D3.js 如何基于形状字段值向D3强制有向图添加两个不同的形状?
我是D3的新手。我使用的是力有向图。我想在节点的位置添加两种不同类型的形状 我的json如下所示:D3.js 如何基于形状字段值向D3强制有向图添加两个不同的形状?,d3.js,force-layout,D3.js,Force Layout,我是D3的新手。我使用的是力有向图。我想在节点的位置添加两种不同类型的形状 我的json如下所示: { "nodes":[ {"name":"00:00:00:00:00:00:00:01","group":0,"shape":1}, {"name":"00:00:00:00:00:00:00:02","group":1,"shape":1}, {"name":"00:00:00:00:00:00:00:03","group":2,"shape":1}, {"
{
"nodes":[
{"name":"00:00:00:00:00:00:00:01","group":0,"shape":1},
{"name":"00:00:00:00:00:00:00:02","group":1,"shape":1},
{"name":"00:00:00:00:00:00:00:03","group":2,"shape":1},
{"name":"00:00:00:00:00:00:00:11","group":0,"shape":0},
{"name":"00:00:00:00:00:00:00:21","group":1,"shape":0},
{"name":"00:00:00:00:00:00:00:31","group":2,"shape":0},
{"name":"00:00:00:00:00:00:00:32","group":2,"shape":0},
{"name":"00:00:00:00:00:00:00:12","group":0,"shape":0},
{"name":"00:00:00:00:00:00:00:22","group":1,"shape":0}
],
"links":[
{ "source": 0, "target": 0, "value": 5 },
{ "source": 1, "target": 1, "value": 5 },
{ "source": 2, "target": 2, "value": 5 },
{ "source": 3, "target": 0, "value": 5 },
{ "source": 4, "target": 1, "value": 5 },
{ "source": 5, "target": 2, "value": 5 },
{ "source": 6, "target": 2, "value": 5 },
{ "source": 7, "target": 0, "value": 5 },
{ "source": 8, "target": 1, "value": 5 }
]
}
若形状值为1,则绘制圆形,若形状值为0,则绘制矩形。
力定向图示例链接为:
我已经尝试了示例链接JSFiddle:您可以这样做,如中所示,例如,对于特定形状,使用and
路径
元素而不是SVG元素。添加形状的代码变为
var node = svg.selectAll(".node")
.data(data.nodes)
.enter().append("path")
.attr("class", "node")
.attr("d", d3.svg.symbol()
.type(function(d) { return d3.svg.symbolTypes[d.s]; }))
.style("fill", function(d) { return color(d.group); })
.call(force.drag);
然后还需要更改勾选
处理程序,以更改路径
元素的转换
属性:
node.attr("transform", function(d) {
return "translate(" + d.x + "," + d.y + ")";
});
完成JSFIDLE。讨论您要寻找的内容。还有,嗨,拉尔斯。。。我试过这么做,但我不知道如何传递圆而不是三角形的参数,而且这些符号和边不相邻。请试着帮我画圆圈。我的JSFiddle链接:这里的答案也适用于你的问题:谢谢Lars…你让我开心。我可以执行类似.enter().append(函数(d){add span或div或任何取决于数据的东西;})的操作吗?I get NotFoundError:未能在“Node”上执行“appendChild”:当我尝试返回使用jQueryIt创建的元素时,新的子元素为空。最好单独问一个问题,您可以提供更多详细信息。d3.v4使用的更正
d3.symbol().size(…).type(…)