Duplicates in object array

Question

EDITED!!

I have an object array:

Array 0:[{id:"a1" , code:"123" , name:"a"},
         {id: "a2", code: "222" , name: "a"},
          {id: "b1", code: "433", name: "b"}]
Array 1:[{id:"a1" , code:"123" , name:"a"},
         {id: "b2", code: "211" , name: "b"},
          {id: "b1", code: "433", name: "b"}]

I want to get the array of object that has no duplicate value for "name". and store it to another array:

Result:
Array 0:{id: "b1", code: "433", name:"b"}
Array 1: {id:"a1" , code:"123" , name:"a"}

How can I get the array of object that has no duplicate value for name ? All I found on some thread is to remove duplicates from array and not get the array with no duplicate. Thanks in advance!

alexpods · Accepted Answer · 2015-01-23 15:24:51Z

1

function filterByName(array) {
    var counts = {};

    array.forEach(function(obj) { 
        counts[obj.name] ? ++counts[obj.name] : (counts[obj.name] = 1);
    });

    return array.filter(function(obj) { 
        return counts[obj.name] === 1;
    });
}

var filteredArray1 = filterByName(array1);
var filteredArray2 = filterByName(array2);

Explanation of function

Create temporary counts object for keeping counts of object with the same name.
Fill this object with data by using standard forEach method of array.
Filter array by using standard filter method

UPD You need this if I understand your question correctly at last:

var filteredArray = arrayOfArrays.map(function(arrayOfObjects) {
  return filterByName(arrayOfObjects)[0];
});

edited Jan 23, 2015 at 15:24

answered Jan 22, 2015 at 3:12

alexpods

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10 Comments

Lew Jason Nuyda Over a year ago

can you explain this sir?

Lew Jason Nuyda Over a year ago

there's an error that says 'Invalid left-hand side in assignment' for the line return !temp[obj.name] && temp[obj.name] = true;

Felix Kling Over a year ago

@LewJasonNuyda: Operator precedence. Use return !temp[obj.name] && (temp[obj.name] = true); instead.

alexpods Over a year ago

@FelixKling Yeah, I forget about precedence

alexpods Over a year ago

@FelixKling Done, explanation is added.

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