这可能是一种将其作为真实有序对象处理的简单方法。不知道它有多慢。也可能更好的while循环。

Object.sortByKeys = function(myObj){
  var keys = Object.keys(myObj)
  keys.sort()
  var sortedObject = Object()
  for(i in keys){
    key = keys[i]
    sortedObject[key]=myObj[key]
   }

  return sortedObject

}

然后我找到了这个逆函数 http://nelsonwells.net/2011/10/swap-object-key-and-values-in-javascript/

Object.invert = function (obj) {

  var new_obj = {};

  for (var prop in obj) {
    if(obj.hasOwnProperty(prop)) {
      new_obj[obj[prop]] = prop;
    }
  }

  return new_obj;
};

So

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15};
var invertedList = Object.invert(list)
var invertedOrderedList = Object.sortByKeys(invertedList)
var orderedList = Object.invert(invertedOrderedList)

找出每个元素的频率，并按频率/值进行排序。

Let response =["苹果"，"橘子"，"苹果"，"香蕉"，"橘子"，"香蕉"，"香蕉"]; 设frequency = {}; response.forEach(函数(项){ 频率[项目]=频率[项目]?频率[项]+ 1:1; })； console.log(频率); let intents = Object.entries(frequency) .sort((a, b) => b[1] - a[1]) . map(函数(x) { 返回x [0]; })； console.log(意图);

输出:

{ apple: 2, orange: 2, banana: 3 }
[ 'banana', 'apple', 'orange' ]

非常简短简单!

var sortedList = {};
Object.keys(list).sort((a,b) => list[a]-list[b]).forEach((key) => {
    sortedList[key] = list[key]; });

下划线。js或Lodash.js用于高级数组或对象排序

var data = { "models": { "LTI": [ "TX" ], "Carado": [ "A", "T", "A(пасс)", "A(груз)", "T(пасс)", "T(груз)", "A", "T" ], "SPARK": [ "SP110C 2", "sp150r 18" ], "Autobianchi": [ "A112" ] } }; var arr = [], obj = {}; for (var i in data.models) { arr.push([i, _.sortBy(data.models[i], function(el) { return el; })]); } arr = _.sortBy(arr, function(el) { return el[0]; }); _.map(arr, function(el) { return obj[el[0]] = el[1]; }); console.log(obj); <script src="https://cdn.jsdelivr.net/npm/lodash@4.17.21/lodash.min.js" integrity="sha256-qXBd/EfAdjOA2FGrGAG+b3YBn2tn5A6bhz+LSgYD96k=" crossorigin="anonymous"></script>

感谢@orad在TypeScript中提供了答案。现在，我们可以在JavaScript中使用下面的代码片断。

function sort(obj,valSelector) { const sortedEntries = Object.entries(obj) .sort((a, b) => valSelector(a[1]) > valSelector(b[1]) ? 1 : valSelector(a[1]) < valSelector(b[1]) ? -1 : 0); return new Map(sortedEntries); } const Countries = { "AD": { "name": "Andorra", }, "AE": { "name": "United Arab Emirates", }, "IN": { "name": "India", }} // Sort the object inside object. var sortedMap = sort(Countries, val => val.name); // Convert to object. var sortedObj = {}; sortedMap.forEach((v,k) => { sortedObj[k] = v }); console.log(sortedObj); //Output: {"AD": {"name": "Andorra"},"IN": {"name": "India"},"AE": {"name": "United Arab Emirates"}}

另一种解决方法:-

var res = [{"s1":5},{"s2":3},{"s3":8}].sort(function(obj1,obj2){ 
 var prop1;
 var prop2;
 for(prop in obj1) {
  prop1=prop;
 }
 for(prop in obj2) {
  prop2=prop;
 }
 //the above two for loops will iterate only once because we use it to find the key
 return obj1[prop1]-obj2[prop2];
});

//res将有结果数组