Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
749 views
in Technique[技术] by (71.8m points)

javascript polar scatter plot using d3.js

I'm trying to do a simple polar scatter plot like this one : http://helpcentral.componentone.com/NetHelp/SpreadNet6/WF/artwork/plottypes-polar-point2.png

I found that answer Polar plots using D3.js but I cannot display the points instead of d3.svg.line.radial. How do I change the line to points?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

If you are using the d3.svg.line.radial() to generate you polar plot, one easy solution would be to parse the generated path to get point coordinates, then add a circle at those coordinates:

var line = d3.svg.line.radial()
  .radius(function(d) {
    return r(d[1]);
  })
  .angle(function(d) {
    return -d[0] + Math.PI / 2;
  });

svg.selectAll("point")
  .data(data)
  .enter()
  .append("circle")
  .attr("class", "point")
  .attr("transform", function(d) {
    // use the line function and parse out the coordinates
    var coors = line([d]).slice(1).slice(0, -1);
    return "translate(" + coors + ")"
  })
  .attr("r", 8);

Here's a full working example:

<!DOCTYPE html>
<meta charset="utf-8">
<style>
  .frame {
    fill: none;
    stroke: #000;
  }
  
  .axis text {
    font: 10px sans-serif;
  }
  
  .axis line,
  .axis circle {
    fill: none;
    stroke: steelblue;
    stroke-dasharray: 4;
  }
  
  .axis:last-of-type circle {
    stroke: steelblue;
    stroke-dasharray: none;
  }
  
  .line {
    fill: none;
    stroke: orange;
    stroke-width: 3px;
  }
</style>

<body>
  <script src="//d3js.org/d3.v3.min.js"></script>
  <script>
    var width = 960,
      height = 500,
      radius = Math.min(width, height) / 2 - 30;

    var r = d3.scale.linear()
      .domain([0, 1])
      .range([0, radius]);

    var line = d3.svg.line.radial()
      .radius(function(d) {
        return r(d[1]);
      })
      .angle(function(d) {
        return -d[0] + Math.PI / 2;
      });

    var svg = d3.select("body").append("svg")
      .attr("width", width)
      .attr("height", height)
      .append("g")
      .attr("transform", "translate(" + width / 2 + "," + height / 2 + ")");

    var gr = svg.append("g")
      .attr("class", "r axis")
      .selectAll("g")
      .data(r.ticks(3).slice(1))
      .enter().append("g");

    gr.append("circle")
      .attr("r", r);

    var ga = svg.append("g")
      .attr("class", "a axis")
      .selectAll("g")
      .data(d3.range(0, 360, 30))
      .enter().append("g")
      .attr("transform", function(d) {
        return "rotate(" + -d + ")";
      });

    ga.append("line")
      .attr("x2", radius);
      
    var color = d3.scale.category20();

    var line = d3.svg.line.radial()
      .radius(function(d) {
        return r(d[1]);
      })
      .angle(function(d) {
        return -d[0] + Math.PI / 2;
      });
      
    var data = [
      [Math.PI / 3, Math.random()],
      [Math.PI / 6, Math.random()],
      [0 * Math.PI, Math.random()],
      [(11 * Math.PI) / 6, Math.random()],
      [(5 * Math.PI / 3), Math.random()],
      [(3 * Math.PI) / 2, Math.random()],
      [(4 * Math.PI / 3), Math.random()],
      [(7 * Math.PI) / 6, Math.random()],
      [Math.PI, Math.random()],
      [(5 * Math.PI) / 6, Math.random()],
      [(2 * Math.PI) / 3, Math.random()],
      [Math.PI / 2, Math.random()]
    ]

    svg.selectAll("point")
      .data(data)
      .enter()
      .append("circle")
      .attr("class", "point")
      .attr("transform", function(d) {
        var coors = line([d]).slice(1).slice(0, -1);
        return "translate(" + coors + ")"
      })
      .attr("r", 8)
      .attr("fill",function(d,i){
        return color(i);
      });

  </script>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...