如何防止力导向图中链接重叠?

13

我正在跟随“力导向图”。如何防止力导向图中链接(线)的重叠。期望的输出应该像这样。当前的JSON数据不包含任何父子关系。

以下是我的代码:

var width = 960,
    height = 500;

var color = d3.scale.category20();

var force = d3.layout.force()
    .charge(-120)
    .linkDistance(30)
    .size([width, height]);

var svg = d3.select("body").append("svg")
    .attr("width", width)
    .attr("height", height);

var graph = {
  "nodes":[
    {"name":"Myriel","group":1},
    {"name":"Napoleon","group":1},
    {"name":"Mlle.Baptistine","group":1},
    {"name":"Mme.Magloire","group":1},
    {"name":"CountessdeLo","group":1},
    {"name":"Geborand","group":1},
    {"name":"Champtercier","group":1},
    {"name":"Cravatte","group":1},
    {"name":"Count","group":1},
    {"name":"OldMan","group":1},
    {"name":"Labarre","group":2},
    {"name":"Valjean","group":2},
    {"name":"Marguerite","group":3},
    {"name":"Mme.deR","group":2},
    {"name":"Isabeau","group":2},
    {"name":"Gervais","group":2},
    {"name":"Tholomyes","group":3},
    {"name":"Listolier","group":3},
    {"name":"Fameuil","group":3},
    {"name":"Blacheville","group":3},
    {"name":"Favourite","group":3},
    {"name":"Dahlia","group":3},
    {"name":"Zephine","group":3},
    {"name":"Fantine","group":3},
    {"name":"Mme.Thenardier","group":4},
    {"name":"Thenardier","group":4},
    {"name":"Cosette","group":5},
    {"name":"Javert","group":4},
    {"name":"Fauchelevent","group":0},
    {"name":"Bamatabois","group":2},
    {"name":"Perpetue","group":3},
    {"name":"Simplice","group":2},
    {"name":"Scaufflaire","group":2},
    {"name":"Woman1","group":2},
    {"name":"Judge","group":2},
    {"name":"Champmathieu","group":2},
    {"name":"Brevet","group":2},
    {"name":"Chenildieu","group":2},
    {"name":"Cochepaille","group":2},
    {"name":"Pontmercy","group":4},
    {"name":"Boulatruelle","group":6},
    {"name":"Eponine","group":4},
    {"name":"Anzelma","group":4},
    {"name":"Woman2","group":5},
    {"name":"MotherInnocent","group":0},
    {"name":"Gribier","group":0},
    {"name":"Jondrette","group":7},
    {"name":"Mme.Burgon","group":7},
    {"name":"Gavroche","group":8},
    {"name":"Gillenormand","group":5},
    {"name":"Magnon","group":5},
    {"name":"Mlle.Gillenormand","group":5},
    {"name":"Mme.Pontmercy","group":5},
    {"name":"Mlle.Vaubois","group":5},
    {"name":"Lt.Gillenormand","group":5},
    {"name":"Marius","group":8},
    {"name":"BaronessT","group":5},
    {"name":"Mabeuf","group":8},
    {"name":"Enjolras","group":8},
    {"name":"Combeferre","group":8},
    {"name":"Prouvaire","group":8},
    {"name":"Feuilly","group":8},
    {"name":"Courfeyrac","group":8},
    {"name":"Bahorel","group":8},
    {"name":"Bossuet","group":8},
    {"name":"Joly","group":8},
    {"name":"Grantaire","group":8},
    {"name":"MotherPlutarch","group":9},
    {"name":"Gueulemer","group":4},
    {"name":"Babet","group":4},
    {"name":"Claquesous","group":4},
    {"name":"Montparnasse","group":4},
    {"name":"Toussaint","group":5},
    {"name":"Child1","group":10},
    {"name":"Child2","group":10},
    {"name":"Brujon","group":4},
    {"name":"Mme.Hucheloup","group":8}
  ],
  "links":[
    {"source":1,"target":0,"value":1},
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  ]
}

//d3.json("miserables.json", function(error, graph) {
  force
      .nodes(graph.nodes)
      .links(graph.links)
      .start();

  var link = svg.selectAll(".link")
      .data(graph.links)
    .enter().append("line")
      .attr("class", "link")
      .style("stroke-width", function(d) { return Math.sqrt(d.value); });

  var node = svg.selectAll(".node")
      .data(graph.nodes)
    .enter().append("circle")
      .attr("class", "node")
      .attr("r", 5)
      .style("fill", function(d) { return color(d.group); })
      .call(force.drag);

  node.append("title")
      .text(function(d) { return d.name; });

  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("cx", function(d) { return d.x; })
        .attr("cy", function(d) { return d.y; });
  });
//});
.node {
  stroke: #fff;
  stroke-width: 1.5px;
}

.link {
  stroke: #999;
  stroke-opacity: .6;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.0/d3.min.js"></script>


我认为并非所有图形都可以实现。但我很感兴趣的是,在所有可能的情况下如何实现它。 - Oliver
请参见 https://dev59.com/7Ijca4cB1Zd3GeqPz6vx。 - Lars Kotthoff
1
您IP地址为143.198.54.68,由于运营成本限制,当前对于免费用户的使用频率限制为每个IP每72小时10次对话,如需解除限制,请点击左下角设置图标按钮(手机用户先点击左上角菜单按钮)。 - myrcutio
2个回答

1
如果您说的重叠是指两个节点之间存在多个链接,则可以通过绘制具有逐渐增加曲率的弧线来在视觉上将它们分开。
如果您说的重叠是指边交叉,则只有当您的图是平面图时才可能发生- https://en.wikipedia.org/wiki/Planarity_testing - 从您的图中看,它不是平面图- 至少有两个子图几乎完全连通。您需要一种专门尝试减少边交叉的图形布局算法,而坏消息是这是一个NP完全问题
交叉数是NP完全的:DOI:10.1137 / 0604033 https://en.wikipedia.org/wiki/Crossing_number_(graph_theory) 一些技术可以用于完整子图,方法是用合并的边或模式替换它们的边集,但这涉及到图形绘制研究:没有简单的答案。
合流图:以平面方式可视化非平面图 DOI:10.1007 / 978-3-540-24595-7_1

主题简化:使用扇形、连接器和团状符号提高网络可视化的可读性 DOI: 10.1145/2470654.2466444

简而言之,不产生边交叉的唯一保证方法是不绘制边缘。也许可以尝试查看矩阵类型的图形可视化?- https://bost.ocks.org/mike/miserables/


0

我想留下评论,但是我的声望不够,所以我“回答”了 - 抱歉这不是一个很好的答案...

你期望输出的例子是一棵树/无环图,这几乎神奇地解决了重叠链接的问题。你可以在 Mike Bostock 的这个例子中看到(http://bl.ocks.org/mbostock/1062288),由于数据是一棵树,所以几乎没有链接重叠。

所以...

如果你的数据是一棵树,你可能会很幸运,链接重叠可能不是一个问题。你会注意到,在 Bostock 的例子中,当兄弟节点具有不同的链接距离时(导致侄女/侄子 - 姑妈/叔叔链接重叠),重叠就会发生。控制链接距离(也许使它们成为树深度的函数,并使根附近的链接更长),我认为将防止您的树中的任何重叠。

如果你的数据有循环(如 Miserables 数据),防止重叠就要复杂得多,而且不会自动解决。像其他人说的那样,我认为任何 D3 布局都不可能实现。


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