问题
Does anyone know of a (lodash if possible too) way 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:
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\'
},
];
I want to make a new array of car objects that\'s grouped by make:
var cars = {
\'audi\': [
{
\'model\': \'r8\',
\'year\': \'2012\'
}, {
\'model\': \'rs5\',
\'year\': \'2013\'
},
],
\'ford\': [
{
\'model\': \'mustang\',
\'year\': \'2012\'
}, {
\'model\': \'fusion\',
\'year\': \'2015\'
}
],
\'kia\': [
{
\'model\': \'optima\',
\'year\': \'2012\'
}
]
}
回答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.
回答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);
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回答3:
You are looking for _.groupBy().
Removing the property you are grouping by from the objects should be trivial if required:
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'},];
var grouped = _.groupBy(cars, function(car) {
return car.make;
});
console.log(grouped);
<script src='https://cdn.jsdelivr.net/lodash/4.17.2/lodash.min.js'></script>
As a bonus, you get even nicer syntax with ES6 arrow functions:
const grouped = _.groupBy(cars, car => car.make);
回答4:
The short version to group an array of objects by a certain key in es6:
result = array.reduce((h, obj) => Object.assign(h, { [obj.key]:( h[obj.key] || [] ).concat(obj) }), {})
The longer version:
result = array.reduce(function(h, obj) {
h[obj.key] = (h[obj.key] || []).concat(obj);
return h;
}, {})
It appear the original question asks how to group cars by make, but omit the make in each group. So the answer would look like this:
result = cars.reduce((h, {model,year,make}) => {
return Object.assign(h, { [make]:( h[make] || [] ).concat({model,year})})
}, {})
回答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);
回答6:
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);
回答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, a) => {
const {
model,
year,
make
} = a;
r[make] = [...r[make] || [], {
model,
year
}];
return r;
}, {});
console.log(cars);
回答8:
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 });
}
回答9:
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);
Да здравствуют высокие показатели мастерства программистов во имя процветания всего человечества! Ура, товарищи!
回答10:
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
回答11:
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));
回答12:
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');
回答13:
Just try this one it works fine for me.
let grouped = _.groupBy(cars, 'make');
回答14:
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"));
回答15:
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);
回答16:
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"
}
]
}
回答17:
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);
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<script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>
回答18:
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.
回答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'));
来源:https://stackoverflow.com/questions/40774697/how-to-group-an-array-of-objects-by-key