如何连接对象数组中的多个数字

Question

几天来，我一直试图找到解决办法，但没有成功。仅举个例子，我需要一个用任何selectedNode来执行此操作的函数。我会用在图表上。

数组：

const edges = [
        {
            "id": "e1-2",
            "from": "1",
            "to": "2"
        },
        {
            "id": "e2-3",
            "from": "2",
            "to": "3"
        },
        {
            "id": "e2-8",
            "from": "2",
            "to": "8"
        },
        {
            "id": "e2-13",
            "from": "2",
            "to": "13"
        },
        {
            "id": "e3-4",
            "from": "3",
            "to": "4"
        },
        {
            "id": "e4-5",
            "from": "4",
            "to": "5"
        },
        {
            "id": "e4-6",
            "from": "4",
            "to": "6"
        },
        {
            "id": "e4-7",
            "from": "4",
            "to": "7"
        },
        {
            "id": "e8-9",
            "from": "8",
            "to": "9"
        },
        {
            "id": "e9-10",
            "from": "9",
            "to": "10"
        },
        {
            "id": "e9-11",
            "from": "9",
            "to": "11"
        },
        {
            "id": "e9-12",
            "from": "9",
            "to": "12"
        },
        {
            "id": "e13-14",
            "from": "13",
            "to": "14"
        },
        {
            "id": "e14-15",
            "from": "14",
            "to": "15"
        },
        {
            "id": "e14-16",
            "from": "14",
            "to": "16"
        },
        {
            "id": "e14-17",
            "from": "14",
            "to": "17"
        },
        {
            "id": "e14-18",
            "from": "14",
            "to": "18"
        },
        {
            "id": "e14-19",
            "from": "14",
            "to": "19"
        }
    ]

我试着使用过滤器和地图，但只捕捉第一个孩子。我需要孩子的孩子等等。

const selectedNode = "2";

const connections = edges
      .filter((edge) => edge.from === selectedNode).map((edge) => edge.to)




console.log(connections)

产出：

日志："3“、"8”、"13“

但我需要所有的孩子。

像这样的

• 输出：

"3“、"8”、"13“、"4”、"5“、"6”、"7“、"9”、"10“、"11”、"12“、"14”、"15“、"16”、"17“、"18”、"19“

• ，但如果selectedNode为"8"，则数组为：

"9“"10”"11“"12"

Answer 1

您需要制定某种递归算法。一种方法是通过from的第一步将边缘重构为映射。然后，当迭代边缘(从selectedNode开始)时，如果边缘的to包含任何其他to，则对每个to执行递归调用。

​

const edges=[{id:"e1-2",from:"1",to:"2"},{id:"e2-3",from:"2",to:"3"},{id:"e2-8",from:"2",to:"8"},{id:"e2-13",from:"2",to:"13"},{id:"e3-4",from:"3",to:"4"},{id:"e4-5",from:"4",to:"5"},{id:"e4-6",from:"4",to:"6"},{id:"e4-7",from:"4",to:"7"},{id:"e8-9",from:"8",to:"9"},{id:"e9-10",from:"9",to:"10"},{id:"e9-11",from:"9",to:"11"},{id:"e9-12",from:"9",to:"12"},{id:"e13-14",from:"13",to:"14"},{id:"e14-15",from:"14",to:"15"},{id:"e14-16",from:"14",to:"16"},{id:"e14-17",from:"14",to:"17"},{id:"e14-18",from:"14",to:"18"},{id:"e14-19",from:"14",to:"19"}];

const toByFrom = {};
for (const edge of edges) {
  toByFrom[edge.from] ??= [];
  toByFrom[edge.from].push(edge.to);
}
const allNodes = [];
const getNodes = (parent, allNodes = []) => {
  const tos = toByFrom[parent];
  if (tos) {
    allNodes.push(...tos);
    for (const to of tos) {
      getNodes(to, allNodes);
    }
  }
  return allNodes;
};
console.log(getNodes('2'));

​

Answer 2

这听起来像是递归任务。这和另一个答案是一样的。只是作为一种功能。

​

const edges = [{ "id": "e1-2", "from": "1", "to": "2" }, { "id": "e2-3", "from": "2", "to": "3" }, { "id": "e2-8", "from": "2", "to": "8" }, { "id": "e2-13", "from": "2", "to": "13" }, { "id": "e3-4", "from": "3", "to": "4" }, { "id": "e4-5", "from": "4", "to": "5" }, { "id": "e4-6", "from": "4", "to": "6" }, { "id": "e4-7", "from": "4", "to": "7" }, { "id": "e8-9", "from": "8", "to": "9" }, { "id": "e9-10", "from": "9", "to": "10" }, { "id": "e9-11", "from": "9", "to": "11" }, { "id": "e9-12", "from": "9", "to": "12" }, { "id": "e13-14", "from": "13", "to": "14" }, { "id": "e14-15", "from": "14", "to": "15" }, { "id": "e14-16", "from": "14", "to": "16" }, { "id": "e14-17", "from": "14", "to": "17" }, { "id": "e14-18", "from": "14", "to": "18" }, { "id": "e14-19", "from": "14", "to": "19" }];

function find_immediate_edges(selected) {
  return edges.filter(item => item.from == selected).map(item => item.to);
}

function find_edges(selected) {
  var result = []
  
  function loop(selected) {
    var immediate = find_immediate_edges(selected);
    result = result.concat(immediate)
    immediate.forEach(loop)
  }
  
  loop(selected);
  return result;
}


console.log("" + find_edges(2));
console.log("" + find_edges(8));

​