1
Q
D3樹方向
A
回答
1
建立此example。
// Compute the new tree layout.
var nodes = tree.nodes(root).reverse(),
links = tree.links(nodes);
// figure out the placement of the "top-most" node at each depth
var nodeMap = {};
nodes.forEach(function(d) {
if (!nodeMap[d.depth] || d.x < nodeMap[d.depth]){
nodeMap[d.depth] = d.x;
}
});
// shift all nodes up
nodes.forEach(function(d) {
d.y = d.depth * 100;
d.x -= nodeMap[d.depth];
});
這產生：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Tree Example</title>
<style>
\t
\t .node {
\t \t cursor: pointer;
\t }
\t .node circle {
\t fill: #fff;
\t stroke: steelblue;
\t stroke-width: 3px;
\t }
\t .node text {
\t font: 12px sans-serif;
\t }
\t .link {
\t fill: none;
\t stroke: #ccc;
\t stroke-width: 2px;
\t }
\t
</style>
</head>
<body>
<!-- load the d3.js library --> \t
<script src="http://d3js.org/d3.v3.min.js"></script>
\t
<script>
var treeData = [
{
"name": "Top Level",
"parent": "null",
"children": [
{
"name": "Level 2: A",
"parent": "Top Level",
"children": [
{
"name": "Son of A",
"parent": "Level 2: A"
},
{
"name": "Daughter of A",
"parent": "Level 2: A"
}
]
},
{
"name": "Level 2: B",
"parent": "Top Level",
"children": [
{
"name": "Son of B",
"parent": "Level 2: B"
},
{
"name": "Daughter of B",
"parent": "Level 2: B"
}
]
}
]
}
];
// ************** Generate the tree diagram \t *****************
var margin = {top: 20, right: 120, bottom: 20, left: 120},
\t width = 960 - margin.right - margin.left,
\t height = 500 - margin.top - margin.bottom;
\t
var i = 0,
\t duration = 750,
\t root;
var tree = d3.layout.tree()
\t .size([height, width]);
var diagonal = d3.svg.diagonal()
\t .projection(function(d) { return [d.y, d.x]; });
var svg = d3.select("body").append("svg")
\t .attr("width", width + margin.right + margin.left)
\t .attr("height", height + margin.top + margin.bottom)
.append("g")
\t .attr("transform", "translate(" + margin.left + "," + margin.top + ")");
root = treeData[0];
root.x0 = height/2;
root.y0 = 0;
update(root);
function update(source) {
// Compute the new tree layout.
var nodes = tree.nodes(root).reverse(),
\t links = tree.links(nodes);
// Normalize for fixed-depth.
var nodeMap = {};
nodes.forEach(function(d) {
if (!nodeMap[d.depth] || d.x < nodeMap[d.depth]){
nodeMap[d.depth] = d.x;
}
});
nodes.forEach(function(d) {
d.y = d.depth * 100;
d.x -= nodeMap[d.depth];
});
// Update the nodes…
var node = svg.selectAll("g.node")
\t .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")
\t .attr("class", "node")
\t .attr("transform", function(d) { return "translate(" + source.y0 + "," + source.x0 + ")"; })
\t .on("click", click);
nodeEnter.append("circle")
\t .attr("r", 1e-6)
\t .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; });
nodeEnter.append("text")
\t .attr("x", function(d) { return d.children || d._children ? -13 : 13; })
\t .attr("dy", ".35em")
\t .attr("text-anchor", function(d) { return d.children || d._children ? "end" : "start"; })
\t .text(function(d) { return d.name; })
\t .style("fill-opacity", 1e-6);
// Transition nodes to their new position.
var nodeUpdate = node.transition()
\t .duration(duration)
\t .attr("transform", function(d) { return "translate(" + d.y + "," + d.x + ")"; });
nodeUpdate.select("circle")
\t .attr("r", 10)
\t .style("fill", function(d) { return d._children ? "lightsteelblue" : "#fff"; });
nodeUpdate.select("text")
\t .style("fill-opacity", 1);
// Transition exiting nodes to the parent's new position.
var nodeExit = node.exit().transition()
\t .duration(duration)
\t .attr("transform", function(d) { return "translate(" + source.y + "," + source.x + ")"; })
\t .remove();
nodeExit.select("circle")
\t .attr("r", 1e-6);
nodeExit.select("text")
\t .style("fill-opacity", 1e-6);
// Update the links…
var link = svg.selectAll("path.link")
\t .data(links, function(d) { return d.target.id; });
// Enter any new links at the parent's previous position.
link.enter().insert("path", "g")
\t .attr("class", "link")
\t .attr("d", function(d) {
\t \t var o = {x: source.x0, y: source.y0};
\t \t return diagonal({source: o, target: o});
\t });
// Transition links to their new position.
link.transition()
\t .duration(duration)
\t .attr("d", diagonal);
// Transition exiting nodes to the parent's new position.
link.exit().transition()
\t .duration(duration)
\t .attr("d", function(d) {
\t \t var o = {x: source.x, y: source.y};
\t \t return diagonal({source: o, target: o});
\t })
\t .remove();
// Stash the old positions for transition.
nodes.forEach(function(d) {
\t d.x0 = d.x;
\t d.y0 = d.y;
});
}
// Toggle children on click.
function click(d) {
if (d.children) {
\t d._children = d.children;
\t d.children = null;
} else {
\t d.children = d._children;
\t d._children = null;
}
update(d);
}
</script>
\t
</body>
</html>
0
不僅節點（X，Y）應該可以互換，但也連接節點的鏈接，即
var diagonal = d3.svg.diagonal()
.projection(function(d) {return [d.y,d.x];});
+0
如果（x，y）的節點被互換，則樹似乎具有的（頂部中心）取向的層次結構而不是（左上方），如果我沒有弄錯的話，那也不是我所需要的。 – william
0
看看這個例子，我做
<!doctype html>
<html>
<head>
<script src="http://d3js.org/d3.v3.min.js"></script>
</head>
<body>
<script>
var data = {
"name" : "A",
"children" : [
{
"name" : "B",
"children" : [
{
"name" : "F"
},
{
"name" : "G"
}
]
},
{
"name" : "C",
"children" : [
{
"name" : "H"
},
{
"name" : "I"
},
{
"name" : "J"
}
]
},
{
"name" : "D",
"children" : [
{
"name" : "K"
},
{
"name" : "L"
}
]
},
{
"name" : "E"
}
]
};
var canvas = d3.select("body")
.append("svg")
.attr("width",600)
.attr("height",500)
.append("g")
.attr("transform","translate(50,50)");
var cluster = d3.layout.cluster()
.size([400,400]);
var nodes = cluster.nodes(data);
var links = cluster.links(nodes);
var node = canvas.selectAll(".node")
.data(nodes)
.enter()
.append("g")
.attr("class","node")
.attr("transform",function(d) {return "translate(" + d.y + "," + d.x + ")"});
node.append("circle")
.attr("r",5)
.attr("fill","silver");
node.append("text")
.text(function(d) {return d.name;});
var diagonal = d3.svg.diagonal()
.projection(function(d) {return [d.y,d.x];});
canvas.selectAll(".link")
.data(links)
.enter()
.append("path")
.attr("class","link")
.attr("fill","none")
.attr("stroke","#ADADAD")
.attr("d",diagonal);
</script>
+0
非常感謝你的例子，但我仍然看到最初的點位於佈局的左側中心，我真正需要的是從頂部中心開始。 – william
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太棒了！謝謝！ – william