문제

I am experimenting with force directed layout using D3.js. What I need is center the layout by root (or other selected node) and return this node to the svg (e.g. canvas) center after the tick function is done (the graph alpha is low). Is it possible? I found an example at

http://bl.ocks.org/1080941

but I am unable to make the root (when using aplha or other custom tick function calculation) return back to the center (center the layout by this particular node).

Any help would be appreciated.

도움이 되었습니까?

해결책

Actually I solved this like this(similar to previous but more sophisticated):

force.on("tick", function(e) {

     node.attr("transform", function(d) { 
         //TODO move these constants to the header section
        //center the center (root) node when graph is cooling down
        if(d.index==0){
            damper = 0.1;
            d.x = d.x + (w/2 - d.x) * (damper + 0.71) * e.alpha;
            d.y = d.y + (h/2 - d.y) * (damper + 0.71) * e.alpha;
        }
        //start is initiated when importing nodes from XML
        if(d.start === true){
            d.x = w/2;
            d.y = h/2;
            d.start = false;
        }

        r = d.name.length;
        //these setting are used for bounding box, see [http://blockses.appspot.com/1129492][1]
        d.x = Math.max(r, Math.min(w - r, d.x));
        d.y = Math.max(r, Math.min(h - r, d.y));

            return "translate("+d.x+","+d.y+")";            

     }
    );

     });   

다른 팁

Try to change the force event handler like this:

force.on("tick", function(e) {
nodes[0].x = w / 2;
nodes[0].y = h / 2;

var k = 0.05 * e.alpha;
          nodes.forEach(function(o, i) {
            o.y += (nodes[0].y - o.y) * k;
            o.x += (nodes[0].x - o.x) * k;
          });

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; });

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