331

Does anyone know of a way (lodash if possible too) to group an array of objects by an object key then create a new array of objects based on the grouping? For example, I have an array of car objects:

const cars = [
    {
        'make': 'audi',
        'model': 'r8',
        'year': '2012'
    }, {
        'make': 'audi',
        'model': 'rs5',
        'year': '2013'
    }, {
        'make': 'ford',
        'model': 'mustang',
        'year': '2012'
    }, {
        'make': 'ford',
        'model': 'fusion',
        'year': '2015'
    }, {
        'make': 'kia',
        'model': 'optima',
        'year': '2012'
    },
];

I want to make a new array of car objects that's grouped by make:

const cars = {
    'audi': [
        {
            'model': 'r8',
            'year': '2012'
        }, {
            'model': 'rs5',
            'year': '2013'
        },
    ],

    'ford': [
        {
            'model': 'mustang',
            'year': '2012'
        }, {
            'model': 'fusion',
            'year': '2015'
        }
    ],

    'kia': [
        {
            'model': 'optima',
            'year': '2012'
        }
    ]
}
double-beep
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Trung Tran
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30 Answers30

468

In plain Javascript, you could use Array#reduce with an object

var cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }],
    result = cars.reduce(function (r, a) {
        r[a.make] = r[a.make] || [];
        r[a.make].push(a);
        return r;
    }, Object.create(null));

console.log(result);
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Nina Scholz
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    how can i iterate `result` results? – Mounir Elfassi Jul 18 '19 at 17:30
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    you could take the entries with [`Object.entries`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/entries) and loop through the key/value pairs. – Nina Scholz Jul 18 '19 at 17:31
  • Is there a way to remove the `make` from data set once grouped? It's taking extra space. – Mercurial Oct 19 '19 at 19:33
  • yes, by [Rest in Object Destructuring](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment#Rest_in_Object_Destructuring). – Nina Scholz Oct 19 '19 at 19:36
  • What does r and a stand for? Would it be correct to assume that r is the accumulator and a the currentValue? – Omar Nov 07 '19 at 23:13
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    The best answer as ever. Why are you passing Object.create(null)? – kartik Jun 14 '20 at 11:38
  • @kartik, it creates an empty object without prototypes. – Nina Scholz Jun 14 '20 at 12:10
  • If you remove prototypes, How will the object has access to object's prototypes method? – kartik Jun 15 '20 at 09:04
  • @kartik, there is no one used here. – Nina Scholz Jun 15 '20 at 09:18
  • What does this line of code do r[a.make] = r[a.make] || []; – Sarah Jul 28 '20 at 01:23
  • @Sarah, it generates a new property `a.make` with an array if it does not exist (the value of not existing properties is `undefined`). if exist, it assign itself. – Nina Scholz Sep 28 '20 at 17:19
  • Why do you need this line Object.create(null)? I read your answer to kartik but I don t understand it. – Aalexander Dec 09 '20 at 10:46
  • @Alex, the reason is to prevent to have for example prototype functions like `toString` as property. for the most application you need not to worry about but you nee to know it, this is to show to use a really emypty object. – Nina Scholz Dec 09 '20 at 10:49
  • Have you a specific example when ` toString ` can be a property? – Aalexander Dec 09 '20 at 10:52
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    @Alex, please have a look here: https://stackoverflow.com/questions/38068424/push-a-array-object-into-another-array-object-if-the-value-not-exists/38068553#38068553 – Nina Scholz Dec 09 '20 at 10:56
  • Its shows following error [Vue warn]: Error in render: "TypeError: Cannot convert object to primitive value" – Sakthi Karthik Mar 23 '22 at 02:59
  • @SakthiKarthik, sorry, i can not help. i have no idea of *Vue*. – Nina Scholz Mar 23 '22 at 07:55
  • @NinaScholz could you help out here by any chance? https://stackoverflow.com/questions/72301893/get-value-of-object-key-and-parent-javascript-and-create-a-new-object/72302136#72302136 – peter flanagan May 19 '22 at 13:56
177

Timo's answer is how I would do it. Simple _.groupBy, and allow some duplications in the objects in the grouped structure.

However the OP also asked for the duplicate make keys to be removed. If you wanted to go all the way:

var grouped = _.mapValues(_.groupBy(cars, 'make'),
                          clist => clist.map(car => _.omit(car, 'make')));

console.log(grouped);

Yields:

{ audi:
   [ { model: 'r8', year: '2012' },
     { model: 'rs5', year: '2013' } ],
  ford:
   [ { model: 'mustang', year: '2012' },
     { model: 'fusion', year: '2015' } ],
  kia: 
   [ { model: 'optima', year: '2012' } ] 
}

If you wanted to do this using Underscore.js, note that its version of _.mapValues is called _.mapObject.

Antoine V
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Jonathan Eunice
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131

You are looking for _.groupBy().

Removing the property you are grouping by from the objects should be trivial if required:

const cars = [{
  'make': 'audi',
  'model': 'r8',
  'year': '2012'
}, {
  'make': 'audi',
  'model': 'rs5',
  'year': '2013'
}, {
  'make': 'ford',
  'model': 'mustang',
  'year': '2012'
}, {
  'make': 'ford',
  'model': 'fusion',
  'year': '2015'
}, {
  'make': 'kia',
  'model': 'optima',
  'year': '2012'
}];

const grouped = _.groupBy(cars, car => car.make);

console.log(grouped);
<script src='https://cdn.jsdelivr.net/lodash/4.17.2/lodash.min.js'></script>
double-beep
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TimoStaudinger
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116

There is absolutely no reason to download a 3rd party library to achieve this simple problem, like the above solutions suggest.

The one line version to group a list of objects by a certain key in es6:

const groupByKey = (list, key) => list.reduce((hash, obj) => ({...hash, [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}), {})

The longer version that filters out the objects without the key:

function groupByKey(array, key) {
   return array
     .reduce((hash, obj) => {
       if(obj[key] === undefined) return hash; 
       return Object.assign(hash, { [obj[key]]:( hash[obj[key]] || [] ).concat(obj)})
     }, {})
}


var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars, 'make'))

NOTE: It appear the original question asks how to group cars by make, but omit the make in each group. So the short answer, without 3rd party libraries, would look like this:

const groupByKey = (list, key, {omitKey=false}) => list.reduce((hash, {[key]:value, ...rest}) => ({...hash, [value]:( hash[value] || [] ).concat(omitKey ? {...rest} : {[key]:value, ...rest})} ), {})

var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars, 'make', {omitKey:true}))
metakungfu
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25

Here is your very own groupBy function which is a generalization of the code from: https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore

function groupBy(xs, f) {
  return xs.reduce((r, v, i, a, k = f(v)) => ((r[k] || (r[k] = [])).push(v), r), {});
}

const cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }];

const result = groupBy(cars, (c) => c.make);
console.log(result);
cdiggins
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    I love this answer due to the fact that you can use nested properties with it as well - very nice. I just changed it for Typescript and it was exactly what I was looking for :) – Alfa Bravo Oct 20 '21 at 13:41
22

var cars = [{
  make: 'audi',
  model: 'r8',
  year: '2012'
}, {
  make: 'audi',
  model: 'rs5',
  year: '2013'
}, {
  make: 'ford',
  model: 'mustang',
  year: '2012'
}, {
  make: 'ford',
  model: 'fusion',
  year: '2015'
}, {
  make: 'kia',
  model: 'optima',
  year: '2012'
}].reduce((r, car) => {

  const {
    model,
    year,
    make
  } = car;

  r[make] = [...r[make] || [], {
    model,
    year
  }];

  return r;
}, {});

console.log(cars);
G.aziz
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  • need urgent help how to store above like [ { {"audi": [ { "model": "r8", "year": "2012" }] },{ {"ford": [ { "model": "r9", "year": "2021" }] } ...] each in object – Ravi Sharma Oct 28 '21 at 10:32
18

Its also possible with a simple for loop:

 const result = {};

 for(const {make, model, year} of cars) {
   if(!result[make]) result[make] = [];
   result[make].push({ model, year });
 }
Jonas Wilms
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  • And probably faster as well, and simpler. I've expanded your snippet to be a bit more dynamic as I had a long list of fields from a db table I didn't want to type in. Also note you will need to replace const with let. `for ( let { TABLE_NAME, ...fields } of source) { result[TABLE_NAME] = result[TABLE_NAME] || []; result[TABLE_NAME].push({ ...fields }); }` – adrien Jul 04 '19 at 07:12
  • TIL, thanks! https://medium.com/@mautayro/es6-variable-declaration-for-loops-why-const-works-in-a-for-in-loop-but-not-in-a-normal-a200cc5467c2 – adrien Jul 04 '19 at 18:57
12

I'd leave REAL GROUP BY for JS Arrays example exactly the same this task here

const inputArray = [ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
];

var outObject = inputArray.reduce(function(a, e) {
  // GROUP BY estimated key (estKey), well, may be a just plain key
  // a -- Accumulator result object
  // e -- sequentally checked Element, the Element that is tested just at this itaration

  // new grouping name may be calculated, but must be based on real value of real field
  let estKey = (e['Phase']); 

  (a[estKey] ? a[estKey] : (a[estKey] = null || [])).push(e);
  return a;
}, {});

console.log(outObject);
ssuperczynski
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SynCap
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11

You can try to modify the object inside the function called per iteration by _.groupBy func. Notice that the source array change his elements!

var res = _.groupBy(cars,(car)=>{
    const makeValue=car.make;
    delete car.make;
    return makeValue;
})
console.log(res);
console.log(cars);
nickxbs
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    While this code may solve the question, [including an explanation](http://meta.stackexchange.com/questions/114762/explaining-entirely-code-based-answers) of how and why this solves the problem would really help to improve the quality of your post. Remember that you are answering the question for readers in the future, not just the person asking now! Please edit your answer to add explanation, and give an indication of what limitations and assumptions apply. – Makyen Nov 24 '16 at 05:33
  • It looks like the best answer to me since you go through the array only once to get the desired result. There's no need to use another function to remove the `make` property, and it is more readable as well. – Carrm Jun 29 '18 at 06:20
8

Just simple forEach loop will work here without any library

var cars = [
{
    'make': 'audi',
    'model': 'r8',
    'year': '2012'
}, {
    'make': 'audi',
    'model': 'rs5',
    'year': '2013'
}, {
    'make': 'ford',
    'model': 'mustang',
    'year': '2012'
}, {
    'make': 'ford',
    'model': 'fusion',
    'year': '2015'
}, {
    'make': 'kia',
    'model': 'optima',
    'year': '2012'
},
];
let ObjMap ={};

  cars.forEach(element => {
    var makeKey = element.make;
     if(!ObjMap[makeKey]) {
       ObjMap[makeKey] = [];
     }

    ObjMap[makeKey].push({
      model: element.model,
      year: element.year
    });
   });
   console.log(ObjMap);
Gautam
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6

Create a method which can be re-used

Array.prototype.groupBy = function(prop) {
      return this.reduce(function(groups, item) {
        const val = item[prop]
        groups[val] = groups[val] || []
        groups[val].push(item)
        return groups
      }, {})
    };

Then below you can group by any criteria

const groupByMake = cars.groupBy('make');
        console.log(groupByMake);

var cars = [
    {
        'make': 'audi',
        'model': 'r8',
        'year': '2012'
    }, {
        'make': 'audi',
        'model': 'rs5',
        'year': '2013'
    }, {
        'make': 'ford',
        'model': 'mustang',
        'year': '2012'
    }, {
        'make': 'ford',
        'model': 'fusion',
        'year': '2015'
    }, {
        'make': 'kia',
        'model': 'optima',
        'year': '2012'
    },
];
  //re-usable method
Array.prototype.groupBy = function(prop) {
   return this.reduce(function(groups, item) {
  const val = item[prop]
  groups[val] = groups[val] || []
  groups[val].push(item)
  return groups
   }, {})
 };
  
 // initiate your groupBy. Notice the recordset Cars and the field Make....
  const groupByMake = cars.groupBy('make');
  console.log(groupByMake);
    
    //At this point we have objects. You can use Object.keys to return an array
Wahinya Brian
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5

For cases where key can be null and we want to group them as others

var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'},
            {'make':'kia','model':'optima','year':'2033'},
            {'make':null,'model':'zen','year':'2012'},
            {'make':null,'model':'blue','year':'2017'},

           ];


 result = cars.reduce(function (r, a) {
        key = a.make || 'others';
        r[key] = r[key] || [];
        r[key].push(a);
        return r;
    }, Object.create(null));
exexzian
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4
function groupBy(data, property) {
  return data.reduce((acc, obj) => {
    const key = obj[property];
    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(obj);
    return acc;
  }, {});
}
groupBy(people, 'age');
sama vamsi
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4

Just try this one it works fine for me.

let grouped = _.groupBy(cars, 'make');

Note: Using lodash lib, so include it.

agravat.in
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    Uncaught ReferenceError: _ is not defined - you should be clear that your solution require to install a 3rd party library just to solve this. – metakungfu Jan 09 '19 at 21:38
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    sorry, i think every one knows. _ stands and mostly used for lodash lib. so you need to use lodash. please read question so you will know he/she is asking for lodash. well thank you. i will remember this. and never forget to write lib. – agravat.in Jan 10 '19 at 09:57
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    You should edit your answer to include you're using a lib. – Rafael Oliveira Oct 05 '20 at 19:31
4

Agree that unless you use these often there is no need for an external library. Although similar solutions are available, I see that some of them are tricky to follow here is a gist that has a solution with comments if you're trying to understand what is happening.

const cars = [{
  'make': 'audi',
  'model': 'r8',
  'year': '2012'
}, {
  'make': 'audi',
  'model': 'rs5',
  'year': '2013'
}, {
  'make': 'ford',
  'model': 'mustang',
  'year': '2012'
}, {
  'make': 'ford',
  'model': 'fusion',
  'year': '2015'
}, {
  'make': 'kia',
  'model': 'optima',
  'year': '2012'
}, ];

/**
 * Groups an array of objects by a key an returns an object or array grouped by provided key.
 * @param array - array to group objects by key.
 * @param key - key to group array objects by.
 * @param removeKey  - remove the key and it's value from the resulting object.
 * @param outputType - type of structure the output should be contained in.
 */
const groupBy = (
  inputArray,
  key,
  removeKey = false,
  outputType = {},
) => {
  return inputArray.reduce(
    (previous, current) => {
      // Get the current value that matches the input key and remove the key value for it.
      const {
        [key]: keyValue
      } = current;
      // remove the key if option is set
      removeKey && keyValue && delete current[key];
      // If there is already an array for the user provided key use it else default to an empty array.
      const {
        [keyValue]: reducedValue = []
      } = previous;

      // Create a new object and return that merges the previous with the current object
      return Object.assign(previous, {
        [keyValue]: reducedValue.concat(current)
      });
    },
    // Replace the object here to an array to change output object to an array
    outputType,
  );
};

console.log(groupBy(cars, 'make', true))
wattry
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4

Another one solution:

var cars = [
    {'make': 'audi','model': 'r8','year': '2012'}, {'make': 'audi','model': 'rs5','year': '2013'}, 
    {'make': 'ford','model': 'mustang','year': '2012'}, {'make': 'ford','model': 'fusion','year': '2015'}, 
    {'make': 'kia','model': 'optima','year': '2012'},
];


const reducedCars = cars.reduce((acc, { make, model, year }) => (
    { 
      ...acc, 
      [make]: acc[make] ? [ ...acc[make], { model, year }] : [ { model, year } ],
    }
 ), {});

console.log(reducedCars);
A1exandr Belan
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3

Prototype version using ES6 as well. Basically this uses the reduce function to pass in an accumulator and current item, which then uses this to build your "grouped" arrays based on the passed in key. the inner part of the reduce may look complicated but essentially it is testing to see if the key of the passed in object exists and if it doesn't then create an empty array and append the current item to that newly created array otherwise using the spread operator pass in all the objects of the current key array and append current item. Hope this helps someone!.

Array.prototype.groupBy = function(k) {
  return this.reduce((acc, item) => ((acc[item[k]] = [...(acc[item[k]] || []), item]), acc),{});
};

const projs = [
  {
    project: "A",
    timeTake: 2,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 4,
    desc: "this is a description"
  },
  {
    project: "A",
    timeTake: 12,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 45,
    desc: "this is a description"
  }
];

console.log(projs.groupBy("project"));
Azayda
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2

You can also make use of array#forEach() method like this:

const cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }];

let newcars = {}

cars.forEach(car => {
  newcars[car.make] ? // check if that array exists or not in newcars object
    newcars[car.make].push({model: car.model, year: car.year})  // just push
   : (newcars[car.make] = [], newcars[car.make].push({model: car.model, year: car.year})) // create a new array and push
})

console.log(newcars);
BlackBeard
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2

I liked @metakunfu answer, but it doesn't provide the expected output exactly. Here's an updated that get rid of "make" in the final JSON payload.

var cars = [
    {
        'make': 'audi',
        'model': 'r8',
        'year': '2012'
    }, {
        'make': 'audi',
        'model': 'rs5',
        'year': '2013'
    }, {
        'make': 'ford',
        'model': 'mustang',
        'year': '2012'
    }, {
        'make': 'ford',
        'model': 'fusion',
        'year': '2015'
    }, {
        'make': 'kia',
        'model': 'optima',
        'year': '2012'
    },
];

result = cars.reduce((h, car) => Object.assign(h, { [car.make]:( h[car.make] || [] ).concat({model: car.model, year: car.year}) }), {})

console.log(JSON.stringify(result));

Output:

{  
   "audi":[  
      {  
         "model":"r8",
         "year":"2012"
      },
      {  
         "model":"rs5",
         "year":"2013"
      }
   ],
   "ford":[  
      {  
         "model":"mustang",
         "year":"2012"
      },
      {  
         "model":"fusion",
         "year":"2015"
      }
   ],
   "kia":[  
      {  
         "model":"optima",
         "year":"2012"
      }
   ]
}
Daniel Vukasovich
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1

With lodash/fp you can create a function with _.flow() that 1st groups by a key, and then map each group, and omits a key from each item:

const { flow, groupBy, mapValues, map, omit } = _;

const groupAndOmitBy = key => flow(
  groupBy(key),
  mapValues(map(omit(key)))
);

const cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }];

const groupAndOmitMake = groupAndOmitBy('make');

const result = groupAndOmitMake(cars);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
<script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>
Ori Drori
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1

Grouped Array of Object in typescript with this:

groupBy (list: any[], key: string): Map<string, Array<any>> {
    let map = new Map();
    list.map(val=> {
        if(!map.has(val[key])){
            map.set(val[key],list.filter(data => data[key] == val[key]));
        }
    });
    return map;
});
Tiago Martins Peres
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1

I love to write it with no dependency/complexity just pure simple js.

const mp = {}
const cars = [
  {
    model: 'Imaginary space craft SpaceX model',
    year: '2025'
  },
  {
    make: 'audi',
    model: 'r8',
    year: '2012'
  },
  {
    make: 'audi',
    model: 'rs5',
    year: '2013'
  },
  {
    make: 'ford',
    model: 'mustang',
    year: '2012'
  },
  {
    make: 'ford',
    model: 'fusion',
    year: '2015'
  },
  {
    make: 'kia',
    model: 'optima',
    year: '2012'
  }
]

cars.forEach(c => {
  if (!c.make) return // exit (maybe add them to a "no_make" category)

  if (!mp[c.make]) mp[c.make] = [{ model: c.model, year: c.year }]
  else mp[c.make].push({ model: c.model, year: c.year })
})

console.log(mp)
1

I made a benchmark to test the performance of each solution that don't use external libraries.

JSBen.ch

The reduce() option, posted by @Nina Scholz seems to be the optimal one.

leonardofmed
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  • 37
1

A proposal that adds Array.prototype.groupBy and Array.prototype.groupByToMap is now on Stage 3!

When it reaches Stage 4 and is implemented on most major browsers, you'll be able to do this:

const cars = [
  { make: 'audi', model: 'r8', year: '2012' },
  { make: 'audi', model: 'rs5', year: '2013' },
  { make: 'ford', model: 'mustang', year: '2012' },
  { make: 'ford', model: 'fusion', year: '2015' },
  { make: 'kia', model: 'optima', year: '2012' }
];

const grouped = cars.groupBy(item => item.make);
console.log(grouped);

This will output:

{
  audi: [
    { make: 'audi', model: 'r8', year: '2012' },
    { make: 'audi', model: 'rs5', year: '2013' }
  ],
  ford: [
    { make: 'ford', model: 'mustang', year: '2012' },
    { make: 'ford', model: 'fusion', year: '2015' }
  ],
  kia: [
    { make: 'kia', model: 'optima', year: '2012' }
  ]
}

Until then, you can use this core-js polyfill:

const cars = [
  { make: 'audi', model: 'r8', year: '2012' },
  { make: 'audi', model: 'rs5', year: '2013' },
  { make: 'ford', model: 'mustang', year: '2012' },
  { make: 'ford', model: 'fusion', year: '2015' },
  { make: 'kia', model: 'optima', year: '2012' }
];

const grouped = cars.groupBy(item => item.make);
//console.log(grouped);

// Optional: remove the "make" property from resulting object
const entriesUpdated = Object
  .entries(grouped)
  .map(([key, value]) => [
    key,
    value.map(({make, ...rest}) => rest)
  ]);
const noMake = Object.fromEntries(entriesUpdated);
console.log(noMake);
<script src="https://unpkg.com/core-js-bundle@3.20.1/minified.js"></script>
double-beep
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0

Building on the answer by @Jonas_Wilms if you do not want to type in all your fields:

    var result = {};

    for ( let { first_field, ...fields } of your_data ) 
    { 
       result[first_field] = result[first_field] || [];
       result[first_field].push({ ...fields }); 
    }

I didn't make any benchmark but I believe using a for loop would be more efficient than anything suggested in this answer as well.

adrien
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0
const reGroup = (list, key) => {
    const newGroup = {};
    list.forEach(item => {
        const newItem = Object.assign({}, item);
        delete newItem[key];
        newGroup[item[key]] = newGroup[item[key]] || [];
        newGroup[item[key]].push(newItem);
    });
    return newGroup;
};
const animals = [
  {
    type: 'dog',
    breed: 'puddle'
  },
  {
    type: 'dog',
    breed: 'labradoodle'
  },
  {
    type: 'cat',
    breed: 'siamese'
  },
  {
    type: 'dog',
    breed: 'french bulldog'
  },
  {
    type: 'cat',
    breed: 'mud'
  }
];
console.log(reGroup(animals, 'type'));
const cars = [
  {
      'make': 'audi',
      'model': 'r8',
      'year': '2012'
  }, {
      'make': 'audi',
      'model': 'rs5',
      'year': '2013'
  }, {
      'make': 'ford',
      'model': 'mustang',
      'year': '2012'
  }, {
      'make': 'ford',
      'model': 'fusion',
      'year': '2015'
  }, {
      'make': 'kia',
      'model': 'optima',
      'year': '2012'
  },
];

console.log(reGroup(cars, 'make'));
AKelley
  • 11
  • 1
0

Here is a solution inspired from Collectors.groupingBy() in Java:

function groupingBy(list, keyMapper) {
  return list.reduce((accummalatorMap, currentValue) => {
    const key = keyMapper(currentValue);
    if(!accummalatorMap.has(key)) {
      accummalatorMap.set(key, [currentValue]);
    } else {
      accummalatorMap.set(key, accummalatorMap.get(key).push(currentValue));
    }
    return accummalatorMap;
  }, new Map());
}

This will give a Map object.

// Usage

const carMakers = groupingBy(cars, car => car.make);
Rahul Sethi
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  • 4
0

Slightly different version of @metakungfus answer, main difference being that it omits the original key from the resulting objects since it's no longer needed on the object itself in some cases since it's now available in the parent object.

const groupBy = (_k, a) => a.reduce((r, {[_k]:k, ...p}) => ({
    ...r, ...{[k]: (
        r[k] ? [...r[k], {...p}] : [{...p}]
    )}
}), {});

Considering your original input object:

console.log(groupBy('make', cars));

Would result in:

{
  audi: [
    { model: 'r8', year: '2012' },
    { model: 'rs5', year: '2013' }
  ],
  ford: [
    { model: 'mustang', year: '2012' },
    { model: 'fusion', year: '2015' }
  ],
  kia: [
    { model: 'optima', year: '2012' }
  ]
}
Nathan F.
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  • 2
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  • 63
0
const groupBy = (array, callback) => {
  const groups = {};
  
  array.forEach((element) => {
    const groupName = callback(element);
    if (groupName in groups) {
      groups[groupName].push(element);
    } else {
      groups[groupName] = [element];
    }
  });
  
  return groups;
};

or for fancy pants:

(() => {
  Array.prototype.groupBy = function (callback) {
    const groups = {};
    this.forEach((element, ...args) => {
      const groupName = callback(element, ...args);
      if (groupName in groups) {
        groups[groupName].push(element);
      } else {
        groups[groupName] = [element];
      }
    });

    return groups;
  };
})();

const res = [{ name: 1 }, { name: 1 }, { name: 0 }].groupBy(({ name }) => name);

// const res = {
//   0: [{name: 0}],
//   1: [{name: 1}, {name: 1}]
// }

This is a polyfill for the MDN Array.groupBy function.

Stan
  • 1,364
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-1

Here is another solution to it. As requested.

I want to make a new array of car objects that's grouped by make:

function groupBy() {
  const key = 'make';
  return cars.reduce((acc, x) => ({
    ...acc,
    [x[key]]: (!acc[x[key]]) ? [{
      model: x.model,
      year: x.year
    }] : [...acc[x[key]], {
      model: x.model,
      year: x.year
    }]
  }), {})
}

Output:

console.log('Grouped by make key:',groupBy())
Eugen Sunic
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