Question

我有一系列正确的分数，如下所示：

var correctScores = [
   {score: "2-2", code: "draw"},
   {score: "2-1", code: "home"},
   {score: "0-1", code: "away"},
   {score: "1-0", code: "home"},
   {score: "3-0", code: "home"},
   {score: "any home win", code: "home"},
   {score: "1-2", code: "away"},
   {score: "0-3", code: "away"},
   {score: "any draw", code: "draw"},
   {score: "0-0", code: "draw"},
   {score: "any away win", code: "away"},
   {score: "0-2", code: "away"},
   {score: "1-1", code: "draw"},
   {score: "2-0", code: "home"}
]

我已设法按home，draw，away对其进行排序，但我也尝试将其扩展到得分，以便它们按照一致的顺序排列。当我说一致的顺序时，我想要对得分进行排序，以便最少数量的目标在阵列中排在第一位。如果目标数量相同，则应首先显示具有更多主目标的游戏。

目前，我的代码如下所示，仅按home，draw，away排序。我发现很难扩展这种排序方法也按分数排序。分数也将不断更新，因此这些分数不一定总是与数组中的分数相同。

let order = { HOME: 1, DRAW: 2, AWAY: 3 };

correctScores.sort(function (a, b) { return order[a.code] - order[b.code]; });

我试图让数组看起来像这样：

var correctScores = [
   {score: "1-0", code: "home"},
   {score: "2-0", code: "home"},
   {score: "3-0", code: "home"},
   {score: "2-1", code: "home"},
   {score: "any home win", code: "home"}

   {score: "0-0", code: "draw"},
   {score: "1-1", code: "draw"},
   {score: "2-2", code: "draw"},
   {score: "any draw", code: "draw"},

   {score: "0-1", code: "away"},
   {score: "0-2", code: "away"},
   {score: "0-3", code: "away"},
   {score: "1-2", code: "away"},
   {score: "any away win", code: "away"}

]

任何人都可以就如何扩展这种排序方法给我任何建议吗？

Answer 1

基本上，您需要订购对象的小键，并且您需要拆分分数以便使用值进行排序。

然后，您可以从索引1开始链接增量，并按索引零的增量排序。

＆＃13;

var correctScores = [{ score: "2-2", code: "draw" }, { score: "2-1", code: "home" }, { score: "0-1", code: "away" }, { score: "1-0", code: "home" }, { score: "3-0", code: "home" }, { score: "1-2", code: "away" }, { score: "0-3", code: "away" }, { score: "0-0", code: "draw" }, { score: "0-2", code: "away" }, { score: "1-1", code: "draw" }, { score: "2-0", code: "home" }],
    order = { home: 1, draw: 2, away: 3 };

correctScores.sort(function (a, b) {
    function getParts(s) { return s.split('-'); }
    var aa = getParts(a.score),
        bb = getParts(b.score);

    return order[a.code] - order[b.code] || aa[1] - bb[1] || aa[0] - bb[0];
});

console.log(correctScores);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

更新问题的解决方案，检查分数是否不可拆分，然后选择Infinite作为值的数组，因为它将排序到底部。

＆＃13;

var correctScores = [{ score: "2-2", code: "draw" }, { score: "2-1", code: "home" }, { score: "0-1", code: "away" }, { score: "1-0", code: "home" }, { score: "3-0", code: "home" }, { score: "any home win", code: "home" }, { score: "1-2", code: "away" }, { score: "0-3", code: "away" }, { score: "any draw", code: "draw" }, { score: "0-0", code: "draw" }, { score: "Any away win", code: "away" }, { score: "0-2", code: "away" }, { score: "1-1", code: "draw" }, { score: "2-0", code: "home" }],
    order = { home: 1, draw: 2, away: 3 };

correctScores.sort(function (a, b) {
    function getParts(s) {
        var t = s.split('-');
        return t.length === 1 ? [Infinity, Infinity] : t;
    }
    var aa = getParts(a.score),
        bb = getParts(b.score);

    return order[a.code] - order[b.code] || aa[1] - bb[1] || aa[0] - bb[0];
});

console.log(correctScores);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

Answer 2

那样

var correctScores = [
   {score: "2-1", code: "home"},
   {score: "1-0", code: "home"},
   {score: "3-0", code: "home"},
   {score: "2-0", code: "home"},

   {score: "1-1", code: "draw"},
   {score: "0-0", code: "draw"},
   {score: "2-2", code: "draw"},

   {score: "0-3", code: "away"},
   {score: "0-2", code: "away"},
   {score: "0-1", code: "away"},
   {score: "1-2", code: "away"}

];

var groups = {};
var scores = [];

correctScores.map(function(e) { 
  groups[e.code] = groups[e.code] || [];
  groups[e.code].push(e);
});

for(var i in groups) {
  groups[i].sort(function (a, b) {
    var scoreA = a.score.split('-').map(Number);
    var scoreB = b.score.split('-').map(Number);
    
    return scoreA[0]+scoreA[1] > scoreB[0]+scoreB[1] ? 1:(
     scoreA[0]+scoreA[1] < scoreB[0]+scoreB[1] ? -1:0);
  });
  scores = scores.concat(groups[i]);
}

console.log(scores);

Answer 3

包括新要求：

  var correctScores = [
  {score: "2-2", code: "draw"},
  {score: "2-1", code: "home"},
  {score: "0-1", code: "away"},
  {score: "1-0", code: "home"},
  {score: "3-0", code: "home"},
  {score: "Any home win", code: "home"},
  {score: "1-2", code: "away"},
  {score: "0-3", code: "away"},
  {score: "any draw", code: "draw"},
  {score: "0-0", code: "draw"},
  {score: "Any away win", code: "away"},
  {score: "0-2", code: "away"},
  {score: "1-1", code: "draw"},
  {score: "2-0", code: "home"}
];

let order = { home: 1, draw: 2, away: 3 };

correctScores.sort(function (a, b) {
  s1 = parseInt(a.score.split("-")[0]);
  s2 = parseInt(b.score.split("-")[0]);
  s1s = s1+parseInt(a.score.split("-")[1]);
  s2s = s2+parseInt(b.score.split("-")[1]);
  if(a.score.toLowerCase().indexOf("any")>-1){s1=999;s1s=999;}
  if(b.score.toLowerCase().indexOf("any")>-1){s2=999;s2s=999;}
  if(order[a.code]<order[b.code]) return -1;
  if(order[a.code]>order[b.code]) return 1;
  if(s1s<s2s) return -1;
  if(s1s>s2s) return 1;
  if(s1>s2) return -1;
  if(s1<s2) return 1;  
});

console.log(correctScores);

小提琴：https://jsfiddle.net/ct6knnqb/1/

根据许多条件对数组中的项进行排序

3 个答案: