javascript如何通过两条规则对对象数组进行排序？ [复制]

Question

我有一个这样的对象列表：

 [
      {"groups":["aaa"]},
      {"groups":["bbb"]},
      {"groups":["ccc"]},
      {"groups":["ddd"]},
      {"groups":["eee"]},
      {"groups":["fff"]},
      {"groups":["ggg"]},
      {"groups":["hhh"]},
      {"groups":["iii"]},
      {"groups":["mmm", "mmm"]}
      {"groups":["lll", "lll"]}
];

我想按字母顺序将具有多个组的所有对象按字母顺序放在列表的末尾，并将其他对象与一组放在列表的开头而不更改原始顺序。

这是我的代码

var list = [
  {"groups":["aaa"]},
  {"groups":["bbb"]},
  {"groups":["ccc"]},
  {"groups":["ddd"]},
  {"groups":["eee"]},
  {"groups":["fff"]},
  {"groups":["ggg"]},
  {"groups":["hhh"]},
  {"groups":["iii"]},
  {"groups":["mmm", "mmm"]},
  {"groups":["lll", "lll"]}
];

list.sort(function(a, b){
        var aIsGroup = (a.groups.length > 1);
        var bIsGroup = (b.groups.length > 1);

        if (aIsGroup && !bIsGroup) {
            return 1;
        } else if (!aIsGroup && bIsGroup) {
            return -1;
        } else if(aIsGroup && bIsGroup){
            return a.groups[0].toLowerCase().localeCompare(b.groups[0].toLowerCase());
        }
        return 0;
});

console.log(list);

在sn -p中可以看到当前的输出，但是预期的输出是：

[
  {
    "groups": [
      "aaa"
    ]
  },
  {
    "groups": [
      "bbb"
    ]
  },
  {
    "groups": [
      "ccc"
    ]
  },
  {
    "groups": [
      "ddd"
    ]
  },
  {
    "groups": [
      "eee"
    ]
  },
  {
    "groups": [
      "fff"
    ]
  },
  {
    "groups": [
      "ggg"
    ]
  },
  {
    "groups": [
      "hhh"
    ]
  },
  {
    "groups": [
      "iii"
    ]
  },
  {
    "groups": [
      "lll",
      "lll"
    ]
  },
  {
    "groups": [
      "mmm",
      "mmm"
    ]
  }
]

Answer 1

正如@deceze 在评论中所说的

Javascript 的排序不能保证稳定的排序。任何“相等”的元素都有未定义的顺序。请参阅副本以实现稳定排序

但您可以这样做以使它们保持与原始顺序相同的顺序（稳定排序）

var list = [
  {"groups":["aaa"]},
  {"groups":["bbb"]},
  {"groups":["ccc"]},
  {"groups":["mmm", "mmm"]},
  {"groups":["ddd"]},
  {"groups":["fff"]},
  {"groups":["eee"]},
  {"groups":["ggg"]},
  {"groups":["hhh"]},
  {"groups":["iii"]},
  {"groups":["lll", "lll"]}
];

list.sort(function(a, b){
        var aIsGroup = (a.groups.length > 1);
        var bIsGroup = (b.groups.length > 1);

        if (aIsGroup && !bIsGroup) {
            return 1;
        } else if (!aIsGroup && bIsGroup) {
            return -1;
        } else if(aIsGroup && bIsGroup) {
            return a.groups[0].toLowerCase().localeCompare(b.groups[0].toLowerCase());
        }
        // keep in the same order
        return (list.indexOf(a) < list.indexOf(b)) ? -1 : 1;
});

console.log(list);