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Match Arrowhead Color to Line Color in D3

I am trying to do the obvious thing of getting the arrowhead colors of my directed graph's links to match the edge colors. Surprisingly I have not found a complete solution for doing this, although this older post seems like an excellent starting point. I would be fine with adapting that solution to work as outlined below, or if there is a superior method for creating arrowheads that achieves this effect I would be most thankful.

First, I have a linear gradient color function to color my edges by property like this:

var gradientColor = d3.scale.linear().domain([0,1]).range(["#08519c","#bdd7e7"]);

Then, like that previous post I have a function for adding markers:

function marker (color) {
  var reference;
  svg.append("svg:defs").selectAll("marker")
    .data([reference]) 
   .enter().append("svg:marker")    
    .attr("id", String)
    .attr("viewBox", "0 -5 10 10")
    .attr("refX", 15)  // This sets how far back it sits, kinda
    .attr("refY", 0)
    .attr("markerWidth", 9)
    .attr("markerHeight", 9)
    .attr("orient", "auto")
    .attr("markerUnits", "userSpaceOnUse")
   .append("svg:path")
    .attr("d", "M0,-5L10,0L0,5")
    .style("fill", color);
  return "url(#" + reference + ")";    };

And then the links definition I have is this one based on the Curved Links example.

var link = svg.selectAll(".link")
  .data(bilinks)
  .enter().append("path")
  .attr("class", "link")
  .style("fill", "none")
  .style("opacity", "0.5")    
  .style("stroke-width", "2")  
  .style("stroke", function(d) { return gradientColor(d[3]); } )
  .attr("marker-end", marker( "#FFCC33" ) );

This DOES NOT work as written; the browser gives me an "Uncaught TypeError: Cannot read property '5' of undefined" (where 'd[5]' refers to the fifth property in a list of properties that the links have). The problem is clearly passing the data function to the marker function in this case. If I feed in a static color like "#FFCC33" then the arrowheads DO change color (now). Unfortunately the person who posted this "marker function" solution 1.5 years ago didn't include the bit about passing the color to the marker function at all.

I don't know how to feed in the link's color properly. Ideally I would be able to use a reference to the color of the link that the arrowhead is attached to rather than inputting the same color function (because eventually I'm going to be coloring the links via different schemes based on button presses).

I've created a JS Fiddle that includes all the necessary bits to see and solve the problem. Currently I'm passing a static color to the markers, but it should be whatever is the color of the link it is attached to. I've also included features for another question on properly positioning the arrowheads and edge tails.

like image 800
Aaron Bramson Avatar asked Oct 06 '15 07:10

Aaron Bramson


1 Answers

I don't believe you're able to define a single SVG marker and change it's colour. Instead you need to define the marker many times (1 for each colour that you need to use). There's a nice example that recently popped up onto the D3 website.

enter image description here

The way this works, is by having lots if different markers, each defining the colour of the marker. Here's a screenshot of all the markers that are defined:

enter image description here

Then this particular example, cycles the CSS classes on the paths. The particular colored marker that each path is using is defined within the CSS class that's being applied to a path at any given time.

I've modified your example to add a new marker per path (and changed the colors slightly in the gradient to prove that it's working). Here's what I've got:

var width = 960,
    height = 500;

var color = d3.scale.category20();
var gradientColor = d3.scale.linear().domain([0, 15]).range(["#ff0000", "#0000ff"]);

var force = d3.layout.force()
    .linkDistance(10)
    .linkStrength(2)
    .size([width, height]);

var svg = d3.select("body").append("svg")
    .attr("width", width)
    .attr("height", height);

var defs = svg.append("svg:defs");

d3.json("http://bost.ocks.org/mike/miserables/miserables.json", function (error, graph) {
    if (error) throw error;


    function marker(color) {

        defs.append("svg:marker")
            .attr("id", color.replace("#", ""))
            .attr("viewBox", "0 -5 10 10")
            .attr("refX", 15) // This sets how far back it sits, kinda
            .attr("refY", 0)
            .attr("markerWidth", 9)
            .attr("markerHeight", 9)
            .attr("orient", "auto")
            .attr("markerUnits", "userSpaceOnUse")
            .append("svg:path")
            .attr("d", "M0,-5L10,0L0,5")
            .style("fill", color);
        
        return "url(" + color + ")";
    };

    var nodes = graph.nodes.slice(),
        links = [],
        bilinks = [];

    graph.links.forEach(function (link) {
        var s = nodes[link.source],
            t = nodes[link.target],
            i = {}, // intermediate node
            linkValue = link.value // for transfering value from the links to the bilinks
            ;
        nodes.push(i);
        links.push({
            source: s,
            target: i
        }, {
            source: i,
            target: t
        });
        bilinks.push([s, i, t, linkValue]);
    });

    force.nodes(nodes)
        .links(links)
        .start();

    var link = svg.selectAll(".link")
        .data(bilinks).enter().append("path")
        .attr("class", "link")
        .style("fill", "none")
        .style("opacity", "0.5")
        .style("stroke-width", "2")
        .each(function(d) {
            var color = gradientColor(d[3]);
            console.log(d[3]);
            d3.select(this).style("stroke", color)
                           .attr("marker-end", marker(color));
        });

    var node = svg.selectAll(".node")
        .data(graph.nodes)
        .enter().append("g")
        .attr("class", "node")
        .call(force.drag);

    node.append("circle")
        .attr("r", function (d) {
        return 2 + d.group;
    })
        .style("opacity", 0.5)
        .style("fill", function (d) {
        return color(d.group);
    });

    node.append("title")
        .text(function (d) {
        return d.name;
    });

    force.on("tick", function () {
        link.attr("d", function (d) {
            return "M" + d[0].x + "," + d[0].y + "S" + d[1].x + "," + d[1].y + " " + d[2].x + "," + d[2].y;
        });
        node.attr("transform", function (d) {
            return "translate(" + d.x + "," + d.y + ")";
        });
    });
    });
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
like image 67
Ian Avatar answered Oct 02 '22 21:10

Ian