我已经在堆栈上查看了不同的链接距离,看起来为了改变链接距离你需要实现一个函数,然后传递它来动态分配链接距离:
function linkDistance(d) {
return d.distance;
}
然后我认为我能够传递给svg而是返回函数错误,而不是现有的linkdistance或距离
var link = svg.selectAll(".link")
.data(bilinks)
.enter().append("path")
.style("stroke", "#6b7071") //gunmetal grey link
.attr("class", "link")
.linkDistance(linkDistance)
.attr("fill", "none")
使用文档中的.linkDistance:https://github.com/d3/d3/blob/master/API.md#forces-d3-force理想情况下,我想使用数据修改链接距离,电荷,力和链接颜色等参数,就像我对点的颜色所做的那样,我相信我对如何正确地做到这一点缺乏了解。例如,在最后一行代码中,如果我要从:改为
.style("stroke", "#6b7071") //gunmetal grey link
.style("stroke", function(d) { return color(d.group);})
链接颜色是一种颜色,而不是基于该组我预期的39种颜色。 此外,我也尝试了
var simulation = d3.forceSimulation()
.force("link", d3.forceLink().distance(function(d) {return d.distance}).strength(0.1))
.force("charge", d3.forceManyBody(30))
.force("center", d3.forceCenter(width / 2, height / 2));
更新: 我不确定长度的数据是在链接和bilinks的数组中,所以我可以引用距离但是似乎无法使用数组中的值,但console.log显示它正确存储
CODE
<!DOCTYPE html>
<meta charset="utf-8">
<style>
.node {
stroke: #fff;
stroke-width: 1.5px;
}
</style>
<svg width="15000" height="15000"></svg>
<script src="https://d3js.org/d3.v4.min.js"></script>
<script>
//change background color to black
backgroundColor = d3.rgb('#000000')
d3.select("body").style("background-color", backgroundColor)
var svg = d3.select("svg"),
width = +svg.attr("width"),
height = +svg.attr("height");
//var color = d3.scaleOrdinal(d3.schemeCategory20);
var simulation = d3.forceSimulation()
.force("link", d3.forceLink().distance(500).strength(0.1))
.force("charge", d3.forceManyBody())
.force("center", d3.forceCenter(width / 2, height / 2));
d3.json("hc7data.json", function(error, graph) {
if (error) throw error;
var nodes = graph.nodes,
nodeById = d3.map(nodes, function(d) { return d.id; }),
links = graph.links,
bilinks = [];
//get graphics to make color scale us scaleOrdinal if every color chosen
var color = d3.scaleOrdinal()
.domain([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40])
.range(["#af1f45", "#be4f5e","#cd767c","#dc9d9e","#ecc9c8","#fbdbe9","#f7bbd5","#f49ac1","#f179ae","#ef509c",
"#e3d4e4","#cdb1cf","#9a5699","#b990ba", "#a973a9","#d6eaf3","#b0daeb","#8acce4","#5ebfde","#00a5db","#6dbe46","#e0efd4",
"#c3e0ae","#a7d48b","#8cc866","#fff2d1","#ffe8a8","#ffdf80","#ffd751","#fecf07","#fee1c9","#fcc79c","#faae74","#f69d58",
"#f7964a","#fde3d9","#fcccbc","#f58870","#f9b4a0","#f79e87"]);
links.forEach(function(link) {
var s = link.source = nodeById.get(link.source),
t = link.target = nodeById.get(link.target),
i = {}, // intermediate node
linkDist = link.distance;
nodes.push(i);
//console.log(linkDist);
links.push({source: s, target: i, linkDist:linkDist}, {source: i, target: t, linkDist:linkDist });
bilinks.push([s, i, t,linkDist]);
});
var link = svg.selectAll(".link")
.data(bilinks)
.enter().append("path")
.style("stroke", "#6b7071") //gunmetal grey
.attr("class", "link")
.attr("fill", "none")
var node = svg.selectAll(".node")
.data(nodes.filter(function(d) { return d.id; }))
.enter().append("circle")
.attr("class", "node")
//change circle size according to new function
.attr("r", function(d) {return d.size})
.attr("fill", function(d) { return color(d.group); })
.style("stroke", "#000000")
//.style("stroke", function(d) { return color(d.group); })
.call(d3.drag()
.on("start", dragstarted)
.on("drag", dragged)
.on("end", dragended));
node.append("title")
.text(function(d) { return d.id; });
simulation
.nodes(nodes)
.on("tick", ticked);
simulation.force("link")
.links(links);
function ticked() {
link.attr("d", positionLink);
node.attr("transform", positionNode);
}
});
function positionLink(d) {
return "M" + d[0].x + "," + d[0].y
+ "S" + d[1].x + "," + d[1].y
+ " " + d[2].x + "," + d[2].y;
}
function positionNode(d) {
return "translate(" + d.x + "," + d.y + ")";
}
function dragstarted(d) {
if (!d3.event.active) simulation.alphaTarget(0.3).restart();
d.fx = d.x, d.fy = d.y;
}
function dragged(d) {
d.fx = d3.event.x, d.fy = d3.event.y;
}
function dragended(d) {
if (!d3.event.active) simulation.alphaTarget(0);
d.fx = null, d.fy = null;
}
</script>
json doc的样本
{
"nodes": [
{
"id": "test1",
"group": 1,
"size": 10
},
{
"id": "test2",
"group": 1,
"size": 10
}
],
"links": [
{
"source": "test1",
"target": "test2",
"value": 1,
"distance": 5
},
{
"source": "test2",
"target": "test1",
"value": 1,
"distance": 5
}
]
}
答案 0 :(得分:4)
我相信你几乎就在那里。使用像function(d) {return d.distance}
这样的自定义函数是正确的方法。但是,您不需要在links数组中输入额外信息,因为链接已包含JSON文件的属性。
正如您所提到的,可以声明模拟使用distance属性,如下所示:
var simulation = d3.forceSimulation()
.force("link", d3.forceLink().distance(function(d) {return d.distance;}).strength(0.1))
并且没有必要将额外信息推送到链接数组,因此您可以删除此行:
links.push({source: s, target: i, linkDist:linkDist}, {source: i, target: t, linkDist:linkDist });
我创建了一个带有一些修改过的JSON的JSFiddle来显示结果here。我添加了一个额外的链接到另一个节点,距离较短,以显示效果。
答案 1 :(得分:0)
您当然可以根据链接改变链接距离。它只需要设置在稍微不同的位置。在您的情况下,您将链接距离设置为500.相反,您可能希望执行以下操作:
var simulation = d3.forceSimulation()
.force("link", d3.forceLink().distance(linkDistance).strength(0.1))
.force("charge", d3.forceManyBody())
.force("center", d3.forceCenter(width / 2, height / 2));
function linkDistance(d) {
return d.distance;
}
https://github.com/d3/d3-force/blob/master/README.md#link_distance