如何将D3树布局的方向改变90度

时间：2013-08-07 09:18:11

标签： javascript svg d3.js tree orientation

我刚开始参与网络可视化，所以我完全是新手。我的目标是显示一个家族树，其中一个根节点将拥有多个父节点和子节点。在寻找解决方案时，我发现了这个例子：http://bl.ocks.org/jdarling/2503502 它很棒，因为它似乎具有我需要的功能。但是，我想改变方向（从上到下）。我尝试使用此示例执行此操作：http://bl.ocks.org/mbostock/3184089但失败了。

我的代码：

var tree = d3.layout.tree()
    .size([height, width]);

var diagonal = d3.svg.diagonal()
    .projection(function(d) {
        return [d.x, d.y];
    });

var elbow = function (d, i){
    var source = calcTop(d.source);
    var target = calcTop(d.target);
    var hx = (target.x-source.x)/2;
    if(d.isRight)
        hx = -hx;
    return  "M" + source.x + "," + source.y
          + "H" + (source.x+hx)
          + "V" + target.y + "H" + target.x;
};

var connector = elbow;

var calcTop = function(d){
    var top = d.x;
    if(!d.isRight){
        top = d.x-halfHeight;
        top = halfHeight - top;
    }
    return {x : top, y : d.y};
};

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

d3.json("tree.json", function(json) {
    root = json;
    root.x0 = height / 2;
    root.y0 = width / 2;
    var t1 = d3.layout.tree()
        .size([halfHeight, width])
        .children(function(d){
            return d.winners;
        });
    var t2 = d3.layout.tree()
        .size([halfHeight, width])
        .children(function(d){
            return d.challengers;
        });
    t1.nodes(root);
    t2.nodes(root);

    var rebuildChildren = function(node){
        node.children = getChildren(node);
        if(node.children)
            node.children.forEach(rebuildChildren);
    }
    rebuildChildren(root);
    root.isRight = false;
    update(root);
});

var toArray = function(item, arr){
    arr = arr || [];
    var i = 0, l = item.children?item.children.length:0;
    arr.push(item);
    for(; i < l; i++){
        toArray(item.children[i], arr);
    }
    return arr;
};

function update(source) {
// Compute the new tree layout.
var nodes = toArray(source);

// Normalize for fixed-depth.
nodes.forEach(function(d) { d.x = d.depth * 180 + halfHeight; });

// Update the nodes…
var node = vis.selectAll("g.node")
    .data(nodes, function(d) { return d.id || (d.id = ++i); });

// Enter any new nodes at the parent's previous position.
var nodeEnter = node.enter().append("g")
    .attr("class", "node")
    .attr("transform", function(d) {
        return "translate(" + source.x0 + "," + source.y0 + ")";
    })
    .on("click", click);

nodeEnter.append("circle")
    .attr("r", 1e-6)
    .style("fill", function(d) {
        return d._children ? "lightsteelblue" : "#fff";
    });

nodeEnter.append("text")
    .attr("dy", function(d) { return d.isRight?14:-8;})
    .attr("text-anchor", "middle")
    .text(function(d) { return d.name; })
    .style("fill-opacity", 1e-6);

// Transition nodes to their new position.
var nodeUpdate = node.transition()
    .duration(duration)
    .attr("transform", function(d) {
        p = calcTop(d);
        return "translate(" + p.x + "," + p.y + ")";
});

nodeUpdate.select("circle")
    .attr("r", 4.5)
    .style("fill", function(d) {
        return d._children ? "lightsteelblue" : "#fff";
    });

nodeUpdate.select("text")
    .style("fill-opacity", 1);

// Transition exiting nodes to the parent's new position.
var nodeExit = node.exit().transition()
    .duration(duration)
    .attr("transform", function(d) {
         p = calcTop(d.parent||source);
         return "translate(" + p.x + "," + p.y + ")";
    })
    .remove();

nodeExit.select("circle")
    .attr("r", 1e-6);

nodeExit.select("text")
    .style("fill-opacity", 1e-6);

// Update the links...
var link = vis.selectAll("path.link")
    .data(tree.links(nodes), function(d) { return d.target.id; });

// Enter any new links at the parent's previous position.
link.enter().insert("path", "g")
    .attr("class", "link")
    .attr("d", function(d) {
        var o = {x: source.x0, y: source.y0};
        return connector({source: o, target: o});
    });

// Transition links to their new position.
link.transition()
    .duration(duration)
    .attr("d", connector);

// Transition exiting nodes to the parent's new position.
link.exit()
    .transition()
    .duration(duration)
    .attr("d", function(d) {
        var o = calcTop(d.source||source);
        if(d.source.isRight)
            o.x -= halfHeight - (d.target.x - d.source.x);
        else
            o.x += halfHeight - (d.target.x - d.source.x);
        return connector({source: o, target: o});
    })
    .remove();

// Stash the old positions for transition.
nodes.forEach(function(d) {
    var p = calcTop(d);
    d.x0 = p.x;
    d.y0 = p.y;
});

// Toggle children on click.
function click(d) {
    if (d.children) {
        d._children = d.children;
        d.children = null;
    } else {
        d.children = d._children;
        d._children = null;
    }
update(source);

} }

真的很感激帮助！

3 个答案:

答案 0 :(得分：6)

您正在查看的示例实际上已经被翻转 - 这可能会让您感到困惑。 D3中的树木是自然自上而下的树木，代码会进行大量的x-y翻转以使树木侧向移动。

更改

  var nodeUpdate = node.transition()
      .duration(duration)
      .attr("transform", function(d) { p = calcLeft(d); return "translate(" + p.y + "," + p.x + ")"; })
      ;

到

  var nodeUpdate = node.transition()
      .duration(duration)
      .attr("transform", function(d) { p = calcLeft(d); return "translate(" + p.x + "," + p.y + ")"; })
      ;

将使节点显示在正确的位置。使用update（）中的任何交换x-y坐标实例进行类似的更改可以解决大多数定位问题。最后一件事是肘关节功能/变量/无论你想要什么，其中

  return "M" + source.y + "," + source.x
         + "H" + (source.y+hy)
         + "V" + target.x + "H" + target.y;

应改为

  return "M" + source.x + "," + source.y
         + "V" + (source.y+hy)
         + "H" + target.x + "V" + target.y;

这会将连接器形状从水平垂直水平更改为垂直水平垂直。请注意，这是一个原始SVG行，而不是d3。我所做的更改（加上交换宽度和高度，以及将AJAX JSON请求更改为硬编码数据 - AJAX很难在小提琴中工作）都在http://jsfiddle.net/Zj3th/2/。

如果你没有d3和SVG的经验，我肯定会在你进一步修改代码之前先看一下像http://blog.pixelingene.com/2011/07/building-a-tree-diagram-in-d3-js/之类的简单例子。

答案 1 :(得分：3)

这是我制作的底部图表的示例。我认为它可能对你有用，也许你会发现一些想法等。

enter image description here

链接到codepen上的代码。随意提出任何问题。

答案 2 :(得分：0)

its very easy to make rotation in d3 collapsible tree TOP-DOWN..

step 1: specify top,bottom,left,right part of our drawing pan and also the width and                            
        height.
Step 2: Make the orientation as top to bottom 
step 3: Reading the json and make the orientation
step 4: Appending the link and circle to the body

<script>
var jdata='${data}'
    var margin = {top: 20, right: 120, bottom: 20, left: 120},
    width = 980 - margin.left - margin.right,
    height = 500 - margin.top - margin.bottom;

    var orientations = {
      "top-to-bottom": {
        size: [width, height],
        x: function(d) { return d.x; },
        y: function(d) { return d.y; }
      }
    };

    var svg = d3.select("body").selectAll("svg")
        .data(d3.entries(orientations))
        .enter().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 + ")");

    d3.json("resources/grap.json", function(json) {
         root = JSON.parse(jdata);

      svg.each(function(orientation) {
        var svg = d3.select(this),
            o = orientation.value;

        // Compute the layout.
        var tree = d3.layout.tree().size(o.size),
            nodes = tree.nodes(root),
            links = tree.links(nodes);

        // Create the link lines.
        svg.selectAll(".link")
            .data(links)
          .enter().append("path")
            .attr("class", "link")
            .attr("d", d3.svg.diagonal().projection(function(d) { return [o.x(d), o.y(d)]; }));

        // Create the node circles.
        svg.selectAll(".node")
            .data(nodes)
         .enter().append("circle")
            .attr("r",1.5)
            .attr("x", o.x)
            .attr("y", o.y)

      });
    });

</script>
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