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