我希望开发一个由节点链接图组成的viz。我有一系列的点,我的位置我不想改变,除非图上有一个碰撞(另一个节点)。在碰撞节点的情况下,我想将它们隔开以使它们不重叠。我的JS代码如下
var chartWidth = 200;
var chartHeight = 200;
var widthPadding = 40;
var heightPadding = 40;
var link, node;
$(function(){
initialize();
});
function initialize() {
var jsonString = '{"nodes":[{"x":40,"y":64,"r":6,"fixed":true},{"x":40,"y":63,"r":6,"fixed":true},{"x":119,"y":53,"r":6,"fixed":true},{"x":119,"y":73,"r":6,"fixed":true},{"x":137,"y":73,"r":6,"fixed":true},{"x":140,"y":140,"r":6,"fixed":true},{"x":68,"y":57,"r":6,"fixed":true},{"x":70,"y":75,"r":6,"fixed":true},{"x":51,"y":59,"r":6,"fixed":true},{"x":51,"y":54,"r":6,"fixed":true},{"x":137,"y":40,"r":6,"fixed":true}],"links":[{"source":0,"target":1},{"source":1,"target":2},{"source":2,"target":3},{"source":3,"target":4},{"source":4,"target":5},{"source":0,"target":1},{"source":1,"target":6},{"source":6,"target":7},{"source":7,"target":4},{"source":4,"target":5},{"source":0,"target":1},{"source":1,"target":8},{"source":8,"target":9},{"source":9,"target":10},{"source":10,"target":5}]}';
drawForceDirectedNodeLink($.parseJSON(jsonString));
}
function drawForceDirectedNodeLink(graph){
var width = chartWidth + (2*widthPadding);
var height = chartHeight + (2*heightPadding);
var q = d3.geom.quadtree(graph.nodes),
i = 0,
n = graph.nodes.length;
while (++i < n) {
q.visit(collide(graph.nodes[i]));
}
var force = d3.layout.force()
.size([width, height])
.gravity(0.05)
.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; })
.attr("r", function(d) { return d.r; });
});
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height);
var link = svg.selectAll(".link"),
node = svg.selectAll(".node");
force
.nodes(graph.nodes)
.links(graph.links)
.start();
link = link.data(graph.links)
.enter().append("line")
.attr("class", "link");
node = node.data(graph.nodes)
.enter().append("circle")
.attr("class", "node");
}
function collide(node) {
var r = node.radius + 16,
nx1 = node.x - r,
nx2 = node.x + r,
ny1 = node.y - r,
ny2 = node.y + r;
return function(quad, x1, y1, x2, y2) {
if (quad.point && (quad.point !== node)) {
var x = node.x - quad.point.x,
y = node.y - quad.point.y,
l = Math.sqrt(x * x + y * y),
r = node.radius + quad.point.radius;
if (l < r) {
l = (l - r) / l * .5;
node.x -= x *= l;
node.y -= y *= l;
quad.point.x += x;
quad.point.y += y;
}
}
return x1 > nx2
|| x2 < nx1
|| y1 > ny2
|| y2 < ny1;
};
}
如您所见,我尝试实现了here提到的碰撞检测逻辑。但有些我怎么也没能让这部分工作。
答案 0 :(得分:2)
请注意,在jsonString
内的initialize()
声明中,每个节点都被赋予r
属性。但是,在collide()
范围内,您正在执行以下操作:
.attr("r", function(d) { return d.radius - 2; })
确保您的节点附加了radius
属性。如果没有,则应进行以下更改:
.attr("r", function(d) { return d.r - 2; })
您可以在Mike Bostock的脚本第30行看到他的节点最初使用radius
属性声明,而不是r
属性。
var nodes = d3.range(200).map(function() { return {radius: Math.random() * 12 + 4}; }),
答案 1 :(得分:1)
<强>更新强>
将node.radius
更改为node.r
,将quad.point.radius
更改为quad.point.r
。它应该工作。看起来这只是一个NaN
问题。