Question

我对D3完全不熟悉并尝试制作类似于第三个here的实时流式广告。

然而，不同之处在于我需要数据积累/积累而不是传递。我试图从here复制代码并简单地注释掉它们翻译和弹出旧数据的部分，但是仍然没有这样做。

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

svg {
  font: 10px sans-serif;
}

.line {
  fill: none;
  stroke: #000;
  stroke-width: 1.5px;
}

.axis path,
.axis line {
  fill: none;
  stroke: #000;
  shape-rendering: crispEdges;
}

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

var n = 40,
    random = d3.random.normal(0, .2),
    data = d3.range(n).map(random);

var margin = {top: 20, right: 20, bottom: 20, left: 40},
    width = 960 - margin.left - margin.right,
    height = 500 - margin.top - margin.bottom;

var x = d3.scale.linear()
    .domain([1, n - 2])
    .range([0, width]);

var y = d3.scale.linear()
    .domain([-1, 1])
    .range([height, 0]);

var line = d3.svg.line()
    .interpolate("basis")
    .x(function(d, i) { return x(i); })
    .y(function(d, i) { return y(d); });

var svg = d3.select("body").append("svg")
    .attr("width", width + margin.left + margin.right)
    .attr("height", height + margin.top + margin.bottom)
  .append("g")
    .attr("transform", "translate(" + margin.left + "," + margin.top + ")");

svg.append("defs").append("clipPath")
    .attr("id", "clip")
  .append("rect")
    .attr("width", width)
    .attr("height", height);

svg.append("g")
    .attr("class", "x axis")
    .attr("transform", "translate(0," + y(0) + ")")
    .call(d3.svg.axis().scale(x).orient("bottom"));

svg.append("g")
    .attr("class", "y axis")
    .call(d3.svg.axis().scale(y).orient("left"));

var path = svg.append("g")
    .attr("clip-path", "url(#clip)")
  .append("path")
    .datum(data)
    .attr("class", "line")
    .attr("d", line);

tick();

function tick() {

  // push a new data point onto the back
  data.push(random());

  // redraw the line, but don't slide it to the left
  path
      .attr("d", line)
      .attr("transform", null)
    .transition()
      .duration(500)
      .ease("linear")
      //.attr("transform", "translate(" + x(0) + ",0)")
      .each("end", tick);

  // don't pop the old data point off the front
  // data.shift();

}

</script>

Answer 1

在您的勾选功能中，您需要添加以下行：

  data.push(random());//generate the random points
  x.domain([1, data.length - 2])//update the x axis domain
  xaxis.call(d3.svg.axis().scale(x).orient("bottom"))//redraw the x axis

工作代码here

累积的直播流图

1 个答案: