js对象数组按照另一个数组排序

Posted 2023-04-04

参考技术A 场景
下拉列表有4个选项
用户按照先后排序后保存
[2, 4, 1, 3]
保存完后，按照顺序进行排序

思路，两个数组，排序有点困难
如果变成一个数组排序，就方便了

我们给每个option新增一个排序属性sordId
这个sordId的值就是 option的id在 showArr中的索引位置

现在我们的options数组已经可以排序了

再写一个对象数组按照属性名排序的方法

最后输出

我们如何根据节点js中的另一个对象数组值对对象数组进行排序

【中文标题】我们如何根据节点js中的另一个对象数组值对对象数组进行排序

【英文标题】：How can we sort the object array based on another object array value in node js

【发布时间】：2018-09-05 20:12:52

【问题描述】：

我们如何根据另一个对象数组值对对象数组进行排序，就像我们有一个对象数组一样：

var _userEnd = ["userID":554,"tEnd":6,"userID":597,"tEnd":3,"userID":605,"tEnd":3,"userID":617,"tEnd":1,"userID":553,"tEnd":1,"userID":616,"tEnd":1,"userID":596,"tEnd":0]

在此索引 1,2,3,4,5 具有相同的键“tEnd”值，即 3 和 1，所以我只需要根据另一个对象数组对这些索引进行排序

var _profsort=["userID":596,"score":100,"userID":616,"score":95,"userID":553,"score":100,"userID":617,"score":85,"userID":605,"score":95,"userID":597,"score":85,"userID":554,"score":100]

这里用户 ID 597,605,617,553,616 的得分值为 85,95,85,100,95

所以根据分数，我想对我的第一个数组进行排序 输出应该是：

["userID":554,"tEnd":6,"userID":605,"tEnd":3,"userID":597,"tEnd":3,"userID":553,"tEnd":1,"userID":616,"tEnd":1,"userID":617,"tEnd":1,"userID":596,"tEnd":0]

【问题讨论】：

看看在developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…上使用 compareFunction 的例子

【参考方案1】：

您可以获取tEnd 的增量并找到score 并获取该增量。

function getScore(uID) 
    return (_profsort.find(( userID ) => userID === uID) ||  score: 0 ).score;


var _userEnd = [ userID: 554, tEnd: 6 ,  userID: 597, tEnd: 3 ,  userID: 605, tEnd: 3 ,  userID: 617, tEnd: 1 ,  userID: 553, tEnd: 1 ,  userID: 616, tEnd: 1 ,  userID: 596, tEnd: 0 ],
    _profsort = [ userID: 596, score: 100 ,  userID: 616, score: 95 ,  userID: 553, score: 100 ,  userID: 617, score: 85 ,  userID: 605, score: 95 ,  userID: 597, score: 85 ,  userID: 554, score: 100 ];
    
_userEnd.sort(function (a, b) 
    return b.tEnd - a.tEnd || getScore(b.userID) - getScore(a.userID);
);

console.log(_userEnd);

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【讨论】：

一如既往的优雅 :) +1

【参考方案2】：

使用sort 和find

var fnGetScore = uid => _profsort.find(s => s.userID == uid).score; //method to get the score value based on userId

var output = _userEnd.sort( (a, b) => 
        (b.tEnd - a.tEnd) || 
           fnGetScore(b.userID) - fnGetScore(b.userID)) //compare tEnd first and if they are same, compare the score value

演示

var _userEnd = [
  "userID": 554,
  "tEnd": 6
, 
  "userID": 597,
  "tEnd": 3
, 
  "userID": 605,
  "tEnd": 3
, 
  "userID": 617,
  "tEnd": 1
, 
  "userID": 553,
  "tEnd": 1
, 
  "userID": 616,
  "tEnd": 1
, 
  "userID": 596,
  "tEnd": 0
];

var _profsort = [
  "userID": 596,
  "score": 100
, 
  "userID": 616,
  "score": 95
, 
  "userID": 553,
  "score": 100
, 
  "userID": 617,
  "score": 85
, 
  "userID": 605,
  "score": 95
, 
  "userID": 597,
  "score": 85
, 
  "userID": 554,
  "score": 100
];

var fnGetScore = uid => _profsort.find(s => s.userID == uid).score;

var output = _userEnd.sort((a, b) => (b.tEnd - a.tEnd) || fnGetScore(b.userID) - fnGetScore(b.userID))

console.log(output);

【参考方案3】：

通过userID 使用Array.reduce 创建score 的对象。按tEnd 排序，如果相等，则按score 排序，您从sortObject 得到userId：

var _userEnd = ["userID":554,"tEnd":6,"userID":597,"tEnd":3,"userID":605,"tEnd":3,"userID":617,"tEnd":1,"userID":553,"tEnd":1,"userID":616,"tEnd":1,"userID":596,"tEnd":0]

var _profsort=["userID":596,"score":100,"userID":616,"score":95,"userID":553,"score":100,"userID":617,"score":85,"userID":605,"score":95,"userID":597,"score":85,"userID":554,"score":100]

var sortObj = _profsort.reduce((r, o) => 
  r[o.userID] = o.score;

  return r;
, Object.create(null));

_userEnd.sort((a, b) => b.tEnd - a.tEnd || sortObj[b.userID] - sortObj[a.userID]);

console.log(_userEnd);