javascript – 丢弃有向图中的传入节点

2024年2月5日 206次阅读

我想丢弃有向图中的传入节点.

tl; dr(跳到第3节)

我有一个图表,我执行BFS以删除所有不相关的节点(相对于候选边缘).

1.样本图数据

这是一个图形数据结构：

哪里：
  – id1是源,id2是目标

      var links = [
        {id1: 1, id2: 2},
        {id1: 2, id2: 3},
        {id1: 9, id2: 3},
        {id1: 6, id2: 8},
        {id1: 3, id2: 4}
      ]

上面的可视化：

《javascript – 丢弃有向图中的传入节点》

2.我成功地丢弃了不相关的节点/边缘

我运行一个BFS(下面的函数)来丢弃相对于edge = 1的所有不相关的节点,这产生了这个：

      var links = [
        {id1: 1, id2: 2},
        {id1: 2, id2: 3},
        {id1: 9, id2: 3},
       // {id1: 6, id2: 8}, Discarded (no connection with 1)
        {id1: 3, id2: 4}
      ]

上述可视化：

《javascript – 丢弃有向图中的传入节点》

3.我现在想要删除传入的节点

现在我想删除所有传入节点,并且只保留引用“朝向”相关节点的节点(缺少更好的单词),从edge = 1开始

例如：

      var links = [
        {id1: 1, id2: 2},
        {id1: 2, id2: 3},
        {id1: 9, id2: 3}, // remove this
        {id1: 3, id2: 4}
      ]

《javascript – 丢弃有向图中的传入节点》

我怎么能这样做？

这是我目前用来删除不相关的节点/边缘的内容：

var filterUnrelated = function(data, candidateId) {
  var toTest = [candidateId];
  var connected = [];

  function addToTest(node) {
    //If the node is not already set to be tested, and have not been tested
    if (connected.indexOf(node) < 0 && toTest.indexOf(node) < 0) {
      toTest.push(node);
    }
  }

  function findAllConnectedNode(node) {
    //We only test connected node, so this node should be connected
    connected.push(node);
    //Find every link with that node
    data.filter(function(d) {
      return (d.id1 === node) || (d.id2 === node);
      //Add the linked node to test
    }).map(function(d) {
      if (d.id1 === node) {
        addToTest(d.id2);
      }
      else { //d.id1 === node
        addToTest(d.id1);
      }
    });
  }

  while (toTest.length > 0) {
    findAllConnectedNode(toTest.shift());
  }

  return data.filter(function(d) {
    return (connected.indexOf(d.id1) >= 0 ||
      connected.indexOf(d.id2) >= 0);
  })
}


var links = [
  {id1: 1, id2: 2},
  {id1: 2, id2: 3},
  {id1: 9, id2: 3},
  {id1: 6, id2: 8},
  {id1: 3, id2: 4}
]

console.log(filterUnrelated(links, 1));

最佳答案 假设您的图形永远不会有分支(即一个节点通向两个节点),您可以从源节点开始并继续向上查找每个子节点.

function removeIncoming(data, source){
    // keep track of the member we are looking for
    var target = source;
    // the resulting graph
    var result = [];
    // iterate through the data, looking for the target
    for(var i = 0; i < data.length; i++){
        // the object in the list
        var piece = data[i];
        // its properties id1 and id2
        var id1 = piece.id1;
        var id2 = piece.id2;

        // when we have found what we are looking for
        if(id1 === target){
            // look for its child
            target = id2;
            // start at the beginning
            i = -1;
            // and add the link to the resulting list
            result.push(piece);
        }
    }

    return result;
}

或者,对于分支节点,您可以跟踪数组中的每个可能节点,然后使用indexOf搜索它们.

function removeIncoming(data, source){
    // copy the data
    var dataCopy = Array.prototype.slice.call(data);
    // keep track of the members we are looking for
    var targets = [source];
    // the resulting graph
    var result = [];
    // iterate through the data, looking for the target
    for(var i = 0; i < dataCopy.length; i++){
        // the object in the list
        var piece = dataCopy[i];
        // its properties id1 and id2
        var id1 = piece.id1;
        var id2 = piece.id2;

        // when we have found what we are looking for
        if(targets.indexOf(id1) >= 0){
            // begin looking for its child
            targets.push(id2);
            // remove the node we just looked at
            dataCopy.splice(i, 1);
            // start at the beginning
            i = -1;
            // and add the link to the resulting list
            result.push(piece);
        }
    }

    return result;
}
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