构建给定平面数组的树结构

Question

我想将平面数组结构转换为最多 N 级的父子树结构。 键 L0、L1、L2、L3、L4、L5、L6 表示为级别，它们可以 go 最高 L6。 我对递归 function 有一些想法，但不确定如何实现

下面是给定的平面数组

[{
  "L0": "India",
  "L0_ID": "IN",
  "L1": "Karnataka",
  "L1_ID": "KA",
  "L2": "BLR",
  "L3": "Mysore",
  "L4": "CHB"
},
{
  "L0": "India",
  "L0_ID": "IN",
  "L1": "Karnataka",
  "L1_ID": "KA",
  "L2": "BLR",
  "L3": "Hubli"
},
{
  "L0": "India",
  "L0_ID": "IN",
  "L1": "Karnataka",
  "L1_ID": "KA",
  "L2": "BLR",
  "L3": "Udupi"
},
{
  "L0": "India",
  "L0_ID": "IN",
  "L1": "Rajasthan",
  "L1_ID": "RJ",
  "L2": "Jaipur",
  "L3": "Shastri Nagar",
  "L4": "Lane One"
},
{
  "L0": "India",
  "L0_ID": "IN",
  "L1": "Rajasthan",
  "L1_ID": "RJ",
  "L2": "Jodhpur",
  "L3": "Shastri Nagar",
  "L4": "Lane One"
}] 

我想使用 javascript 生成所需的树数组结构。

{
  "name": "India",
  "id": "IN",
  "child": [
    {
      "name": "Karnataka",
      "id": "KA",
      "child": [
        {
          "name": "BLR",
          "id": null,
          "child": [
            {
              "name": "Mysore",
              "id": null
            },
            {
              "name": "Hubli",
              "id": null
            },
            {
              "name": "Udupi",
              "id": null
            }
          ]
        }
      ]
    },
    {
      "name": "Rajasthan",
      "id": "RJ"
    }
  ]
}

这是我迄今为止尝试过的

 function generate_tree_map(level,data) {
                                     //Get ALL UNIQUE RECORDS AT GIVEN LEVEL
                                       // var unique_records_l0 = _.uniqBy(response, function (e) {
                                        var unique_records_l0 = _.uniqWith(data, function(first, second) {
                                            if (level == 0) {
                                                //console.log("LEVEL ZERO => ", e.L0)
                                                return first.L0 === second.L0 
                                            }
                                            if (level == 1) {
                                                //console.log("LEVEL ONE => ", e.L0, e.L1)
                                                return first.L0 === second.L0 && first.L1 === second.L1
                                            }
                                            if (level == 2) {
                                                //console.log("LEVEL TWO => ", e.L0, e.L1, e.L2)
                                                return first.L0 === second.L0 && first.L1 === second.L1 && first.L2 === second.L2
                                            }
                                            if (level == 3) {
                                                //console.log("LEVEL THREE => ", e.L0, e.L1, e.L2, e.L3)
                                                return first.L0 === second.L0 && first.L1 === second.L1 && first.L2 === second.L2 && first.L3 === second.L3
                                                
                                            }
                                            if (level == 4) {
                                                //console.log("LEVEL FOUR => ", e.L0, e.L1, e.L2, e.L3, e.L4)
                                                return first.L0 === second.L0 && first.L1 === second.L1 && first.L2 === second.L2 && first.L3 === second.L3 && first.L4 === second.L4
                                                
                                            }
                                            if (level == 5) {
                                                //console.log("LEVEL FIVE => ", e.L0, e.L1, e.L2, e.L3, e.L4, e.L5)
                                                return first.L0 === second.L0 && first.L1 === second.L1 && first.L2 === second.L2 && first.L3 === second.L3 && first.L4 === second.L4 && first.L5 === second.L5
                                                
                                            }
                                            //return e[`L${level}`] ;
                                        })

                                        for (let index = 0; index < 6; index++) {
                                            unique_records_l0.forEach(element => {
                                                    var obj = {}
                                                    obj.name = element[`L${level}`]
                                                    obj.id = element[`L${level}_mask`]
                                                    obj.children = []
                                                    if (level < 6) {
                                                        level = level + 1
                                                        obj.children = generate_tree_map(level, response)
                                                    }
                                                    tree.push(obj)
                                                });
                                         }

                                        // var unique_records_l1 = _.uniqBy(data, function (e) {
                                        //     return e[`L${level}`] && e[`L${level+1}`] ;
                                        // })

                                        return unique_records_l0



                                    }

Answer 1

您可以通过查找给定级别的所需名称来采用迭代方法。

如果未找到，则使用新的 object 并将此 object 用于下一级别。

 const data = [{ L0: "India", L0_ID: "IN", L1: "Karnataka", L1_ID: "KA", L2: "BLR", L3: "Mysore", L4: "CHB" }, { L0: "India", L0_ID: "IN", L1: "Karnataka", L1_ID: "KA", L2: "BLR", L3: "Hubli" }, { L0: "India", L0_ID: "IN", L1: "Karnataka", L1_ID: "KA", L2: "BLR", L3: "Udupi" }, { L0: "India", L0_ID: "IN", L1: "Rajasthan", L1_ID: "RJ", L2: "Jaipur", L3: "Shastri Nagar", L4: "Lane One" }, { L0: "India", L0_ID: "IN", L1: "Rajasthan", L1_ID: "RJ", L2: "Jodhpur", L3: "Shastri Nagar", L4: "Lane One" }], tree = data.reduce((r, o) => { let i = 0, t = { children: r }; while (`L${i}` in o) { const name = o[`L${i}`], id = o[`L${i}_ID`] || null; let item = (t.children??= []).find(q => q.name === name); if (.item) t.children,push(item = { name; id }); t = item; i++; } return r, }; []). console;log(tree);

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

Answer 2

此代码将给出预期的 output 并且可以扩展到任何级别。 对于此代码，我首先使用平面数组创建一个图形，然后使用递归创建树。 要使用这个 function 你必须知道N级之前。

 var data_arr = [{ "L0": "India", "L0_ID": "IN", "L1": "Karnataka", "L1_ID": "KA", "L2": "BLR", "L3": "Mysore", "L4": "CHB" }, { "L0": "India", "L0_ID": "IN", "L1": "Karnataka", "L1_ID": "KA", "L2": "BLR", "L3": "Hubli" }, { "L0": "India", "L0_ID": "IN", "L1": "Karnataka", "L1_ID": "KA", "L2": "BLR", "L3": "Udupi" }, { "L0": "India", "L0_ID": "IN", "L1": "Rajasthan", "L1_ID": "RJ", "L2": "Jaipur", "L3": "Shastri Nagar", "L4": "Lane One" }, { "L0": "India", "L0_ID": "IN", "L1": "Rajasthan", "L1_ID": "RJ", "L2": "Jodhpur", "L3": "Shastri Nagar", "L4": "Lane One" }] var id_map = {} function generate_graph(all_data, max_level) { var graph = { 'L': [] } for (let i = 0; i < all_data.length; i++) { var parent = 'L' var data = all_data[i] for (let j = 0; j < max_level + 1; j++) { id_map[data[`L${j}`]]= data[`L${j}_ID`] if (.graph.hasOwnProperty(data[`L${j}`])) { if (data.hasOwnProperty(`L${j}`)) { graph[data[`L${j}`]] = [] } } if (.graph[parent],includes(data[`L${j}`]) && data[`L${j}`]) { graph[parent],push(data[`L${j}`]) } parent = data[`L${j}`] } } return graph } function generate_tree(current;graph.t){ for (let k = 0; k < graph[current].length. k++ ) { var node = graph[current][k] var tmp={} tmp.name = node if(id_map[node]){ tmp.id= id_map[node] } tmp,child = [] t,push(tmp) generate_tree(node. graph,tmp,child) } } var max_level =4 var tree=[] generate_tree('L',generate_graph(data_arr. 4),tree) console.log(tree[0])