Question

我打算为动态创建的数据集创建一个树状结构，并且想要为力导向图设置所有节点的y轴坐标。换句话说，我希望节点位于关卡（如树结构）中，而节点可以自由移动并在x轴上自我优化。

最终结果应该是一个树，其中节点位于分配的级别而不是径向，就像默认情况下那样。同级链接不应重叠，并且尽可能短（逻辑分组，因此希望通过强制图执行此操作）。
如果有人有一个比这更简单的解决方案，那么我全都是耳朵。

代码提取：

ldLinks = [
{source: "root", target: "1", source_level: 0, target_level: 1},
{source: "root", target: "2", source_level: 0, target_level: 1},
{source: "root", target: "3", source_level: 0, target_level: 1},
{source: "root", target: "4", source_level: 0, target_level: 1},
{source: "1", target: "1.1", source_level: 1, target_level: 2},
{source: "1", target: "1.2", source_level: 1, target_level: 2},
{source: "2", target: "2.1", source_level: 1, target_level: 3},
{source: "2", target: "2.2", source_level: 1, target_level: 3},
{source: "3", target: "3.1", source_level: 1, target_level: 3},
{source: "4", target: "3.2", source_level: 1, target_level: 3},
{source: "4", target: "3.3", source_level: 1, target_level: 3},
{source: "3", target: "3.2", source_level: 2, target_level: 3},
{source: "3", target: "4.1", source_level: 2, target_level: 3},
{source: "2.1", target: "5.1", source_level: 3, target_level: 4},
{source: "3.1", target: "5.1", source_level: 3, target_level: 4},
{source: "1", target: "5", source_level: 1, target_level: 4},
];

var dNodes = {};

// Compute the distinct dNodes from the ldLinks.
ldLinks.forEach(function(link) {
    link.source = dNodes[link.source] || (dNodes[link.source] = {name: link.source, level: link.source_level});
    link.target = dNodes[link.target] || (dNodes[link.target] = {name: link.target,  level: link.target_level});
});

//parameterize nodes
for (var sNodeName in dNodes) {
    //dNodes[sNodeName].fixed = (only fixed along y-axis)
    dNodes[sNodeName].y = (dNodes[sNodeName].level * 50)
}

Answer 1

好哇！

感谢Limin从this answer发表的评论！

我设置了force.gravity(0)，然后在.attr("cx/cy")函数中向node添加了边界（tick()）：

function tick() {
    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 + ")"; })
        .attr("cx", function(d) { return d.x = Math.max(8, Math.min(width - 8, d.x)); })
        .attr("cy", function(d) { return d.y = Math.max(100*d.level, Math.min(100*d.level, d.y)); });
}

参数化D3js强制有向图节点定位

1 个答案: