Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
341 views
in Technique[技术] by (71.8m points)

javascript - How to layout a non-tree hierarchy with D3

D3 has a variety of layouts for directed graphs that are strict trees, such as the following:

A
|
B C
 / 
D   E

I need to draw a hierarchy of nodes that is not a tree, but is a directed acyclic graph. This is a problem for a tree layout, because several of the branches converge:

A
|
B C
 |
  D

Does anyone know of a D3 layout for general hierarchies? Or alternatively, some clever hack to the existing treelayout? I've noticed GraphVis handles this situation well, but D3 produces a graph that better suits the requirements here.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You could create your own code without having to rely on a D3 layout in order to get it done.

I've provided an example in a jsFiddle. The example is pretty simplistic and would need to be worked a little bit to accommodate more complex examples.

The example could be re-worked to process hierarchical data as well with relatively little effort.

Here is the code I have used in the jsFiddle:

 // Sample data set
var json = {
    nodes: [{
        name: 'A'},
    {
        name: 'B'},
    {
        name: 'C'},
    {
        name: 'D'}],
    links: [{
        source: 'A',
        target: 'B'},
    {
        source: 'A',
        target: 'C'},
    {
        source: 'B',
        target: 'D'},
    {
        source: 'C',
        target: 'D'}
                                                                                   ]

};

var vis = d3.select('#vis').attr('transform', 'translate(20, 20)');

// Build initial link elements - Build first so they are under the nodes
var links = vis.selectAll('line.link').data(json.links);
links.enter().append('line').attr('class', 'link').attr('stroke', '#000');

// Build initial node elements
var nodes = vis.selectAll('g.node').data(json.nodes);
nodes.enter().append('g').attr('class', 'node').append('circle').attr('r', 10).append('title').text(function(d) {
    return d.name;
});

// Store nodes in a hash by name
var nodesByName = {};
nodes.each(function(d) {
    nodesByName[d.name] = d;
});

// Convert link references to objects
links.each(function(link) {
    link.source = nodesByName[link.source];
    link.target = nodesByName[link.target];
    if (!link.source.links) {
        link.source.links = [];
    }
    link.source.links.push(link.target);
    if (!link.target.links) {
        link.target.links = [];
    }
    link.target.links.push(link.source);
});

// Compute positions based on distance from root
var setPosition = function(node, i, depth) {
    if (!depth) {
        depth = 0;
    }
    if (!node.x) {
        node.x = (i + 1) * 40;
        node.y = (depth + 1) * 40;
        if (depth <= 1) {
            node.links.each(function(d, i2) {
                setPosition(d, i2, depth + 1);
            });
        }

    }

};
nodes.each(setPosition);

// Update inserted elements with computed positions
nodes.attr('transform', function(d) {
    return 'translate(' + d.x + ', ' + d.y + ')';
});

links.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;
});

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...