Question

我创建了这个点击展开 - 折叠网络 - http://jsfiddle.net/5Lv8gkqv/

var width = 960,
    height = 500,
    root = {
 "name": "Chocolate", "tag":"class",
 "children": [
  {
   "name": "Wafer", "tag":"subclass",
   "children": [
    {
      "name": "Nestle", "tag":"company",
     "children": [
      {"name": "KitKat", "tag":"product"}
     ]
    }
   ]
  },

  {
   "name": "White", "tag":"subclass",
   "children": [
    {
      "name": "Nestle", "tag":"company",
     "children": [
      {"name": "Milkybar", "tag":"product"}
     ]
    }
   ]
  },

    {
   "name": "Caramel", "tag":"subclass",
   "children": [
    {
      "name": "Nestle", "tag":"company",
     "children": [
      {"name": "BarOne", "tag":"product"}
     ]
    }
   ]
  },    
    {
   "name": "Milk", "tag":"subclass",
   "children": [
    {
      "name": "Nestle", "tag":"company",
     "children": [
      {"name": "Nestle Milk", "tag":"product"}
     ]
    },  {
      "name": "Cadbury", "tag":"company",
     "children": [
      {"name": "Dairy Milk", "tag":"product"}
     ]
    }
   ]
  }




 ]
};

var force = d3.layout.force()
    .linkDistance(150)
    .charge(-120)
    .gravity(.05)
    .size([width, height])
    .on("tick", tick);

var svg = d3.select("body").append("svg")
    .attr("width", width)
    .attr("height", height);

var link = svg.selectAll(".link"),
    node = svg.selectAll(".node");

flatten(root); //to set ids
setParents(root, null);
collapseAll(root);
root.children = root._children;
root._children = null;
update();

function update() {
  var nodes = flatten(root),
      links = d3.layout.tree().links(nodes);
  // Restart the force layout.
  force
      .nodes(nodes)
      .links(links)
      .start();

  // Update links.
  link = link.data(links, function(d) { return d.target.id; });

  link.exit().remove();

  link.enter().insert("line", ".node")
      .attr("class", "link");

  // Update nodes.
  node = node.data(nodes, function(d) { return d.id; });

  node.exit().remove();

  var nodeEnter = node.enter().append("g")
      .attr("class", "node")
      .on("click", click)
      .call(force.drag);

  nodeEnter.append("circle")
      .attr("r", function(d) { return Math.sqrt(d.size) / 10 || 4.5; });

  nodeEnter.append("text")
      .attr("dy", ".35em")
      .text(function(d) { return d.name; });

  node.select("circle")
      .style("fill", color);
}

function tick() {
  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("transform", function(d) { return "translate(" + d.x + "," + d.y + ")"; });
}

function color(d) {
  return d._children ? "#3182bd" // collapsed package
      : d.children ? "#c6dbef" // expanded package
      : "#fd8d3c"; // leaf node
}

// Toggle children on click.
function click(d) {
  if (d3.event.defaultPrevented) return; // ignore drag
  if (d.children) {
      collapseAll(d);
  } else {
      if (d._parent){
          d._parent.children.forEach(function(e){
              if (e != d){
                  collapseAll(e);
              }
          });
      }
    d.children = d._children;
    d._children = null;
  }
  update();
}

function collapseAll(d){
    if (d.children){
        d.children.forEach(collapseAll);
        d._children = d.children;
        d.children = null;
    }
    else if (d._childred){
        d._children.forEach(collapseAll);
    }
}

// Returns a list of all nodes under the root.
function flatten(root) {
  var nodes = [], i = 0;

  function recurse(node) {
    if (node.children) node.children.forEach(recurse);
    if (!node.id) node.id = ++i;
    nodes.push(node);
  }
  recurse(root);
  return nodes;
}

function setParents(d, p){
    d._parent = p;
  if (d.children) {
      d.children.forEach(function(e){ setParents(e,d);});
  } else if (d._children) {
      d._children.forEach(function(e){ setParents(e,d);});
  }
}

现在的事情是，我想知道是否可能将节点大小与子节点数成比例。因此父节点将是最大的圆，叶节点将是最小的，而中间节点大小将取决于每个子节点的数量。

Answer 1

您可以使用d3.scale.linear计算与子项数成比例的节点半径。 d3.scale也有助于找到范围之间的半径。这是更新的fiddle

 var minRadius = 10;
 var maxRadius = 15;
 var scale = d3.scale.linear().range([minRadius,maxRadius]);
 nodeEnter.append("circle")
    .attr("r", function(d) { 
        if(d.children)
            return scale(d.children.length);
        else if(d._children)
            return scale(d._children.length);
        else
            return minRadius;
    });

Answer 2

已更新：现在使用来自@ Gilsha的回答中的想法，考虑未展示的儿童。

当然，您可以将每个<circle>的半径设置为与他们拥有的孩子数量成比例：

node.select("circle")
    .attr("r", function(d){
         var numKids = 0;
         if (d.children) numKids += d.children.length;
         if (d._children) numKids += d._children.length;
         return 10 * (numKids + 1);
})

其中r是半径。使用r=10并在"wafer"节点上切换，您会得到： enter image description here

Answer 3

对于完整的解决方案，您实际上想要使用递归函数来首先计算每个节点的总子项（而不仅仅是一级子项）。例如：

var bubble_up_total_children = function(node) {
  var child, _i, _len, _ref;
  if (node.children && node.children.length > 0) {
    _ref = node.children;
    for (_i = 0, _len = _ref.length; _i < _len; _i++) {
      child = _ref[_i];
      bubble_up_total_children(child);
    }
    node.total_children = node.children.length + node.children.reduce(function(a, b) {
      return a.total_children + b.total_children;
    });
  } else {
    node.total_children = 0;
  }
};

bubble_up_total_children(root);

从那里你现在可以使用d3.scale，如@Gilsha的答案中所述，根据每个节点上的新total_children属性来计算大小。

节点大小与D3中的子节点数成比例

3 个答案: