Question

我有一个像下面这样的数组

 [
   {
      "key":"Human Resources",
      "data":[
         {
            "id":"1HwahYXr5hs3IDTn7RtC",
            "element":{
               "id":"1HwahYXr5hs3IDTn7RtC",
               "ca":"1x"

            },
            "groupBy":"Human Resources_Human Resources"
         },
         {
            "id":"I7kCoWl7JLCWvgSdZm6p",
            "element":{
               "id":"I7kCoWl7JLCWvgSdZm6p",
               "ca":"2x"

            },
            "groupBy":"Human Resources"
         }
      ]
   },
   {
      "key":"Marketing",
      "data":[
         {
            "id":"Ltgk4vvKNILwYry1thXL",
            "element":{
               "id":"Ltgk4vvKNILwYry1thXL",
               "ca":"5x"
            },
            "groupBy":"Marketing_Marketing"
         },
         {
            "id":"aA43supw5IYmVLkniC8c",
            "element":{
               "id":"aA43supw5IYmVLkniC8c",
               "ca":"6x"
            },
            "groupBy":"Marketing"
         }
      ]
   },
   {
      "key":"Sales",
      "data":[
         {
            "id":"WwYl7FelUSVo8i5eMYMk",
            "element":{
               "id":"WwYl7FelUSVo8i5eMYMk",
               "ca":"8x"
            },
            "groupBy":"Sales"
         }
      ]
   }
]

我还有另一个数组，存储上面数组中键的排名，如下所示：

[
{
"id":1,
"name":"Marketing",
"rank":2
},
{
"id":1,
"name":"Human Resources",
"rank":1
}
]

我需要根据第二个数组中的键排位对第一个数组进行排序，如果没有第二个键，它将排在最前面。

请注意，有任何inbuit函数可以像lodash中的任何函数一样执行此操作。

谢谢

Answer 1

您可以将排名或-Infinity排在未排序的排名前。

var array = [{ key: "Human Resources" }, { key: "Marketing" }, { key: "Sales" }],
    ranking = [{ id: 1, name: "Marketing", rank: 2 }, { id: 1, name: "Human Resources", rank: 1 }],
    order = ranking.reduce((o, { name, rank }) => (o[name] = rank, o), Object.create(null));

array.sort(({ key: a }, { key: b }) => (order[a] || -Infinity) - (order[b] || -Infinity));

console.log(array);

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Answer 2

您可以针对每个排序，过滤数据并添加属性范围，然后对数据进行排序

ranking.forEach(x=>{
  data.filter(d=>d.key==d.name).forEach(df=>{
       df.rank=x.rank; //Add a new property
     }
  })

ranking.sort((a,b)=>(a.rank || 0)-(b.rank || 0));

我的排名是数组宽度的排名，数据是数组宽度的数据

Answer 3

这应该有效，

arr.sort((a,b) => {
  rank1 = rank.find(e => e.name ===a.key);
  rank2 = rank.find(e => e.name ===b.key);
  if (!rank1 || !rank2) {
    return 1;
  }
  if (rank1.rank < rank2.rank) {
    return -1;
 } else {
   return 1;
}
})

角排序数组作为另一个数组

3 个答案: