如何减少除相邻节点外的所有节点的不透明度

Question

我有一个功能齐全的d3.js强制定向图。有多种与图表相关的过滤器。

我现在正尝试使用突出显示功能，即，如果双击该节点，则该节点及其所有邻居的不透明度都高于该图的其余部分。再次双击任何一个节点，整个图形将变为可见。以下是我尝试过的代码

 //Toggle stores whether the highlighting is on
    var toggle = 0;
    //Create an array logging what is connected to what
    var linkedByIndex = {};
    for (i = 0; i < graph.nodes.length; i++) {
        linkedByIndex[i + "," + i] = 1;
    };
    graph.links.forEach(function (d) {
        linkedByIndex[d.source.index + "," + d.target.index] = 1;
    });
    //This function looks up whether a pair are neighbours
    function neighboring(a, b) {
        return linkedByIndex[a.index + "," + b.index];
    }
    function connectedNodes() {
        if (toggle == 0) {
            //Reduce the opacity of all but the neighbouring nodes
            d = d3.select(this).node().__data__;
            node.style("opacity", function (o) {
                return neighboring(d, o) | neighboring(o, d) ? 1 : 0.1;
            });
            link.style("opacity", function (o) {
                return d.index==o.source.index | d.index==o.target.index ? 1 : 0.1;
            });
            //Reduce the op
            toggle = 1;
        } else {
            //Put them back to opacity=1
            node.style("opacity", 1);
            link.style("opacity", 1);
            toggle = 0;
        }
    }