同一方向多链路的d3网络
同一方向多链路的d3网络
提问于 2017-06-01 19:34:05
我试图改变移动专利诉讼的例子，以允许多个链接在一个方向。
我有数据(是的，我知道吉姆其实不是帕姆的老板)：
source target relationship
Michael Scott Jan Levenson pro
Jan Levenson Michael Scott personal
Jim Halpert Pam Beasley pro
Jim Halpert Pam Beasley personalMobil专利示例的多路径功能允许正确显示前两行(两条弧)。然而，最后两行只显示了一个混合弧。
问:如何允许具有相同方向性的链接显示为多个弧线而不是单个弧线？
以下是我的弧形代码(直接从移动专利示例中摘录)：
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 + ")";
}任何帮助都将是非常感谢的。谢谢!
回答 1
Stack Overflow用户
回答已采纳
发布于 2017-06-01 20:30:19
对于这一点，可能有几种可能的方法，很快就会浮现出来:为节点之间的每种类型的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(在您的问题中)，并使用它来设置路径对齐。
在下面的片段中，我检查绘制了什么关系，并将个人关系中的弧半径与职业关系弧半径相比减少了50%。有关的部分是：
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;
}
}下面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>
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/44315707
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