Treemap by Count

Forked This example demonstrates how to create a treemap where each leaf node is treated as having equal value, rather than being sized according to a quantitative property.

In the flare.csv file, leaf nodes (representing class source files) have a non-zero value, while internal nodes (representing packages) have a zero value. Thus, when calling node.sum, any non-zero value can be treated as one:

root.sum(function(d) { return d.value ? 1 : 0; });

If you instead give all nodes a value of one, then one unit of additional space will be reserved for the parent.

<!DOCTYPE html>
<meta charset="utf-8">
<style>

body {
  font: 10px sans-serif;
  position: relative;
}

.node {
  box-sizing: border-box;
  position: absolute;
  overflow: hidden;
}

.node-label {
  padding: 4px;
  line-height: 1em;
  white-space: pre;
}

.node-value {
  color: rgba(0,0,0,0.8);
  font-size: 9px;
  margin-top: 1px;
}

</style>
<body>
<script src="//d3js.org/d3.v4.min.js"></script>
<script>

var width = 960,
    height = 1060;

var format = d3.format(",d");

var color = d3.scaleOrdinal()
    .range(d3.schemeCategory10
        .map(function(c) { c = d3.rgb(c); c.opacity = 0.6; return c; }));

var stratify = d3.stratify()
    .parentId(function(d) { return d.id.substring(0, d.id.lastIndexOf(".")); });

var treemap = d3.treemap()
    .size([width, height])
    .padding(1)
    .round(true);

d3.csv("flare.csv", type, function(error, data) {
  if (error) throw error;

var root = stratify(data)
      .sum(function(d) { return d.value ? 1 : 0; })
      .sort(function(a, b) { return b.height - a.height || b.value - a.value; });

treemap(root);

d3.select("body")
    .selectAll(".node")
    .data(root.leaves())
    .enter().append("div")
      .attr("class", "node")
      .attr("title", function(d) { return d.id; })
      .style("left", function(d) { return d.x0 + "px"; })
      .style("top", function(d) { return d.y0 + "px"; })
      .style("width", function(d) { return d.x1 - d.x0 + "px"; })
      .style("height", function(d) { return d.y1 - d.y0 + "px"; })
      .style("background", function(d) { while (d.depth > 1) d = d.parent; return color(d.id); })
    .append("div")
      .attr("class", "node-label")
      .text(function(d) { return d.id.substring(d.id.lastIndexOf(".") + 1).split(/(?=[A-Z][^A-Z])/g).join("\n"); });
});

function type(d) {
  d.value = +d.value;
  return d;
}

</script>

https://d3js.org/d3.v4.min.js