在d3js中指定群集力布局的每个集群中的多个节点

Question

我正在尝试使用集群力布局III示例来设计一个信息图表，我几乎按照我想要的方式来设计，但是我需要指定每个集群中的多个节点。集群1-10节点、集群2-7节点、集群3-11节点、集群4-18节点、集群5-16节点、集群6-19节点、集群7-42节点、集群8-14节点。我有正确的节点总数和正确的集群总数，并且使所有节点的大小相同。现在我只需要推开圆圈，用数据来表示。我的密码在下面。

var width = 175,
    height = 175,
    padding = 1.5, // separation between same-color nodes
    clusterPadding = 6, // separation between different-color nodes
    maxRadius = 4;

var n = 137, // total number of nodes
    m = 8; // number of distinct clusters

var color = d3.scale.category10()
    .domain(d3.range(m));

// The largest node for each cluster.
var clusters = new Array(m);

var nodes = d3.range(n).map(function() {
  var i = Math.floor(Math.random() * m),
      r = maxRadius,
      d = {
        cluster: i,
        radius: r,
        x: Math.cos(i / m * 2 * Math.PI) * 200 + width / 2 + Math.random(),
        y: Math.sin(i / m * 2 * Math.PI) * 200 + height / 2 + Math.random()
      };
  if (!clusters[i] || (r > clusters[i].radius)) clusters[i] = d;
  return d;
});

var force = d3.layout.force()
    .nodes(nodes)
    .size([width, height])
    .gravity(.02)
    .charge(0)
    .on("tick", tick)
    .start();

var svg = d3.select(".wpd3-1042-0").append("svg")
    .attr("width", width)
    .attr("height", height);

var node = svg.selectAll("circle")
    .data(nodes)
  .enter().append("circle")
    .style("fill", function(d) { return color(d.cluster); })
    .call(force.drag);

node.transition()
    .duration(750)
    .delay(function(d, i) { return i * 5; })
    .attrTween("r", function(d) {
      var i = d3.interpolate(0, d.radius);
      return function(t) { return d.radius = i(t); };
    });

function tick(e) {
  node
      .each(cluster(10 * e.alpha * e.alpha))
      .each(collide(.5))
      .attr("cx", function(d) { return d.x; })
      .attr("cy", function(d) { return d.y; });
}

// Move d to be adjacent to the cluster node.
function cluster(alpha) {
  return function(d) {
    var cluster = clusters[d.cluster];
    if (cluster === d) return;
    var x = d.x - cluster.x,
        y = d.y - cluster.y,
        l = Math.sqrt(x * x + y * y),
        r = d.radius + cluster.radius;
    if (l != r) {
      l = (l - r) / l * alpha;
      d.x -= x *= l;
      d.y -= y *= l;
      cluster.x += x;
      cluster.y += y;
    }
  };
}

// Resolves collisions between d and all other circles.
function collide(alpha) {
  var quadtree = d3.geom.quadtree(nodes);
  return function(d) {
    var r = d.radius + maxRadius + Math.max(padding, clusterPadding),
        nx1 = d.x - r,
        nx2 = d.x + r,
        ny1 = d.y - r,
        ny2 = d.y + r;
    quadtree.visit(function(quad, x1, y1, x2, y2) {
      if (quad.point && (quad.point !== d)) {
        var x = d.x - quad.point.x,
            y = d.y - quad.point.y,
            l = Math.sqrt(x * x + y * y),
            r = d.radius + quad.point.radius + (d.cluster === quad.point.cluster ? padding : clusterPadding);
        if (l < r) {
          l = (l - r) / l * alpha;
          d.x -= x *= l;
          d.y -= y *= l;
          quad.point.x += x;
          quad.point.y += y;
        }
      }
      return x1 > nx2 || x2 < nx1 || y1 > ny2 || y2 < ny1;
    });
  };
}

Answer 1

是!您可以通过这样的操作来实现这一点：

var clusterNumber = [10, 7, 11, 18, 16, 19, 42, 14];//your cluster number of node array
var n = d3.sum(clusterNumber, function (d) {
    return d
}); /// total number of nodes

var m = clusterNumber.length;//total number of clusters
var color = d3.scale.category10()
    .domain(d3.range(m));

// The largest node for each cluster.
var clusters = new Array(m);
var nodes = [];
clusterNumber.forEach(function (cn, i) {
//this will make a cluster
    var r = maxRadius;
    for (var j = 0; j < cn; j++) {
//this loop will make all the nodes
        var d = {
            cluster: i,
            radius: r,
            x: Math.cos(i / m * 2 * Math.PI) * 200 + width / 2 + Math.random(),
            y: Math.sin(i / m * 2 * Math.PI) * 200 + height / 2 + Math.random()
        };
        if (!clusters[i] || (r > clusters[i].radius)) clusters[i] = d;
        nodes.push(d);
    }

});

全工作代码这里

希望这能有所帮助！