Detect if object in array has one same property but different other property to other objects

Question

Take this array:

[
  {"color": "blue","type": 1},
  {"color": "red","type": 1},
  {"color": "blue","type": 1},
  {"color": "green","type": 1},
  {"color": "green","type": 1},
  {"color": "red","type": 2},
  {"color": "red","type": 1},
  {"color": "green","type": 2},
  {"color": "red","type": 3},
];

How would I go about finding which "color" has a different "type" (than all other objects with the same "name") in the array?

I want to be able to loop through this array and create a second array that would look like this:

{red, green}

Notice blue is ommited because all of the objects with "color":"blue" have the same "type"

The closest I have gotten is this: https://jsfiddle.net/0wgjs5zh/ but it adds all colors into the array appended with the different types:

arr.forEach(function(item){
  if(newArr.hasOwnProperty(item.color+ '-' +item.type)) {
   // newArr[item.color+ '-' +item.type].push(item);
  }
  else {
    newArr[item.color+ '-' +item.type] = item;
  }
});

// RESULT
{blue-1, green-1, green-2, red-1, red-2, red-3}

1. Just to confirm, the resulting array is a set? It doesn't allow duplicates? 2. How do you establish the base type? In other words, would the result array for { blue: 1, blue: 2, blue: 2} and { blue: 2, blue: 1, blue: 1 } be the same? Or do you know 1 is the base type? — nem035
– nem035, Commented May 25, 2016 at 14:54

Nina Scholz · Accepted Answer · 2016-05-25 14:57:08Z

4

You may use two passes, one for the collection and one for generating the result array.

var array = [{ "color": "blue", "type": 1 }, { "color": "red", "type": 1 }, { "color": "blue", "type": 1 }, { "color": "green", "type": 1 }, { "color": "green", "type": 1 }, { "color": "red", "type": 2 }, { "color": "red", "type": 1 }, { "color": "green", "type": 2 }, { "color": "red", "type": 3 }, ],
    result,
    object = Object.create(null);

array.forEach(function (a) {
    object[a.color] = object[a.color] || {};
    object[a.color][a.type] = true;
});

result = Object.keys(object).filter(function (k) {
    return Object.keys(object[k]).length > 1;
});

console.log(result);

answered May 25, 2016 at 14:57

Nina Scholz

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1 Comment

Mcestone Over a year ago

Exactly what I needed. Thanks for you quick response and clean code.

Redu · Accepted Answer · 2016-05-25 15:19:23Z

0

My solution would be like as follows

var cls = [
  {"color": "blue","type": 1},
  {"color": "red","type": 1},
  {"color": "blue","type": 1},
  {"color": "green","type": 1},
  {"color": "green","type": 1},
  {"color": "red","type": 2},
  {"color": "red","type": 1},
  {"color": "green","type": 2},
  {"color": "red","type": 3},
],

    map = cls.reduce((p,c) => (p[c.color] ? !~p[c.color].indexOf(c.type) && p[c.color].push(c.type)
                                          : p[c.color] = [c.type], p),{}),
 result = Object.keys(map).reduce((p,c) => map[c].length > 1 && p.concat(c) || p,[]);

console.log(result);

answered May 25, 2016 at 15:19

Redu

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