Convert array of objects to one Object using ramda.js

Question

I have an array:

var a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
]

And I want to get transform it to:

var b = {
    1: 'a',
    2: 'b',
    3: 'c',
    4: 'd',
}

Actually I'm using pure js:

var b = a.reduce(
    (ac, pr) => ({
      ...ac,
      [pr.id]: pr.val,
    }),
    {}
  );

But maybe Ramda.js have something special for that purpose?

Note that id is coerced to String during the transformation. Type coercion can lead to subtle bugs. Instead of using Object's as dictionaries, you should use a proper dictionary: Map, where keys can actually be Numbers. — user6445533
– user6445533, Commented Nov 27, 2017 at 9:17
If you choose to go with a Map rather than a plain object, you can instantiate it using var b = new Map(a.map(({id, val}) => [id, val])) — user3297291
– user3297291, Commented Nov 27, 2017 at 16:25

cнŝdk · Accepted Answer · 2017-11-27 08:55:49Z

7

You are looking for Ramda's .mergeAll() method:

var b = R.mergeAll(a.map(function(o) {
  return {
    [o.id]: o.val
  }
}));

The .map()call will return the custom object from each item, taking only the values, then .mergeAll() will merge the array into one object.

mergeAll Documentation:

Merges a list of objects together into one object.

Demo:

var a = [{
    id: 1,
    val: 'a'
  },
  {
    id: 2,
    val: 'b'
  },
  {
    id: 3,
    val: 'c'
  },
  {
    id: 4,
    val: 'd'
  },
];


var b = R.mergeAll(a.map(function(o) {
  return {
    [o.id]: o.val
  }
}));
console.log(b);

<script src="https://cdn.jsdelivr.net/ramda/0.18.0/ramda.min.js"></script>

edited Nov 27, 2017 at 8:55

answered Nov 27, 2017 at 8:32

cнŝdk

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5 Comments

Nina Scholz Over a year ago

where does ramda knows, which property is the key and which the value?

Nick Over a year ago

@NinaScholz source

Nina Scholz Over a year ago

it looks like the equivalent of Object.assign, but the rest is missing for the problem.

cнŝdk Over a year ago

@Nick Yes I updated the answer, it just needs mapping the itesm before calling .mergeAll().

cнŝdk Over a year ago

@NinaScholz Well pointed out, thanks. I missed the fact that there are two properties.

Fredric Waadeland · Accepted Answer · 2021-06-10 09:21:50Z

6

If anyone still passes by here, it does indeed:

R.indexBy(R.prop('id'), someArray);

See indexBy in Ramda's documentation

EDIT: Bennet is correct. If we want val as the only value per key, we can "pluck" it out after:

const createValDict = R.pipe(
  R.indexBy(R.prop('id')),
  R.pluck('val')
)

const valDict = createValDict(myArr)

Pluck works on objects too

edited Jun 10, 2021 at 9:21

answered Sep 30, 2019 at 13:40

Fredric Waadeland

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1 Comment

Bennett McElwee Over a year ago

This is good but doesn't give exactly the result the OP wanted...

Ori Drori · Accepted Answer · 2017-12-01 07:29:26Z

3

Get the ordered values from each object by mapping with R.props, and use R.fromPairs to create an object:

var a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
];

var result = R.compose(R.fromPairs, R.map(R.props(['id', 'val'])));

console.log(result(a));

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>

edited Dec 1, 2017 at 7:29

answered Nov 28, 2017 at 20:32

Ori Drori

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Comments

Nina Scholz · Accepted Answer · 2017-11-27 08:33:41Z

2

With plain Javascript, you could use a combination with Object.assign, spread syntax ..., Array#map, destructuring assignment and short hand properties.

var a = [{ id: 1, val: 'a' }, { id: 2, val: 'b' }, { id: 3, val: 'c' }, { id: 4, val: 'd' }],
    result = Object.assign(...a.map(({ id, val }) => ({ [id]: val })));

console.log(result);

answered Nov 27, 2017 at 8:33

Nina Scholz

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Comments

vicky patel · Accepted Answer · 2017-11-27 09:49:00Z

0

var a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
]

var result = {};
for (var i=0; i<a.length; i++) {
  result[a[i].id] = a[i].val;
}

console.log(result);

answered Nov 27, 2017 at 9:49

vicky patel

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Comments

Scott Sauyet · Accepted Answer · 2017-11-27 14:23:58Z

0

If you wanted something point-free, you could write:

const combine = compose(mergeAll, map(lift(objOf)(prop('id'), prop('val'))))

const {compose, mergeAll, map, lift, objOf, prop} = R;

const combine = compose(mergeAll, map(lift(objOf)(prop('id'), prop('val'))))

var a = [{id:1, val:'a'}, {id:2, val:'b'}, {id:3, val:'c'}, {id:4, val:'d'}]
console.log(combine(a));

<script src="https://cdn.jsdelivr.net/ramda/0.18.0/ramda.min.js"></script>

edited Nov 27, 2017 at 14:23

answered Nov 27, 2017 at 14:11

Scott Sauyet

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Comments

nullqube · Accepted Answer · 2017-12-01 07:13:37Z

0

Here it works like a charm :

var a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
];
// var b = R.fromPairs( a.map(Object.values) );
// Perhaps this is the more general and order independent way:
var b = R.fromPairs(a.map( ({id,val})=>[id,val] ));
console.log( b );

<script src="//cdn.jsdelivr.net/npm/ramda@latest/dist/ramda.min.js"></script>

edited Dec 1, 2017 at 7:13

answered Nov 27, 2017 at 20:26

nullqube

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2 Comments

user3297291 Over a year ago

Note that this solution relies on the order in which the objects' properties are defined. It will break if, for example, you add { val: 'e', id: 5 }.

nullqube Over a year ago

@user3297291 Yes you're right it's case specific , so I've edited the code.

Bennett McElwee · Accepted Answer · 2021-03-31 04:48:15Z

0

This might be the simplest way:

pipe(map(props(['id', 'val'])), fromPairs)(a)

@spflow's answer is simpler but not guaranteed to work on all platforms. Ramda code golf is always fun!

const { fromPairs, map, pipe, props } = R

const a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
]

const result = pipe(map(props(['id', 'val'])), fromPairs)(a)

console.log(result)

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.min.js"></script>

answered Mar 31, 2021 at 4:48

Bennett McElwee

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Comments

v.bortsov · Accepted Answer · 2021-04-13 15:33:54Z

0

Yet one approach:

const { indexBy, prop, pipe, pluck } = R
const a = [
    {id: 1, val: 'a'},
    {id: 2, val: 'b'},
    {id: 3, val: 'c'},
    {id: 4, val: 'd'},
]
const result = pipe(indexBy(prop('id')), pluck('val'))(a)
console.log(result)

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.min.js"></script>

answered Apr 13, 2021 at 15:33

v.bortsov

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