javascript - 从 d3 中的节点中删除文本

Question

我有一个 d3 力图，每次删除节点时都会重写(覆盖？)文本。我认为解决此问题的解决方案是删除 update() 函数中的文本标签。但是，如果屏幕上有很多节点，我不确定性能是否会很重。我也不确定为什么会发生这种情况。

以下是我认为与该问题有关的代码片段:

function update() {

    // refresh list of selected nodes
    selectedNodes = nodes.filter(function(d) { return d.selected; });

    // Update link data based on edges array.
    link = link.data(edges);

    // Create new links
    link.enter().append("line")
            .attr("class", "link")
            .style("stroke-width", 1.5);

    // Delete removed links
    link.exit().remove();

    // Update node data based on nodes array.
    node = node.data(nodes);

    // Create new nodes
    node.enter().append("g")
            .attr("class", "node")
            .attr("id", function(d) { return d.data['id'] })
        //.attr("fixed", function(d) { return d.fixed=true })
            .call(force.drag)
            .on('mouseover', connectedNodes)
            .on('mouseleave', restore)
        //.on('dblclick', highlight)
            .on('dblclick', highlight);

    // Delete removed nodes
    node.exit().remove();

    node.append("circle").attr("r", 11);
    node.classed("selected", function(d) { return d === d.selected; })

    // Node behavior for checking if selected otherwise colors nodes to color given from JSON.
    node.style("fill", function(d) {
        if (d.selected === false) {
            console.log("Not Highlighting " + d.data['id'] + " selected is " + d.selected);
            return d.data['color']
            update();
        }
        else {
            console.log("Highlighting " + d.data['id'] + " selected is " + d.selected);
            return "yellow";
            update();
        }
    }).select("circle").style("stroke", "black");

    // Link color based on JSON data.
    link.style("stroke", function(d) { return d.data['color'] });

    // Adds text to nodes
    node.append("text")
            .attr("dx", 12)
            .attr("dy", ".35em")
            .style("fill", "black")
            .text(function (d) { return d.data['label']; });

    // Creates an index used to figure out neighbor nodes.
    root.edges.forEach(function (d) {
        linkedByIndex[d.data.source + "," + d.data.target] = 1;
    });

    // responsive behavior for graph based on window.
    window.addEventListener('resize', resize);

    force.on("tick", function() {
        link.attr("x1", function(d) { return d.source.x; })
                .attr("y1", function(d) { return d.source.y; })
                .attr("x2", function(d) { return d.target.x; })
                .attr("y2", function(d) { return d.target.y; });

        node.attr("transform", function(d) { return "translate(" + d.x + "," + d.y + ")"; });
    });
    force.start();
}

// Delete node with prompt
function deleteNode() {
    console.log("Prompted to delete selected nodes.");
    if (confirm("Deleting selected element(s) will remove them from the graph entirely.\nAre you sure? (This cannot be undone).")) {
        if (selectedNodes.length > 0) {
            for (var i = 0; i < selectedNodes.length; i++) {
                selectedNodes[i].removed = true;
                nodes.splice(nodes.indexOf(selectedNodes[i]), 1);
                spliceLinksForNode(selectedNodes[i]);
            }
        }
        else alert("No node(s) selected.");
        update();
    }
}

function spliceLinksForNode(node) {
    toSplice = edges.filter(
            function(e) {
                return (e.source === node) || (e.target === node); });
    toSplice.map(
            function(e) {
                edges.splice(edges.indexOf(e), 1); });
}

这是删除节点后的屏幕截图。

Answer 1

我也遇到了同样的问题，据我了解，是该小组造成了问题。

对我来说，显式删除元素解决了问题。旁边:

// Exit any old node
node.exit().remove();

我也打电话:

// Remove any old circle
svg.selectAll("circle").remove();
// Remove any old text
svg.selectAll("text").remove();
// Remove any old title
svg.selectAll("title").remove();

在更新功能中。

我希望这个答案对任何人都有帮助，我花了半天时间才完成。