vis.js - 删除单个或具有 2 度 visjs 的节点

Question

我有一个网络图

现在我有一些连接的节点，你可以看到大多数节点只有一个连接节点，它们的度数为 1。现在我想删除这些节点以清除混乱。自过去 2 天以来无法找到如何操作。visjs 文档中没有这样的辅助函数。非常感谢帮助。

Answer 1

I believe the algorithm suggested by the 1st answer -by macramole- (before updates) would actually hide the non-connected nodes (degree 0), instead of the ones with degree 1. I would probably just iterate over all the edges in the network while keeping 'degree' counters for each node that is an endpoint in the edge you are visiting (you can obtain these nodes by grabbing the edge.from and edge.to values, as shown above). You would increment the degree counter for a node, whenever the node is 'hit' in this search through the edges. Eventually you'll end up with the degree value for each node in the network, at which point you can decide which ones to hide. Updating this answer now to include my suggested code (note: nodes and edges are vis DataSet instances):

Example code:

var nodeToDegrees = {}; // keeps a map of node ids to degrees
var nodeFrom, nodeTo;
for (edge in edges) {
    nodeFrom = edge.from;
    nodeTo = edge.to;
    nodeToDegrees[nodeFrom] = nodeToDegrees[nodeFrom] ? nodeToDegrees[nodeFrom] + 1 : 0;
    nodeToDegrees[nodeTo] = nodeToDegrees[nodeTo] ? nodeToDegrees[nodeTo] + 1 : 0;
}
for (node in nodes) {
    if (nodeToDegrees[node.id] = 1) nodes.update([{node.id, hidden: true}]);
}

Answer 2

这可能有效：

var DEGREES_HIDDEN = 1;

for ( var node of nodes ) {
    node.cantLinks = 0;

    for ( var link of links ) {
        if ( link.from == node.id || link.to == node.id ) {
            node.cantLinks++;
        }
    }
}
for ( var node of nodes ) {
   if ( node.cantLinks <= DEGREES_HIDDEN ) {
       node.hidden = true;
   }  
}

节点和链接是数组而不是 vis.DataSet，我在这样做之后创建了后者。

看起来不是很好的性能，但它确实完成了工作。希望你觉得它有用。