Question

我的Node服务器中有一个包含数千个对象的对象，每个对象都是一个人，并且他有一个总分数。它看起来像这样：

var mainObject = { //it's unsorted, but no need to sort it anyway
    john: {totalScore: 5, info: "No"},
    james: {totalScore: 3, info: "No"},
    lee: {totalScore: 55, info: "No"}
}

现在，我想在mainObject中保留1000个对象，但问题是我需要在发送到客户端之前使用delete mainObject [name]擦除最低分数（除了delete之外的其他解决方案是可接受的）。

mainObject中每个对象的索引不应该相互切换（有一个原因），只需要得到最低分数，列表应该作为对象发送到客户端并且确切地说在1000长度。

Answer 1

您正在尝试将基于数组的概念（如长度和排序）应用于Object。这根本不会。

要实现您的目标，您必须将对象转换为数组。你可以这样做的一种方式是：

var mainObject = {
    john: {totalScore: 5, info: "No"},
    james: {totalScore: 3, info: "No"},
    lee: {totalScore: 55, info: "No"},
};

var mainArray = Object.keys(mainObject).map(function (name) {
  return Object.assign({}, mainObject[name], {name});
});

console.log(mainArray);

然后，您可以按分数对数组进行排序，并仅返回N个最高条目：

var mainArray = [
  {
    "totalScore": 5,
    "info": "No",
    "name": "john"
  },
  {
    "totalScore": 3,
    "info": "No",
    "name": "james"
  },
  {
    "totalScore": 55,
    "info": "No",
    "name": "lee"
  }
]

mainArray.sort(function (a, b) {
  return b.totalScore - a.totalScore;
});

console.log(mainArray.slice(0, 1000));

如果由于某种原因你的客户绝对需要来取回一个Object而不是一个对象数组，你总是可以转换回来：

var mainArray = [
  {
    "totalScore": 5,
    "info": "No",
    "name": "john"
  },
  {
    "totalScore": 3,
    "info": "No",
    "name": "james"
  },
  {
    "totalScore": 55,
    "info": "No",
    "name": "lee"
  }
];

var mainObject = mainArray.reduce(function (prev, curr) {
  var name = curr.name;
  delete curr.name;
  prev[name] = curr;
  return prev;
}, {});

console.log(mainObject);

但是在这一点上，你已经做了相当荒谬的工作来实现你的目标。它暗示的足够的工作实际上是an XY Problem，你应该重新考虑导致你到这一步的决定。

Answer 2

使用lodash

var desiredAmountOfObjects = 2

var mainObject = { //it's unsorted, but no need to sort it anyway
    john: {totalScore: 5, info: "No"},
    james: {totalScore: 3, info: "No"},
    lee: {totalScore: 55, info: "No"}
}

var mainArray = _.map(mainObject, (value, key) => Object.assign(value, { key }))

mainArray.sort(function (a, b) {
  return b.totalScore - a.totalScore;
});

var choppingBlock = mainArray.slice(desiredAmountOfObjects).map(chop => chop.key)

console.log(choppingBlock);

var desiredObject = _.omitBy(mainObject, (value, key) => choppingBlock.indexOf(key) > -1)

console.log(desiredObject)

http://jsbin.com/gijuduzole/1/edit?js,console

将desiredAmountOfObject设置为1000或其他。

Answer 3

您可以创建一个删除分数最低的新对象 Object.keys和Array迭代方法：

const mainObject = {
    john: {totalScore: 5, info: "No"},
    james: {totalScore: 3, info: "No"},
    lee: {totalScore: 55, info: "No"}
};

const keys = Object.keys(mainObject);

const lowest = keys.map(key => mainObject[key].totalScore).sort()[0];

const removed = keys.reduce((result, key) => {
    if(mainObject[key].totalScore > lowest){
        result[key] = mainObject[key];
    }
    return result;
}, Object.create(null));

console.log('without objects with lowest score');
console.log(removed);

// EDIT -- Or, to retain keys:

const removed2 = keys.reduce((result, key) => {
    result[key] = mainObject[key].totalScore > lowest
        ? mainObject[key]
        : undefined;
    return result;
}, Object.create(null));

console.log('retaining all keys');
console.log(removed2);

参考
Object.keys
Array.prototype.map
Array.prototype.reduce

Answer 4

根据评论 - 从初始对象创建一个可以排序的数组，然后获取除x结果之外的所有结果，并从初始对象中删除这些结果。

// object
var mainObject = { //it's unsorted, but no need to sort it anyway
        john: {totalScore: 5, info: "No"},
        james: {totalScore: 3, info: "No"},
        lee: {totalScore: 55, info: "No"}
    }

// array
var array = [];
for (var i in mainObject)
array.push({ key: i, score: mainObject[i].totalScore });

// sort
array.sort(function(a,b){ return a.score < b.score });

// keep top x
var remove = array.slice(2);

// delete all the others
for (var i=0; i<remove.length; i++)
delete mainObject[remove[i].key];

console.log(mainObject);

小提琴：https://jsfiddle.net/xrc8hpya/1/

Answer 5

有很多图书馆可以帮助你解决这类问题（下划线，lodash，ramda）。
使用ramda的一个例子：

const mainObject = { //it's unsorted, but no need to sort it anyway
    john: {totalScore: 5, info: "No"},
    james: {totalScore: 3, info: "No"},
    lee: {totalScore: 55, info: "No"}
}

const obj = R.filter((person) => person.totalScore > 50, mainObject);

这样您就不必担心转换为数组，只需应用过滤器即可获得对象。

JS - 删除对象列表中不必要的对象

5 个答案: