动态更改Javascript D3布局模拟

时间:2011-09-14 23:22:23

标签: javascript d3.js visualization force-layout

我正试图在D3https://github.com/mbostock/d3/blob/master/examples/force/force-multi-foci.html示例中动态更改模拟详细信息。我输入一个复选框,然后动态分配滴答处理程序,如下所示(http://pastebin.com/k4P0uzHK处的完整代码):

$("#chkBox").change(function(){
  if ($(this).is(':checked')) {
    force.on("tick", forceTick);
  } else {
    force.on("tick", forceTick2);
  }
});

forceTick = function(e) {
  // Push different nodes in different directions for clustering.
  var ky = 400 * e.alpha;
  var kx = 20 * e.alpha;
  hLinks.forEach(function(hlink) {
    var yB = hlink.source.y, yT = hlink.target.y;
    if (yB<(yT+20)) { hlink.source.y += Math.min(ky,yT+20-yB); hlink.target.y -= Math.min(ky,yT+20-yB);}
    var xB = hlink.source.x, xT = hlink.target.x;
    if (xB<(xT-20)) { hlink.source.x += Math.min(kx,xT-20-xB); hlink.target.x -= Math.min(kx,xT-20-xB);}
    if (xB>(xT+20)) { hlink.source.x -= Math.min(kx,xB-xT-20); hlink.target.x += Math.min(kx,xB-xT-20);}

  });

  node.attr("cx", function(d) { return d.x; })
      .attr("cy", function(d) { return d.y; });

  link.attr("x1", function(d) { return d.source.x; })
       .attr("y1", function(d) { return d.source.y; })
       .attr("x2", function(d) { return d.target.x; })
       .attr("y2", function(d) { return d.target.y; });
};


forceTick2 = function(e) {

  node.attr("cx", function(d) { return d.x; })
      .attr("cy", function(d) { return d.y; });

  link.attr("x1", function(d) { return d.source.x; })
       .attr("y1", function(d) { return d.source.y; })
       .attr("x2", function(d) { return d.target.x; })
       .attr("y2", function(d) { return d.target.y; });
};

但实际上似乎只有首先给出的处理程序才有效。有没有办法动态控制模拟?

2 个答案:

答案 0 :(得分:4)

强制布局的on运算符(以及使用d3.dispatch的任何其他内容)都会添加一个事件侦听器。它不会替换现有的事件侦听器。强制布局当前不会公开删除或替换现有事件侦听器的机制。

这是一个错误。我打算使布局的on运算符与选择一致,这允许您通过多次调用on来添加,替换和删除事件侦听器。如果使用名称空间(例如“tick.foo”和“tick.bar”),您仍然可以注册多个侦听器。

与此同时,简单的解决方法是使用单个方法作为tick侦听器,但是然后使用一些全局布尔值来确定每个tick的两个行为中的哪一个。在你的情况下,像:

if (checked) {
  … clustering …
}
… update link positions …
… update node positions …

而且,这消除了代码重复。 :)

答案 1 :(得分:1)

你可以在这里看到d3.v4的一个很好的例子: https://bl.ocks.org/steveharoz/8c3e2524079a8c440df60c1ab72b5d03

您可以在此处查看重要功能:

// add forces to the simulation
function initializeForces() {
  // add forces and associate each with a name
  simulation
    .force("link", d3.forceLink())
    .force("charge", d3.forceManyBody())
    .force("collide", d3.forceCollide())
    .force("center", d3.forceCenter())
    .force("forceX", d3.forceX())
    .force("forceY", d3.forceY());
  // apply properties to each of the forces
  updateForces();
}

// apply new force properties
function updateForces() {
  // get each force by name and update the properties
  simulation.force("center")
    .x(width * forceProperties.center.x)
    .y(height * forceProperties.center.y);
  simulation.force("charge")
    .strength(forceProperties.charge.strength * forceProperties.charge.enabled)
    .distanceMin(forceProperties.charge.distanceMin)
    .distanceMax(forceProperties.charge.distanceMax);
  simulation.force("collide")
    .strength(forceProperties.collide.strength * forceProperties.collide.enabled)
    .radius(forceProperties.collide.radius)
    .iterations(forceProperties.collide.iterations);
  simulation.force("forceX")
    .strength(forceProperties.forceX.strength * forceProperties.forceX.enabled)
    .x(width * forceProperties.forceX.x);
  simulation.force("forceY")
    .strength(forceProperties.forceY.strength * forceProperties.forceY.enabled)
    .y(height * forceProperties.forceY.y);
  simulation.force("link")
    .id(function(d) {return d.id;})
    .distance(forceProperties.link.distance)
    .iterations(forceProperties.link.iterations)
    .links(forceProperties.link.enabled ? graph.links : []);

  // updates ignored until this is run
  // restarts the simulation (important if simulation has already slowed down)
  simulation.alpha(1).restart();
}