How can I group an array of objects by key?

Question

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'
        }
    ]
}

Landed

Answer 1

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.

Residuary

Answer 2

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);

.as-console-wrapper { max-height: 100% !important; top: 0; }

UPDATE 2023

Now with Object.groupBy. It takes an iterable and a function for grouping.

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 = Object.groupBy(cars, ({ make }) => make);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Lapotin

how can i iterate result results? – Devalue

you could take the entries with Object.entries and loop through the key/value pairs. – Lapotin

Is there a way to remove the make from data set once grouped? It's taking extra space. – Boyle

yes, by Rest in Object Destructuring. – Lapotin

What does r and a stand for? Would it be correct to assume that r is the accumulator and a the currentValue? – Hijoung

The best answer as ever. Why are you passing Object.create(null)? – Sung

@kartik, it creates an empty object without prototypes. – Lapotin

If you remove prototypes, How will the object has access to object's prototypes method? – Sung

@kartik, there is no one used here. – Lapotin

What does this line of code do r[a.make] = r[a.make] || []; – Hickerson

@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. – Lapotin

Why do you need this line Object.create(null)? I read your answer to kartik but I don t understand it. – Pantin

@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. – Lapotin

Have you a specific example when ` toString ` can be a property? – Pantin

@Alex, please have a look here: #38068924 – Lapotin

Its shows following error [Vue warn]: Error in render: "TypeError: Cannot convert object to primitive value" – Lyndialyndon

@SakthiKarthik, sorry, i can not help. i have no idea of Vue. – Lapotin

@NinaScholz could you help out here by any chance? #72302393 – Flin

may I know how to understand this r[a.make] = r[a.make] || []; ? seems like assiging boolean value – Lapidify

@NinaScholz May I know how Object Destructuring could remove the make field? – Lapidify

@AeLeung, ad 1: its a logical OR ||, not a bitwise OR |. ad 2: yes with a Rest in Object Destructuring ... and mapping this object. – Lapotin

Is there a way to make this use the order of the items in the array? I have sorted my array of objects by year descending e.g 2023, 2022, 2021, but this code example returns them in ascending order. – Inquire

@MrPablo, objects have a given order, for positive 32 bit integer its sorted and coming first, then by insertation order. – Lapotin

@NinaScholz How can I exclude "make" from the groupBy callback to avoid displaying "make": "" in the return result? – Delbert

@PennyLiu, Object.groupBy does not change the data. for removing a certrain property in the result, you need to take another iteration. there, you could map new objects (kepp original data set) or itterate the values and delete unwanted property (mutating original data set). – Lapotin

@NinaScholz Vielen Dank! – Delbert

Answer 3

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>

Memorable

And if you want it shorter still, var grouped = _.groupBy(cars, 'make'); No need for a function at all, if the accessor is a simple property name. – Residuary

What '_' stands for? – Magdalenemagdalenian

@AdrianGrzywaczewski it was the default convention for name-spacing 'lodash' or 'underscore'. Now that the librairies are modular it's no longer required ie. npmjs.com/package/lodash.groupby – Mocambique

And how can I interate in the result? – Girandole

I believe that would be with Object.keys(grouped) – Amulet

Answer 4

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:

var groupByKey = (list, key) => list.reduce((map, obj) => {
 const group = obj[key];
if(map.has(key)) {
  map.get(group).push(obj);
} else { 
  map.set(group, [obj])
}
return map
}, new Map())


 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}))

Oceania

this is definitely not es5 – Aria

I liked both of your answers, but I see they both provide "make" field as a member of each "make" array. I've provided an answer based on yours where the delivered output matches the expected output. Thanks! – Sobersided

@Jeevan reduce has as its argument a callback and an initial value. the callback has two args, previous and current value. Here, previous value is called hash. Maybe someone can explain more about its use here. It seems that reduce here reduces the array by a property which is extracted. – Esbjerg

To anyone considering using this, be careful. The code is not optimised for large datasets, if you have a large array you want to group by, you're going to run into problems. I had an array of 33000 items that took almost 30 seconds to group with this. I'm pretty sure the problem is mainly in the use of {...hash} in the reducer, which creates lots of unused objects that waste space, leading to excessive memory use and garbage collection – Puerperium

This option has very bad performance characteristics (quadratic time) and isn’t particularly readable. It should literally never be used. – Alleen

Answer 5

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);

Brooklet

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 :) – Heterotopia

shorter version: groupBy=(x,f,r={})=>(x.forEach(v=>(r[f(v)]??=[]).push(v)),r) – Resound

Answer 6

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 });
 }

Hombre

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 }); } – Foreleg

TIL, thanks! medium.com/@mautayro/… – Foreleg

Answer 7

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);

Positron

need urgent help how to store above like [ { {"audi": [ { "model": "r8", "year": "2012" }] },{ {"ford": [ { "model": "r9", "year": "2021" }] } ...] each in object – Aoudad

Answer 8

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);

Statuette

Answer 9

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);

Waterman

While this code may solve the question, including an explanation 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. – Beechnut

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. – Remove

Answer 10

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);

Sohn

Answer 11

A proposal that adds Object.groupBy() and Map.groupBy() has reached Stage 4!

It has already been implemented on most major browsers (see caniuse) and so you are 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 = Object.groupBy(cars, item => item.make);
console.log(grouped);

which 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' }
  ]
}

You can also 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 = Object.groupBy(cars, 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/[email protected]/minified.js"></script>

Anabelanabella

Answer 12

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));

Gabionade

Answer 13

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);

Showboat

Answer 14

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');

Knickknack

Answer 15

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))

Spilt

Answer 16

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);

Debutant

Answer 17

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"));

Nuncupative

Answer 18

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"
      }
   ]
}

Sobersided

Answer 19

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'));

Antwanantwerp

Answer 20

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;
});

Esque

This looks inefficient as you do a search for each key. The search has most likely a complexity of O(n). – Chongchoo

With Typescript you already have groupBy method. You can use by your_array.groupBy(...) – Pluralism

Answer 21

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)

Tangier

Answer 22

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.

Jeanene

Answer 23

letfinaldata=[]

let data =[{id:1,name:"meet"},{id:2,name:"raj"},{id:1,name:"hari"},{id:3,name:"hari"},{id:2,name:"ram"}]

data = data.map((item)=> 
{
    return {...item,
        name: [item.name]
    }
}) // Converting the name key from string to array


let temp = [];

for(let i =0 ;i<data.length;i++)
{
    const index = temp.indexOf(data[i].id) // Checking if the object id is already present
    if(index>=0)
    {
        letfinaldata[index].name = [...letfinaldata[index].name,...data[i].name] // If present then append the name to the name of that object
    }
    else{
        temp.push(data[i].id); // Push the checked object id
        letfinaldata.push({...data[i]}) // Push the object
    }
}

console.log(letfinaldata)

Output

[ { id: 1, name: [ 'meet', 'hari' ] },
  { id: 2, name: [ 'raj', 'ram' ] },
  { id: 3, name: [ 'hari' ] } ]

Attendance

Answer 24

2023

Object.groupBy has arrived in native JavaScript -

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(Object.groupBy(
  cars,
  car => car.make,
))

In supported browsers, as of October 2023 -

{
  "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"
    }
  ]
}

Browser	Version
Chrome	117	✅
Edge	117	✅
Firefox	119	✅
Opera	103	✅
Safari	TP	⚠️

Caiaphas

Answer 25

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>

Progestational

Answer 26

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.

Foreleg

A for...of loop is the least efficient. Use a for(i...) for a forEach loop – Spilt

Answer 27

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);

Breadroot

Answer 28

While it isn't in the interest of answering your question with the most optimal or efficient code, I thought this a fun oportunity to turn this into a Code Golf exercise.

This is a slightly altered 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.

Additionally, it heavily utilizes object, parameter and array expansion along with the reduce JavaScript function.

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

Considering your original input object:

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(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' }
  ]
}

Overload

very unreadable code – Griseous

Answer 29

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.

Liebermann

Answer 30

This is a generic function which will return Array groupBy its own key.

const getSectionListGroupedByKey = < T > (
  property: keyof T,
  List: Array < T >
): Array < {
  title: T[keyof T];data: Array < T >
} > => {
  const sectionList: Array < {
    title: T[keyof T];data: Array < T >
  } > = [];

  if (!property || !List ? .[0] ? .[property]) {
    return [];
  }

  const groupedTxnListMap: Map < T[keyof T], Array < T >> = List.reduce((acc, cv) => {
    const keyValue: T[keyof T] = cv[property];

    if (acc.has(keyValue)) {
      acc.get(keyValue) ? .push(cv);
    } else {
      acc.set(keyValue, [cv]);
    }

    return acc;
  }, new Map < T[keyof T], Array < T >> ());

  groupedTxnListMap.forEach((value, key) => {
    sectionList.push({
      title: key,
      data: value
    });
  });

  return sectionList;
};


// Example
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 = getSectionListGroupedByKey('make', cars);
console.log('result: ', result)

Blankly

Answer 31

Try

groupBy= (a,f) => a.reduce( (x,c) => (x[f(c)]??=[]).push(c)&&x, {} )

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 groupBy= (a,f) => a.reduce( (x,c) => (x[f(c)]??=[]).push(c)&&x, {} )

console.log('gr', groupBy(cars, o=>o.make));

This answer is inspired by cdiggins answer and Endless comment (without key remove in final objects). The improvement is that we have same small size but function interface groupBy(a,f) not contains additional redundant variables. To get array of grouped arrays you can use Object.values(groupBy(cars, o=>o.make))

Roentgenoscope

Answer 32

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())

Abortive