【问题标题】:d3 Network with Multiple Links in the Same Direction具有同一方向的多个链接的 d3 网络
【发布时间】:2017-06-01 19:34:05
【问题描述】:

我正在尝试更改 Mobile Patent Suits 示例,以允许在一个方向上有多个链接。

我有数据(是的,我知道 Jim 实际上并不是 Pam 的老板):

source          target          relationship          
Michael Scott   Jan Levenson    pro
Jan Levenson    Michael Scott   personal
Jim Halpert     Pam Beasley     pro
Jim Halpert     Pam Beasley     personal

美孚专利诉讼示例的多路径功能允许正确显示前两行(两条弧线)。但是,最后两行仅显示一个混合弧。

问题:如何让具有相同方向性的链接显示为多个弧而不是单个弧?

这是我的 arc 代码(直接来自移动专利示例):

function tick() {
  path.attr("d", linkArc);
  circle.attr("transform", transform);
  text.attr("transform", transform);
}

function linkArc(d) {
  var dx = d.target.x - d.source.x,
      dy = d.target.y - d.source.y,
      dr = Math.sqrt(dx * dx + dy * dy);
  return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + d.target.x + "," + d.target.y;
}

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

任何帮助都将不胜感激。谢谢!

【问题讨论】:

    标签: javascript d3.js


    【解决方案1】:

    可能有几种可能的方法,很快就会想到一个:为节点之间的每种类型关系使用不同的路径生成器。您必须有一个属性来指示关系的性质(您在问题中有),并使用它来设置路径对齐。

    在下面的 sn-p 中,我检查了正在绘制什么关系,并且与职业关系弧半径相比,将个人关系中的弧半径减少了 50%。相关部分是:

    function linkArc(d) {
    
      var dx = d.target.x - d.source.x,
          dy = d.target.y - d.source.y,
          dr = Math.sqrt(dx * dx + dy * dy);
      if(d.relationship == "pro") { 
         return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + d.target.x + "," + d.target.y;
      }
      else {
        return "M" + d.source.x + "," + d.source.y + "A" + (dr * 0.3) + "," + (dr * 0.3) + " 0 0,1 " + d.target.x + "," + d.target.y;
      }
    }
    

    这是实践中的全部内容:

    var links = [
      { source: "Michael Scott",
        target:"Jan Levenson",
        relationship: "pro"
      },
      { source:"Jan Levenson",
        target:"Michael Scott",
        relationship: "Personal"
      },
      { source: "Jim Halpert",
        target: "Pam Beasley",
        relationship: "pro"
      },
      {
        source: "Jim Halpert",
        target: "Pam Beasley",
        relationship: "Personal" 
      }
      ]
      
      var nodes = {};
    
    // Compute the distinct nodes from the links.
    links.forEach(function(link) {
      link.source = nodes[link.source] || (nodes[link.source] = {name: link.source});
      link.target = nodes[link.target] || (nodes[link.target] = {name: link.target});
    });
    
    var width = 960,
        height = 500;
    
    var force = d3.layout.force()
        .nodes(d3.values(nodes))
        .links(links)
        .size([width, height])
        .linkDistance(60)
        .charge(-300)
        .on("tick", tick)
        .start();
    
    var svg = d3.select("body").append("svg")
        .attr("width", width)
        .attr("height", height);
    
    // Per-type markers, as they don't inherit styles.
    svg.append("defs").selectAll("marker")
        .data(["suit", "licensing", "resolved"])
      .enter().append("marker")
        .attr("id", function(d) { return d; })
        .attr("viewBox", "0 -5 10 10")
        .attr("refX", 15)
        .attr("refY", -1.5)
        .attr("markerWidth", 6)
        .attr("markerHeight", 6)
        .attr("orient", "auto")
      .append("path")
        .attr("d", "M0,-5L10,0L0,5");
    
    var path = svg.append("g").selectAll("path")
        .data(force.links())
      .enter().append("path")
        .attr("class", function(d) { return "link " + d.type; })
        .attr("marker-end", function(d) { return "url(#" + d.type + ")"; });
    
    var circle = svg.append("g").selectAll("circle")
        .data(force.nodes())
      .enter().append("circle")
        .attr("r", 6)
        .call(force.drag);
    
    var text = svg.append("g").selectAll("text")
        .data(force.nodes())
      .enter().append("text")
        .attr("x", 8)
        .attr("y", ".31em")
        .text(function(d) { return d.name; });
    
    // Use elliptical arc path segments to doubly-encode directionality.
    function tick() {
      path.attr("d", linkArc);
      circle.attr("transform", transform);
      text.attr("transform", transform);
    }
    
    function linkArc(d) {
    
      var dx = d.target.x - d.source.x,
          dy = d.target.y - d.source.y,
          dr = Math.sqrt(dx * dx + dy * dy);
      if(d.relationship == "pro") { 
         return "M" + d.source.x + "," + d.source.y + "A" + dr + "," + dr + " 0 0,1 " + d.target.x + "," + d.target.y;
      }
      else {
        return "M" + d.source.x + "," + d.source.y + "A" + (dr * 0.3) + "," + (dr * 0.3) + " 0 0,1 " + d.target.x + "," + d.target.y;
      }
    }
    
    function transform(d) {
      return "translate(" + d.x + "," + d.y + ")";
    }
    .link {
      fill: none;
      stroke: #666;
      stroke-width: 1.5px;
    }
    
    #licensing {
      fill: green;
    }
    
    .link.licensing {
      stroke: green;
    }
    
    .link.resolved {
      stroke-dasharray: 0,2 1;
    }
    
    circle {
      fill: #ccc;
      stroke: #333;
      stroke-width: 1.5px;
    }
    
    text {
      font: 10px sans-serif;
      pointer-events: none;
      text-shadow: 0 1px 0 #fff, 1px 0 0 #fff, 0 -1px 0 #fff, -1px 0 0 #fff;
    }
    <script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>

    【讨论】:

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