有人知道一种方法(lodash如果可能的话)通过对象键分组对象数组，然后根据分组创建一个新的对象数组吗?例如，我有一个汽车对象数组:
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'
},
];
我想创建一个新的汽车对象数组，由make分组:
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'
}
]
}
我喜欢写它没有依赖/复杂性，只是纯粹的简单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)
同意除非经常使用这些库，否则不需要外部库。虽然有类似的解决方案，但我发现其中一些很难遵循。如果您试图理解正在发生的事情，这里有一个带有注释的解决方案的要点。
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))
提莫的答案是我会怎么做。简单的_。groupBy，并允许在分组结构中的对象中有一些重复。
然而，OP还要求删除重复的make键。如果你想从头到尾:
var grouped = _.mapValues(_.groupBy(cars, 'make'),
clist => clist.map(car => _.omit(car, 'make')));
console.log(grouped);
收益率:
{ audi:
[ { model: 'r8', year: '2012' },
{ model: 'rs5', year: '2013' } ],
ford:
[ { model: 'mustang', year: '2012' },
{ model: 'fusion', year: '2015' } ],
kia:
[ { model: 'optima', year: '2012' } ]
}
如果你想使用Underscore.js来实现这个功能，请注意它的_. js版本。mapValues被称为_.mapObject。
你也可以像这样使用数组#forEach()方法:
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);
