将多个路径转换为一个儿童数组Javascript

Question

我想将具有多个路径的倍数数组转换为一个平面数组的儿童。这样做的最好方法是什么？我尝试使用对象引用来创建它，但它创建了一个无限循环。我可以用其他方式做到这一点吗？

感谢您的帮助

这是输入（因为有需要结合多个路径，也有认为是相同的多个节点例如A到B发生两次。）：

arry = [ 
[ 
    {"title":"abc", "from":"", "current": "A", "count": 1}, 
    {"title":"abc", "from":"A", "current": "B", "count": 3}, 
    {"title":"abc", "from":"B", "current": "C", "count": 4}, 
    {"title":"abc", "from":"C", "current": "D", "count": 1}, 
], 
[ 
    {"title":"abc", "from":"", "current": "A", "count": 1}, 
    {"title":"abc", "from":"A", "current": "B", "count": 1}, 
    {"title":"abc", "from":"B", "current": "D", "count": 1}, 
], 
[ 
    {"title":"abc", "from":"", "current": "A", "count": 1}, 
    {"title":"abc", "from":"A", "current": "J", "count": 1}, 
    {"title":"abc", "from":"J", "current": "C", "count": 2}, 
    {"title":"abc", "from":"C", "current": "D", "count": 3}, 
], 
[ 
    {"title":"abc", "from":"", "current": "A", "count": 1}, 
    {"title":"abc", "from":"A", "current": "B", "count": 3}, 
    {"title":"abc", "from":"B", "current": "F", "count": 1}, 
    {"title":"abc", "from":"F", "current": "D", "count": 1}, 
], 
//THIS COULD BE INFINITE 
]//for testing 

所需的输出

{ 
"title": "A", 
"count": 1, 
"children": [ 
{ 
    "value": "B", 
    "count": 4, //The count is sum of the nodes A->B = 3 and C->B 1 
    "children": [ 
     { 
     "value": "C", 
     "count": 6, 
     "children": [...] 
    }, 
    { 
     "value": "D" , 
     "count": 6, 
    "children": [...] 
    }, 
    { 
     "value": "F" , 
     "count": 6, 
    "children": [...] 
    } 
    ], 
    { 
     "value": "C" , 
     "count": 6, 
    "children": [...] 
    }, 
    { 
     "value": "J" , 
     "count": 6, 
    "children": [...] 
    }, 
] 
.. 
} 

这是我到目前为止有：

getTreeData(){ 
var map = {}, node, roots = []; 
for (var i = data.nodes.length - 1; i >= 0; i--) { 
    for (var j = data.nodes[i].length - 1; j >=0 ; j --) { 
     node = data.nodes[i][j]; 
     node.children = []; 
     if(!(data.nodes[i][j].current in map)){ 
     map[data.nodes[i][j].current] = node; // use map to look-up the parents 
     }else{ 
     node = map[data.nodes[i][j].current] 
     } 
     if (node.from != null && node.from !== "") { 
      // if(map[node.from] != null && !(data.nodes[i][j].current in map[data.nodes[i][j].from].children)){ 
      // console.log() 
      if(map[node.from] != null && ! map[data.nodes[i][j].from].children.some(function (el) { return el.current === data.nodes[i][j].current; })){ 
      map[data.nodes[i][j].from].children.push(node); 
      // map[data.nodes[i][j].from].children[data.nodes[i][j].current] = node; 
      } 
     } else if(roots.length <= 0){ 
      roots.push(node); 
     } 
    } 
} 
return roots; 
} 

Answer 1

创建树：

var arr = [ 
    {"title":"abc", "from":"", "current": "A", "count": 1}, 
    {"title":"abc", "from":"A", "current": "B", "count": 3}, 
    {"title":"abc", "from":"B", "current": "C", "count": 4}, 
    {"title":"abc", "from":"C", "current": "D", "count": 1}, 
]; 

arr.reduceRight((a, b) => { 
    b.child = a; 
    return b; 
});