如果我有一个JavaScript对象,如:

var list = {
  "you": 100, 
  "me": 75, 
  "foo": 116, 
  "bar": 15
};

是否有一种方法可以根据值对属性进行排序?最后得到

list = {
  "bar": 15, 
  "me": 75, 
  "you": 100, 
  "foo": 116
};

当前回答

打印稿

下面的函数根据值或值的属性对对象进行排序。如果你不使用TypeScript,你可以删除类型信息,将其转换为JavaScript。

/**
 * Represents an associative array of a same type.
 */
interface Dictionary<T> {
  [key: string]: T;
}

/**
 * Sorts an object (dictionary) by value or property of value and returns
 * the sorted result as a Map object to preserve the sort order.
 */
function sort<TValue>(
  obj: Dictionary<TValue>,
  valSelector: (val: TValue) => number | string,
) {
  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);
}

使用

var list = {
  "one": { height: 100, weight: 15 },
  "two": { height: 75, weight: 12 },
  "three": { height: 116, weight: 9 },
  "four": { height: 15, weight: 10 },
};

var sortedMap = sort(list, val => val.height);

JavaScript对象中键的顺序是不保证的,所以我将排序并将结果返回为一个保留排序顺序的Map对象。

如果你想把它转换回Object,你可以这样做:

var sortedObj = {} as any;
sortedMap.forEach((v,k) => { sortedObj[k] = v });

其他回答

我为此做了一个插件,它接受1个arg,这是一个未排序的对象,并返回一个对象,已排序的道具值。这将适用于所有二维对象,如{"Nick": 28, "Bob": 52}…

var sloppyObj = {
    'C': 78,
    'A': 3,
    'B': 4
};

// Extend object to support sort method
function sortObj(obj) {
    "use strict";

    function Obj2Array(obj) {
        var newObj = [];
        for (var key in obj) {
            if (!obj.hasOwnProperty(key)) return;
            var value = [key, obj[key]];
            newObj.push(value);
        }
        return newObj;
    }

    var sortedArray = Obj2Array(obj).sort(function(a, b) {
        if (a[1] < b[1]) return -1;
        if (a[1] > b[1]) return 1;
        return 0;
    });

    function recreateSortedObject(targ) {
        var sortedObj = {};
        for (var i = 0; i < targ.length; i++) {
            sortedObj[targ[i][0]] = targ[i][1];
        }
        return sortedObj;
    }
    return recreateSortedObject(sortedArray);
}

var sortedObj = sortObj(sloppyObj);

alert(JSON.stringify(sortedObj));

下面是该函数按预期工作的演示 http://codepen.io/nicholasabrams/pen/RWRqve?editors=001

function sortObjByValue(list){
 var sortedObj = {}
 Object.keys(list)
  .map(key => [key, list[key]])
  .sort((a,b) => a[1] > b[1] ? 1 : a[1] < b[1] ? -1 : 0)
  .forEach(data => sortedObj[data[0]] = data[1]);
 return sortedObj;
}
sortObjByValue(list);

Github Gist Link

以防万一,有人正在寻找保持对象(键和值),使用@Markus R和@James Moran注释的代码引用,只需使用:

var list = {"you": 100, "me": 75, "foo": 116, "bar": 15};
var newO = {};
Object.keys(list).sort(function(a,b){return list[a]-list[b]})
                 .map(key => newO[key] = list[key]);
console.log(newO);  // {bar: 15, me: 75, you: 100, foo: 116}

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)

如果我有一个这样的对象,

var dayObj = {
              "Friday":["5:00pm to 12:00am"] ,
              "Wednesday":["5:00pm to 11:00pm"],
              "Sunday":["11:00am to 11:00pm"], 
              "Thursday":["5:00pm to 11:00pm"],
              "Saturday":["11:00am to 12:00am"]
           }

我想按天排序,

我们应该先有daySorterMap,

var daySorterMap = {
  // "sunday": 0, // << if sunday is first day of week
  "Monday": 1,
  "Tuesday": 2,
  "Wednesday": 3,
  "Thursday": 4,
  "Friday": 5,
  "Saturday": 6,
  "Sunday": 7
}

初始化一个单独的对象sortedDayObj,

var sortedDayObj={};
Object.keys(dayObj)
.sort((a,b) => daySorterMap[a] - daySorterMap[b])
.forEach(value=>sortedDayObj[value]= dayObj[value])

你可以返回sortedDayObj