Question

我正在修改path transition（code）上第二次可视化的代码，以向路径的末尾添加一个圆，以便与转换过渡一起，圆的“ cy”可以也可以过渡。

但是，我无法获得圆滑的过渡，而是出现了生涩的行为。我知道是因为tick函数每500毫秒被调用一次。我无法理解在转换过渡期间如何更新圆的“ cy”属性。有人可以帮我实现这一目标。

代码：

<!DOCTYPE html>
<meta charset="utf-8">
<style>
.line {
  fill: none;
  stroke: #000;
  stroke-width: 1.5px;
}
</style>
<svg width="960" height="500"></svg>
<script src="https://d3js.org/d3.v5.min.js"></script>
<script>
var n = 40,
    random = d3.randomNormal(0, .2),
    data = d3.range(n).map(random);
var svg = d3.select("svg"),
    margin = {top: 20, right: 20, bottom: 20, left: 40},
    width = +svg.attr("width") - margin.left - margin.right,
    height = +svg.attr("height") - margin.top - margin.bottom,
    g = svg.append("g").attr("transform", "translate(" + margin.left + "," + margin.top + ")");
var x = d3.scaleLinear()
    .domain([0, n - 1])
    .range([0, width]);
var y = d3.scaleLinear()
    .domain([-1, 1])
    .range([height, 0]);
var line = d3.line()
    .x(function(d, i) { return x(i); })
    .y(function(d, i) { return y(d); });
g.append("defs").append("clipPath")
    .attr("id", "clip")
  .append("rect")
    .attr("width", width)
    .attr("height", height);
g.append("g")
    .attr("class", "axis axis--x")
    .attr("transform", "translate(0," + y(0) + ")")
    .call(d3.axisBottom(x));
g.append("g")
    .attr("class", "axis axis--y")
    .call(d3.axisLeft(y));
///////////////Modified Code////////////////// 
var circle = g.append("circle")
    .attr("r", 6)
    .attr("cx", x(n-1));

var path =  g.append("g")
            .attr("clip-path", "url(#clip)")
        .append("path")
            .datum(data)
            .attr("class", "line")
        .transition()
            .duration(500)
            .ease(d3.easeLinear)
            .on("start", tick);
//////////////////////////////////////////////
function tick() {

    data.push(random());
  d3.select(this)
      .attr("d", line)
      .attr("transform", null);
///////////////Modified Code////////////////// 
  circle.transition()
    .attr("cy", getPoint(path).y)
//////////////////////////////////////////////
  d3.active(this)
      .attr("transform", "translate(" + x(-1) + ",0)")
    .transition()
      .on("start", tick);
  data.shift();
}
///////////////Modified Code////////////////// 
function getPoint(path){
    var length = path.node().getTotalLength();
    return path.node().getPointAtLength(length)
}
//////////////////////////////////////////////
</script>

Answer 1

如果匹配线路的难易程度和持续时间，则可以使其平滑：

circle.transition()
  .duration(500)
  .ease(d3.easeLinear)
  .attr("cy", getPoint(path).y);

<!DOCTYPE html>
<meta charset="utf-8">
<style>
  .line {
    fill: none;
    stroke: #000;
    stroke-width: 1.5px;
  }
</style>
<svg width="500" height="500"></svg>
<script src="https://d3js.org/d3.v5.min.js"></script>
<script>
  var n = 40,
    random = d3.randomNormal(0, .2),
    data = d3.range(n).map(random);
  var svg = d3.select("svg"),
    margin = {
      top: 20,
      right: 20,
      bottom: 20,
      left: 40
    },
    width = +svg.attr("width") - margin.left - margin.right,
    height = +svg.attr("height") - margin.top - margin.bottom,
    g = svg.append("g").attr("transform", "translate(" + margin.left + "," + margin.top + ")");
  var x = d3.scaleLinear()
    .domain([0, n - 1])
    .range([0, width]);
  var y = d3.scaleLinear()
    .domain([-1, 1])
    .range([height, 0]);
  var line = d3.line()
    .x(function(d, i) {
      return x(i);
    })
    .y(function(d, i) {
      return y(d);
    });
  g.append("defs").append("clipPath")
    .attr("id", "clip")
    .append("rect")
    .attr("width", width)
    .attr("height", height);
  g.append("g")
    .attr("class", "axis axis--x")
    .attr("transform", "translate(0," + y(0) + ")")
    .call(d3.axisBottom(x));
  g.append("g")
    .attr("class", "axis axis--y")
    .call(d3.axisLeft(y));
  ///////////////Modified Code////////////////// 
  var circle = g.append("circle")
    .attr("r", 6)
    .attr("cx", x(n - 1));

  var path = g.append("g")
    .attr("clip-path", "url(#clip)")
    .append("path")
    .datum(data)
    .attr("class", "line")
    .transition()
    .duration(500)
    .ease(d3.easeLinear)
    .on("start", tick);
  //////////////////////////////////////////////
  function tick() {

    data.push(random());
    d3.select(this)
      .attr("d", line)
      .attr("transform", null);
    ///////////////Modified Code////////////////// 
    circle.transition()
      .duration(500)
      .ease(d3.easeLinear)
      .attr("cy", getPoint(path).y)
      //////////////////////////////////////////////
    d3.active(this)
      .attr("transform", "translate(" + x(-1) + ",0)")
      .transition()
      .on("start", tick);
    data.shift();
  }
  ///////////////Modified Code////////////////// 
  function getPoint(path) {
    var length = path.node().getTotalLength();
    return path.node().getPointAtLength(length)
  }
  //////////////////////////////////////////////
</script>

在D3中的变换过渡期间更新圆的位置

1 个答案: