Question

我有一系列对象，我需要＆＃34;排序＆＃34;并根据给出的回调函数放入另一个对象。例如：

var list = [{id: "102", name: "Alice"},
            {id: "205", name: "Bob", title: "Dr."},
            {id: "592", name: "Clyde", age: 32}];


groupBy(list, function(i) { return i.name.length; });

应该返回：

  {
    "3": [{id: "205", name: "Bob", title: "Dr."}],
    "5": [{id: "102", name: "Alice"},
      {id: "592", name: "Clyde", age: 32}]

这是我到目前为止所得到的但是我被卡住了......

function groupBy(arr, cb) {
let result = {};
arr.forEach(function (obj) {
  var group = [];
  group.push(cb(i))
  result[group].push(obj);
}
  return result;
)}

Answer 1

我将如何做到这一点：

 function groupBy(arr, cb){
   const res = {};
   for(const el of arr){
     const key = cb(el);
     (res[key] || (res[key] = [])).push(el);
   }
   return res;
 }

为什么它不起作用：

 function groupBy(arr, cb) {
  let result = {};
  arr.forEach(function (obj) {
    //That creates one group for each obj?
    var group = [];
   //group should contain the obj, not the key?
   group.push(cb(i))
   // Now you are accessing the result by an array as index??
   // And try to push the object to it which will obvipusly fails as you never set result[group] to an array
   result[group].push(obj);
   }
 return result;
//Typo??
)}

Answer 2

您需要从return value获取callback并对此做出决定

var list = [{id: "102", name: "Alice"},
            {id: "205", name: "Bob", title: "Dr."},
            {id: "592", name: "Clyde", age: 32}];

 function groupBy(arr, cb) {
        let result = {};
        arr.forEach(function (obj) {
           var group = [];
           const length = cb(obj)
           // if result[length] is undefined then create a new array with the current object 
           // else push into the existing array
           result[length] ? result[length].push(obj) : result[length] = [obj];   
        })
        return result;

}

console.log(groupBy(list, function(i) { return i.name.length}))

Answer 3

使用数组reduce功能。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

const list = [{id: "102", name: "Alice"},
            {id: "205", name: "Bob", title: "Dr."},
            {id: "592", name: "Clyde", age: 32}];
            
const result = list.reduce((acc, curr) => {
  const { name } = curr;
  acc[name.length] = (acc[name.length] || []).concat(curr);
  return acc;
}, {});


console.log(result);

基于回调函数的组对象

3 个答案: