Is there a more compact way to map values of specified keys, remove all the undefined values from the resultant array and get unique values from it

Question

I have an object array:

var objects = [
  {'key1': '1', 'key2': '2'},
  {'key1': '3'},
  {'key1': '4', 'key2': '5', 'key3': '6'},
  {'key1': '1', 'key2': '2'},
  {'key1': '3'},
  {'key1': '4', 'key2': '5', 'key3': '6'},
];

I want an array with unique values of key2 but without any undefined values in it. Something like this:

['2', '5']

Right now, I'm using:

lodash.without(lodash.uniq(lodash.map(objects, 'key2')), undefined)

My Question:

Is there a better and more compact way to do this?

if your code is working maybe it should go on https://codereview.stackexchange.com/ — A. L, Jun 27 '17 at 09:39
This is a use case I currently have. I won't be able to share the working code. — SiddAjmera, Jun 27 '17 at 09:43
You could make it more readable using chaining: `let result = _.chain(objects).map('key2').uniq().without(undefined).value()`. Looks better when vertically stacked :) — Gruff Bunny, Jun 27 '17 at 11:40
@GruffBunny you can remove `lodash#chain` and then replace `lodash#without` with `lodash#compact`. `_(objects).map('key2').uniq().compact().value()` — ryeballar, Jun 27 '17 at 13:44

score 2 · Accepted Answer · answered Jun 27 '17 at 09:43

2

objects.reduce(function(m,d){
    if(d.key2 != undefined && m.indexOf(d.key2)<0){
        m.push(d.key2)
    }
    return m;
},[])

answered Jun 27 '17 at 09:43

Chirag Ravindra

4,760
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How do we compare the performance of this approach against mine? – SiddAjmera Jun 27 '17 at 09:56
1

There are many ways to measure your code's performance. You can use profilers, log timestamps and on to more advanced options. You can start here: https://www.sitepoint.com/measuring-javascript-functions-performance/ https://stackoverflow.com/questions/111368/how-do-you-performance-test-javascript-code – Chirag Ravindra Jun 27 '17 at 10:10
Great. That helps. Unlike Callum's comment that doesn't. – SiddAjmera Jun 27 '17 at 10:14
Hey Chirag, thanks. This is Infact, faster than my approach. – SiddAjmera Jun 27 '17 at 15:18

Nina Scholz · Answer 2 · 2017-06-27T09:58:44.917

1

You could use Set and filter the object before making the values unique.

var objects = [{ key1: '1', key2: '2' }, { key1: '3' }, { key1: '4', key2: '5', key3: '6' }, { key1: '1', key2: '2' }, { key1: '3' }, { key1: '4', key2: '5', key3: '6' }],
    unique = [...new Set(objects.filter(o => 'key2' in o).map(o => o.key2))];

console.log(unique);

ES5

var objects = [{ key1: '1', key2: '2' }, { key1: '3' }, { key1: '4', key2: '5', key3: '6' }, { key1: '1', key2: '2' }, { key1: '3' }, { key1: '4', key2: '5', key3: '6' }],
    values = objects.reduce(function (r, o) {
        ('key2' in o) && (r[o.key2] = true);
        return r;
    }, Object.create(null));
    unique = Object.keys(values);

console.log(unique);

edited Jun 27 '17 at 09:58

answered Jun 27 '17 at 09:43

Nina Scholz

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1

I don't shure that ecma 2015 solution is suitable if pointed that lodash is used. – YD1m Jun 27 '17 at 09:52
@YD1m why not? lodash also uses ES2015: [just one example](https://github.com/lodash/lodash/blob/master/.internal/baseFlatten.js#L28) – Thomas Jun 27 '17 at 10:46

Is there a more compact way to map values of specified keys, remove all the undefined values from the resultant array and get unique values from it

2 Answers2