

You start at zero and ask, is x object here? Which it is. So, it gets spliced. By the time you get to the input you want to remove you have spliced out all the input that you wanted to keep.

var datetimes = [];
for(var i = 1; i < obj.temperature.length; i++) {
  if($.inArray(obj.temperature[i].datetime, datetimes)) {
    obj.temperature.splice(i,1);
  }else {
    datetimes.push(obj.temperature[i].datetime);
  }
}

answered Mar 31, 2016 at 18:21

Lea Fox

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Comments

vorillaz · Accepted Answer · 2016-03-31 18:22:44Z

Take a look at this snippet:

var items = [{
  "datetime": "2011-01-27T11:40:50.000Z",
  "value": 15
}, {
  "datetime": "2011-01-27T11:40:50.000Z",
  "value": 16
}, {
  "datetime": "2011-01-27T11:41:00.000Z",
  "value": 14
}, {
  "datetime": "2011-01-27T11:41:10.000Z",
  "value": 15
}, {
  "datetime": "2011-01-27T11:41:10.000Z",
  "value": 15
}, {
  "datetime": "2011-01-27T11:41:10.000Z",
  "value": 14
}, {
  "datetime": "2011-01-27T11:41:20.000Z",
  "value": 16
}];

var dedup = items.reduce(function(a, b) {
  function indexOfProperty(a, b) {
    for (var i = 0; i < a.length; i++) {
      if (a[i].value == b.value && a[i].datetime == b.datetime) {
        return i;
      }
    }
    return -1;
  }

  if (indexOfProperty(a, b) < 0) a.push(b);
  return a;
}, []);

Kiril Stoilkov · Accepted Answer · 2016-03-31 19:02:10Z

0

Well, there is one simple solution without jQuery:

var obj = {
"temperature": [
  {"datetime":"2011-01-27T11:40:50.000Z", "value":15},
  {"datetime":"2011-01-27T11:40:50.000Z", "value":16},
  {"datetime":"2011-01-27T11:41:00.000Z", "value":14},
  {"datetime":"2011-01-27T11:41:10.000Z", "value":15},
  {"datetime":"2011-01-27T11:41:10.000Z", "value":15},
  {"datetime":"2011-01-27T11:41:10.000Z", "value":14},  
  {"datetime":"2011-01-27T11:41:20.000Z", "value":16}
 ]
};

var temp = [];

for (var i = 0; i < obj.temperature.length; i++) {
    if (temp.indexOf(obj.temperature[i].datetime) === -1) {
        temp.push(obj.temperature[i].datetime);
        obj.temperature.splice(i + 1, 1)
    };
};

console.log(obj);

edited Mar 31, 2016 at 19:02

answered Mar 31, 2016 at 18:36

Kiril Stoilkov

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