如果我想用d3层次来创建一棵树,我就必须使用d3-传家宝,因为我所使用的库使用了这个结构。我正在使用的API返回一个响应如下:
{
"Nodes": [
{
"Name": "node-1",
"Type": "Model"
},
{
"Name": "node-2",
"Type": "Job"
},
{
"Name": "node-3",
"Type": "Data"
},
{
"Name": "node-4",
"Type": "Data"
},
],
"Edges": [
{
"Source": "node-4",
"Destination": "node-2"
},
{
"Source": "node-3",
"Destination": "node-2"
},
{
"Source": "node-2",
"Destination": "node-1"
},
]}我试图获得如下输出,可以与d3层次结构一起使用:
[
{
Name: 'node-1',
Type: 'Model',
children: [
{
Name: 'node-2',
Type: 'Job',
children: [
{
Name: 'node-3',
Type: 'Data'
},
{
Name: 'node-4',
Type: 'Data'
},
]
},
]
}
]任何帮助都将不胜感激!
发布于 2022-09-06 18:27:23
因此,为了便于访问,我首先按节点名称分组。然后构建连接(作为父级子连接)。就这样。我有图表和所有可能的入口点。我选择第一个,因为它有希望是树的根。
var input={Nodes:[{Name:"node-1",Type:"Model"},{Name:"node-2",Type:"Job"},{Name:"node-3",Type:"Data"},{Name:"node-4",Type:"Data"},],Edges:[{Source:"node-4",Destination:"node-2"},{Source:"node-3",Destination:"node-2"},{Source:"node-2",Destination:"node-1"},]}
// first obj of nodes grouped by name
var obj_nodes = input.Nodes.reduce(function(agg, item) {
agg[item.Name] = { ...item, children: [] };
return agg;
}, {})
// console.log(obj_nodes)
// connecting edges (child parent relations)
input.Edges.forEach(function(item) {
var source = obj_nodes[item.Source];
var destination = obj_nodes[item.Destination];
destination.children.push(source);
}, {})
var trees = Object.values(obj_nodes);
var result = trees[0]
console.log(result).as-console-wrapper {max-height: 100% !important}
https://stackoverflow.com/questions/73625482
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