Question

我正在尝试使这个d3树形图可缩放，但似乎正在努力解决它。我一直得到“treemap.nodes不是函数”错误。我试图在这里遵循Mike Bostock的例子，https://bost.ocks.org/mike/treemap/

这是我的代码：

 var data = 
    {
        "name" : "Max",
        "value": 100,
        "children": 
        [
            {
             "name": "A",
             "value": 75,
             "children": [
                 { "name": "Alpha", "result": 20},
                 { "name": "Aplha 2", "result": 40},
                 { "name": "Aplha 3", "result": 35}
             ]
            },
            {
             "name": "B",
             "value": 75,
             "children": [
                 {"name": "Bravo", "result": 80},
                 {"name": "Bravo 2", "result": 20},
                 {"name": "Bravo 3", "result": 33}
             ]
            },
            {
             "name": "C",
             "value": 75,
             "children": [
                 {"name": "Charle", "result": 84},
                 {"name": "Charle 2", "result": 43},
                 {"name": "Charle 3", "result": 24}
              ]
            }
        ]
    }
    var margin = {top: 20, right: 0, bottom: 0, left: 0},
        width = 960,
        height = 500 - margin.top - margin.bottom,
        formatNumber = d3.format(",d"),
        transitioning;

    var x = d3.scale.linear()
        .domain([0, width])
        .range([0, width]);

    var y = d3.scale.linear()
        .domain([0, height])
        .range([0, height]);

    var color = d3.scale.category10();
    var svg = d3.select("#chart").append("svg")
        .attr("width", 1000)
        .attr("height", 1000);

    var treemap =  d3.layout.treemap()
        .size([1000, 1000])
        .children(function(d, depth) { return depth ? null : d.children; })
        .sort(function(a, b) {return a.value - b.value; })
        .ratio(height / width * 0.5 * (1 + Math.sqrt(5)))
        .round(false)
        .nodes(data);

    var cells = svg.selectAll(".cell")
        .data(treemap)
        .enter()
        .append("g")
        .attr("class", "cell")

    cells.append("rect")
        .attr("x", function(d){ return d.x})
        .attr("y", function(d){ return d.y})
        .attr("width", function(d){ return d.dx})
        .attr("height", function(d){ return d.dy})
        .attr("stroke", "#fff")
        .attr("fill", function(d){ return d.children ? null : color(d.parent.name);})

    cells.append("text")
        .attr("x", function (d) { return d.x + d.dx /2})
        .attr("y", function (d) { return d.y + d.dy /2})
        .text(function(d) { return d.children ? null : d.name + d.value;})

    var grandparent = svg.append("g")
        .attr("class", "grandparent");

    grandparent.append("rect")
        .attr("y", -margin.top)
        .attr("width", width)
        .attr("height", margin.top);

    grandparent.append("text")
        .attr("x", 6)
        .attr("y", 6 - margin.top)
        .attr("dy", ".75em"); 

   var root = sample_data;
   initialize(root);
    accumulate(root);
    layout(root);
    display(root);

    function initialize(root) {
        root.x = root.y = 0;
        root.dx = width;
        root.dy = height;
        root.depth = 0;
    }

    function accumulate(d) {
        return d.children
            ? d.value = d.children.reduce(function(p, v) { return p + accumulate(v); }, 0)
            : d.value;
    }

    function layout(d) {
        if (d.children) {
          treemap.nodes({children: d.children});
          d.children.forEach(function(c) {
            c.x = d.x + c.x * d.dx;
            c.y = d.y + c.y * d.dy;
            c.dx *= d.dx;
            c.dy *= d.dy;
            c.parent = d;
            layout(c);
          });
        }
      }

    function display(d) {
        grandparent
           .datum(d.parent)
           .on("click", transition)
          .select("text")
            .text(name(d));

    var g1 = svg.insert("g", ".grandparent")
        .datum(d)
        .attr("class", "depth");

    var g = g1.selectAll("g")
        .data(d.children)
      .enter().append("g");

    g.filter(function(d) { return d.children; })
        .classed("children", true)
        .on("click", transition);

    g.selectAll(".child")
        .data(function(d) { return d.children || [d]; })
      .enter().append("rect")
        .attr("class", "child")
        .call(rect);

    g.append("rect")
        .attr("class", "parent")
        .call(rect)
      .append("title")
        .text(function(d) { return formatNumber(d.value); });

    g.append("text")
        .attr("dy", ".75em")
        .text(function(d) { return d.name; })
        .call(text);

    function transition(d) {
      if (transitioning || !d) return;
      transitioning = true;

        var g2 = display(d),
            t1 = g1.transition().duration(750),
            t2 = g2.transition().duration(750);

        x.domain([d.x, d.x + d.dx]);
        y.domain([d.y, d.y + d.dy]);

        svg.style("shape-rendering", null);

        svg.selectAll(".depth").sort(function(a, b) { return a.depth - b.depth; });

        g2.selectAll("text").style("fill-opacity", 0);

        t1.selectAll("text").call(text).style("fill-opacity", 0);
        t2.selectAll("text").call(text).style("fill-opacity", 1);
        t1.selectAll("rect").call(rect);
        t2.selectAll("rect").call(rect);

        t1.remove().each("end", function() {
        svg.style("shape-rendering", "crispEdges");
        transitioning = false;
        });
    }

    return g;
  }

  function text(text) {
    text.attr("x", function(d) { return x(d.x) + 6; })
        .attr("y", function(d) { return y(d.y) + 6; });
  }

  function rect(rect) {
    rect.attr("x", function(d) { return x(d.x); })
        .attr("y", function(d) { return y(d.y); })
        .attr("width", function(d) { return x(d.x + d.dx) - x(d.x); })
        .attr("height", function(d) { return y(d.y + d.dy) - y(d.y); });
  }

  function size(d) {
    return d.value;
  }

  function name(d) {
    return d.parent
        ? name(d.parent) + "." + d.name
        : d.name;
  }

任何帮助都将不胜感激。

Answer 1

.nodes()函数不会返回调用它的树形图（就像许多其他函数一样），而是返回节点数组，因此你不能拥有treemap= d3.layout.treemap(). [...] .nodes(data)。请改用以下内容：

 var treemap =  d3.layout.treemap()
    .size([1000, 1000])
    .children(function(d, depth) { return depth ? null : d.children; })
    .sort(function(a, b) {return a.value - b.value; })
    .ratio(height / width * 0.5 * (1 + Math.sqrt(5)))
    .round(false);
 var nodes = treemap.nodes(data);

D3树形图可缩放

1 个答案: