javascript - ES6 groupBy，分区，排序对象列表

Question

我试图在 ES6 中找到一种优雅的方式来根据指定的值对对象数组进行排序。场景如下:

const list = [
{
"name": "john",
"lastName": "smith"
}, {
"name": "tony",
"lastName": "smith"
}, {
"name": "tony",
"lastName": "grey"
}, {
"name": "mary",
"lastName": "smith"
}, {
"name": "john",
"lastName": "x"
}, {
"name": "tom",
"lastName": "y"
}
, {
"name": "mary",
"lastName": "x"
}
]

let orderList = [{"name":["john","mary"]}, {"lastName":["x"]}];

因此，基本上按姓名 (John, Mary) 对结果进行排序，然后按 lastName(x) 对结果进行排序，但姓名排序仍然优先。结果应如下所示:

[
  {
   "name": "john",
   "lastName": "x"
  }, {
   "name": "john",
   "lastName": "smith"
  }, {
    "name": "mary",
   "lastName": "x"
  }, {
   "name": "mary",
   "lastName": "smith"
  }, {
   "name": "tony",
   "lastName": "smith"
   }, {
    "name": "tony",
    "lastName": "grey"
   }, {
    "name": "tom",
    "lastName": "y"
   }
 ]

我已经尝试过使用 group by 做一些事情，但这是针对每个名字和姓氏的手动过程。

_.groupBy(列表, {"name": "john"});

我也尝试过使用 array reduce 进行试验，但似乎找不到好的动态解决方案。

const sortArr = ['john', 'mary'];
const sortedList= list.reduce((result, element) => {
   let index = sortArr.findIndex(x => x === element.name);
   result[index !== -1
      ? index
      : result.length - 1].push(element); 
     return result;
       },[  [], [], [] ]);

感谢任何帮助。谢谢

Answer 1

你可以使用 Array#sort并采用迭代方法首先按 name 排序，然后按 lastName 排序。

该提案检查属性是否在数组中，并通过获取索引的增量将这些值排序到顶部。重复此操作，直到增量不为零或订单数组结束为止。

为了获得增量，一个 Array#indexOf进行搜索，如果未找到项的值 -1 被替换为 Infinity，因为必须将此项排序到数组的末尾。找到索引的项目根据索引排序。

为了加快排序速度，具有单个键/值对的对象被转换为具有键和值的数组。

var list = [{ name: "john", lastName: "smith" }, { name: "tony", lastName: "smith" }, { name: "tony", lastName: "grey" }, { name: "mary", lastName: "smith" }, { name: "john", lastName: "x" }, { name: "tom", lastName: "y" }, { name: "mary", lastName: "x" }],
    orderList = [{ name: ["john", "mary"] }, { lastName: ["x"] }],
    order = orderList.map(o => (k => [k, o[k]])(Object.keys(o)[0]));

list.sort((a, b) => {
    var d;
    order.some(([k, v]) =>
        d = (v.indexOf(a[k]) + 1 || Infinity) - (v.indexOf(b[k]) + 1 || Infinity)
    );
    return d;
});

console.log(list);

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