如何调整d3中散点图矩阵的刷牙?

时间:2018-04-26 17:32:48

标签: d3.js

我正在使用Mike Bostok的Block来刷涂散点图矩阵。

我想使用不透明度数字刷一下矩阵的对角线图;刷子范围两侧的画笔范围中间opacity = 1opacity value = low

除了this question之外,我无法在网上找到提示,它执行相同的操作,但使用dc.js(借助pretransition属性)。

这可以在d3中实现吗?

Running fiddle

1 个答案:

答案 0 :(得分:6)

有趣的问题。

这是一段代码片段,它会根据所选圈子的距离在所选圈子上设置不透明度。它在刷子端完成了这个过程。以下是相关方法:

function brushend(p) {

    // reset opacity
    svg.selectAll("circle").style('opacity', 1);

    // if no brush then bail
    if (brush.empty()) {
      svg.selectAll(".hidden").classed("hidden", false);
      return;
    }

    // some calculations
    var e = brush.extent(),
        xC = (e[1][0] + e[0][0]) / 2,
        yC = (e[1][1] + e[0][1]) / 2,
        // maxD is the maximum of the distance to the two edges
        maxD = Math.max( Math.sqrt((xC - e[0][0])**2), Math.sqrt ((yC - e[0][1])**2) );

    // those circles not hidden
    svg.selectAll("circle:not(.hidden)")
      .transition()
      .duration(1000)
      .style("opacity", function(d) {
        // distance from furthest edge
        var xD = Math.sqrt( (xC - d[p.x])**2 + (yC - d[p.y])**2 );
        // set opacity as percent of distance
        return (maxD - xD) / maxD;
    });

  }

运行代码:



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

svg {
  font: 10px sans-serif;
  padding: 10px;
}

.axis,
.frame {
  shape-rendering: crispEdges;
}

.axis line {
  stroke: #ddd;
}

.axis path {
  display: none;
}

.cell text {
  font-weight: bold;
  text-transform: capitalize;
}

.frame {
  fill: none;
  stroke: #aaa;
}

circle {
  fill-opacity: .7;
}

circle.hidden {
  fill: #ccc !important;
}

.extent {
  fill: #000;
  fill-opacity: .125;
  stroke: #fff;
}

</style>
<body>
<script src="//d3js.org/d3.v3.min.js"></script>
<script>

var width = 960,
    size = 230,
    padding = 20;

var x = d3.scale.linear()
    .range([padding / 2, size - padding / 2]);

var y = d3.scale.linear()
    .range([size - padding / 2, padding / 2]);

var xAxis = d3.svg.axis()
    .scale(x)
    .orient("bottom")
    .ticks(6);

var yAxis = d3.svg.axis()
    .scale(y)
    .orient("left")
    .ticks(6);

var color = d3.scale.category10();

d3.csv("https://gist.githubusercontent.com/mbostock/4063663/raw/13276d1a3d8e99e74c9fe33685da3afa6ed3a926/flowers.csv", function(error, data) {
  if (error) throw error;

  var domainByTrait = {},
      traits = d3.keys(data[0]).filter(function(d) { return d !== "species"; }),
      n = traits.length;

  traits.forEach(function(trait) {
    domainByTrait[trait] = d3.extent(data, function(d) { return d[trait]; });
  });

  xAxis.tickSize(size * n);
  yAxis.tickSize(-size * n);

  var brush = d3.svg.brush()
      .x(x)
      .y(y)
      .on("brushstart", brushstart)
      .on("brush", brushmove)
      .on("brushend", brushend);

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

  svg.selectAll(".x.axis")
      .data(traits)
    .enter().append("g")
      .attr("class", "x axis")
      .attr("transform", function(d, i) { return "translate(" + (n - i - 1) * size + ",0)"; })
      .each(function(d) { x.domain(domainByTrait[d]); d3.select(this).call(xAxis); });

  svg.selectAll(".y.axis")
      .data(traits)
    .enter().append("g")
      .attr("class", "y axis")
      .attr("transform", function(d, i) { return "translate(0," + i * size + ")"; })
      .each(function(d) { y.domain(domainByTrait[d]); d3.select(this).call(yAxis); });

  var cell = svg.selectAll(".cell")
      .data(cross(traits, traits))
    .enter().append("g")
      .attr("class", "cell")
      .attr("transform", function(d) { return "translate(" + (n - d.i - 1) * size + "," + d.j * size + ")"; })
      .each(plot);

  // Titles for the diagonal.
  cell.filter(function(d) { return d.i === d.j; }).append("text")
      .attr("x", padding)
      .attr("y", padding)
      .attr("dy", ".71em")
      .text(function(d) { return d.x; });

  cell.call(brush);

  function plot(p) {
    var cell = d3.select(this);

    x.domain(domainByTrait[p.x]);
    y.domain(domainByTrait[p.y]);

    cell.append("rect")
        .attr("class", "frame")
        .attr("x", padding / 2)
        .attr("y", padding / 2)
        .attr("width", size - padding)
        .attr("height", size - padding);

    cell.selectAll("circle")
        .data(data)
      .enter().append("circle")
        .attr("cx", function(d) { return x(d[p.x]); })
        .attr("cy", function(d) { return y(d[p.y]); })
        .attr("r", 4)
        .style("fill", function(d) { return "#4682b4"; });
  }

  var brushCell;

  // Clear the previously-active brush, if any.
  function brushstart(p) {
    if (brushCell !== this) {
      d3.select(brushCell).call(brush.clear());
      x.domain(domainByTrait[p.x]);
      y.domain(domainByTrait[p.y]);
      brushCell = this;
    }
  }

  // Highlight the selected circles.
  function brushmove(p) {
    var e = brush.extent();
    svg.selectAll("circle").classed("hidden", function(d) {
      return e[0][0] > d[p.x] || d[p.x] > e[1][0]
          || e[0][1] > d[p.y] || d[p.y] > e[1][1];
    });
  }

  function brushend(p) {
    
    // reset opacity
    svg.selectAll("circle").style('opacity', 1);
    
    // if no brush then bail
    if (brush.empty()) {
      svg.selectAll(".hidden").classed("hidden", false);
      return;
    }
    
    // some calculations
    var e = brush.extent(),
        xC = (e[1][0] + e[0][0]) / 2,
        yC = (e[1][1] + e[0][1]) / 2,
        // maxD is the maximum of the distance to the two edges
        maxD = Math.max( Math.sqrt((xC - e[0][0])**2), Math.sqrt ((yC - e[0][1])**2) );
    
    // those circles not hidden
    svg.selectAll("circle:not(.hidden)")
      .transition()
      .duration(1000)
      .style("opacity", function(d) {
        // distance from furthest edge
        var xD = Math.sqrt( (xC - d[p.x])**2 + (yC - d[p.y])**2 );
        // set opacity as percent of distance
        return (maxD - xD) / maxD;
    });
    
  }
});

function cross(a, b) {
  var c = [], n = a.length, m = b.length, i, j;
  for (i = -1; ++i < n;) for (j = -1; ++j < m;) c.push({x: a[i], i: i, y: b[j], j: j});
  return c;
}

</script>
&#13;
&#13;
&#13;

相关问题