'How can I group an array of objects by key?
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'
}
]
}
Solution 1:[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.
Solution 2:[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; }Solution 3:[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>Solution 4:[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:
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}))Solution 5:[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);Solution 6:[6]
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);Solution 7:[7]
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 });
}
Solution 8:[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);Solution 9:[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);
Solution 10:[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);Solution 11:[11]
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 arraySolution 12:[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));
Solution 13:[13]
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');
Solution 14:[14]
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))Solution 15:[15]
Just try this one it works fine for me.
let grouped = _.groupBy(cars, 'make');
Note: Using lodash lib, so include it.
Solution 16:[16]
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);Solution 17:[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"));
Solution 18:[18]
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);Solution 19:[19]
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"
}
]
}
Solution 20:[20]
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>Solution 21:[21]
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;
});
Solution 22:[22]
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)Solution 23:[23]
I made a benchmark to test the performance of each solution that don't use external libraries.
The reduce() option, posted by @Nina Scholz seems to be the optimal one.
Solution 24:[24]
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/[email protected]/minified.js"></script>Solution 25:[25]
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.
Solution 26:[26]
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'));
Solution 27:[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);Solution 28:[28]
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' }
]
}
Solution 29:[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.
Solution 30:[30]
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())
Sources
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Source: Stack Overflow