Get rid of duplicates in array of objects

Question

I'm trying to get rid of duplicates in array of objects. I want my code linear so I'm using hashtable:

output = Array.from(new Set(output.map(e => e.company)))
  .map(company => { return output.find(e => e.company === company) })

This code works properly for this array of objects:

let output = [
  { company: 'sony'},
  { company: 'sony'},
  { company: 'apple'}
]

but thing is that I want to reuse this code so I would like to move it to function which would take array output and company string. I don't know how to do that, I tried like this:

const uniq = (array, key) => {
  return Array.from(new Set(array.map( e => e[key] )))
    .map(key => { return array.find(e => e[key] === key) })
}

but this return array of two undefined.

Also second thing is that it's pretty complicated code for such trivial task. Isn't there some easier way to do same without losing performance?

georg · Accepted Answer · 2019-09-23 15:41:25Z

3

This one-liner works with arbitrary objects, not necessarily key:value pairs:

let uniqBy = (a, key) => [...new Map(a.map(x => [x[key], x]).reverse()).values()];

//

let data = [
    {company: 'sony', more: 1},
    {company: 'apple', more: 1},
    {company: 'sony', more: 2},
    {company: 'sony', more: 3},
    {company: 'apple', more: 3},
]

console.log(uniqBy(data, 'company'))

If you don't care about the order of the result, you can also skip .reverse().

answered Sep 23, 2019 at 15:41

georg

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2 Comments

BT101 Over a year ago

I think this and @Nina Scholz are my personal favourite solutions

BT101 Over a year ago

Also it looks like it's just slightly faster but rather insignificant jsperf.com/filter-arr-of-obj-2/1

Nina Scholz · Accepted Answer · 2019-09-23 15:36:40Z

2

You could map the unique values and create a new object.

const 
    unique = (array, key) => Array.from(new Set(array.map(o => o[key])), value => ({ [key]: value }));

var output = [{ company: 'sony'}, { company: 'sony'}, { company: 'apple'}];

console.log(unique(output, "company"));

answered Sep 23, 2019 at 15:36

Nina Scholz

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Comments

Taki · Accepted Answer · 2019-09-23 15:35:12Z

1

You have naming conflicts, just rename the second key:

let input = [{ company: "sony" }, { company: "sony" }, { company: "apple" }];

const uniq = (array, key) => {
  return Array.from(new Set(array.map(e => e[key]))).map(key1 => {
    //                                                   ^^^^
    return array.find(e => e[key] === key1);
    //                                ^^^^
  });
};

const result = uniq(input, "company");

console.log(result);

answered Sep 23, 2019 at 15:35

Taki

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Comments

ADyson · Accepted Answer · 2019-09-23 15:35:14Z

1

I think the issue is you are overwriting the definition of key in the .map() callback. This version - with a separate variable name for the callback parameter - seems to do what you need:

let output = [
  { company: 'sony'},
  { company: 'sony'},
  { company: 'apple'}
]
const uniq = (array, key) => {
  return Array.from(new Set(array.map( e => e[key] )))
    .map(x => { return array.find(e => e[key] === x) })
}

console.log(uniq(output, "company"));

answered Sep 23, 2019 at 15:35

ADyson

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Comments

Djaouad · Accepted Answer · 2019-09-23 15:35:30Z

1

You need to rename your second map parameter to something other than key, otherwise the key inside find will not be the parameter key that is passed to the function:

let output = [
  { company: 'sony'},
  { company: 'sony'},
  { company: 'apple'}
]

const uniq = (array, key) => {
  return Array.from(new Set(array.map( e => e[key] )))
    .map(k => { return array.find(e => e[key] === k) })
}

console.log(uniq(output, 'company'));

answered Sep 23, 2019 at 15:35

Djaouad

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Comments

karthikdivi · Accepted Answer · 2019-09-23 15:33:25Z

0

If you are allowed to use Lodash there is a simple way of eliminating duplicates

_.uniqBy(your_data, 'company');

answered Sep 23, 2019 at 15:33

karthikdivi

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