D3力布图绘制--节点间的多条关系连接线的方法

在项目中遇到这样的场景，在使用D3.js绘制力布图的过程中，需要在2个节点间绘制多条连接线，找到一个不错的算法，在此分享下。
效果图：

HTML中要连接

    // 下面是JS部分的代码，使用前请连接`https://d3js.org/d3.v3.min.js` 和 `https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.5.0/lodash.min.js`
    var nodes = [{}, {}, {}, {}, {}, {}];
    var links = [      
        // one link
        { source: 0, target: 1 },
      
        // two links
        { source: 2, target: 1 }, 
        { source: 1, target: 2 },
      
        //three links
        { source: 3, target: 2 }, 
        { source: 2, target: 3 }, 
        { source: 2, target: 3 },
      
        // four links
        { source: 3, target: 4 }, 
        { source: 3, target: 4 }, 
        { source: 3, target: 4 }, 
        { source: 3, target: 4 },
      
        // five links
        { source: 4, target: 5 }, 
        { source: 4, target: 5 }, 
        { source: 4, target: 5 }, 
        { source: 4, target: 5 }, 
        { source: 4, target: 5 }
    ];

    // DATA FORMATTING
    _.each(links, function(link) {
        var same = _.where(links, {
            'source': link.source,
            'target': link.target
        });
        var sameAlt = _.where(links, {
            'source': link.target,
            'target': link.source
        });
        var sameAll = same.concat(sameAlt);

        _.each(sameAll, function(s, i) {
            s.sameIndex = (i + 1);
            s.sameTotal = sameAll.length;
            s.sameTotalHalf = (s.sameTotal / 2);
            s.sameUneven = ((s.sameTotal % 2) !== 0);
            s.sameMiddleLink = ((s.sameUneven === true) && (Math.ceil(s.sameTotalHalf) === s.sameIndex));
            s.sameLowerHalf = (s.sameIndex <= s.sameTotalHalf);
            s.sameArcDirection = s.sameLowerHalf ? 0 : 1;
            s.sameIndexCorrected = s.sameLowerHalf ? s.sameIndex : (s.sameIndex - Math.ceil(s.sameTotalHalf));
        });

        let sameStandard = sameAll[0];
        let sourceStandard = sameStandard.source;
        let targetStandard = sameStandard.target;
        _.each(sameAll,function(s,i){
          if(s.source === targetStandard && s.target === sourceStandard && s.sameTotal > 1){
            s.sameArcDirection = s.sameArcDirection === 0 ? 1 : 0
          }
        })
    });  

    var maxSame = _.chain(links)
        .sortBy(function(x) {
            return x.sameTotal;
        })
        .last()
        .value().sameTotal;

    _.each(links, function(link) {
        link.maxSameHalf = Math.floor(maxSame / 2);
    });    

    var width = 960,
        height = 500;
    var force = d3.layout.force()
        .nodes(nodes)
        .links(links)
        .size([width, height])
        .linkDistance(100)
        .charge(-200)
        .on('tick', tick)
        .start();

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

    var path = svg.append("g").selectAll("path")
        .data(force.links())
        .enter().append("path")
        .style("stroke", function(d) {
            return d3.scale.category20().range()[d.sameIndex - 1];
        });

    var circle = svg.append("g").selectAll("circle")
        .data(force.nodes())
        .enter().append("circle")
        .attr("r", 8)
        .call(force.drag);

    function tick(d) {
        circle.attr("transform", function(d) {
            return "translate(" + d.x + "," + d.y + ")";
        });
        path.attr("d", linkArc);
    }

    function linkArc(d) {
        var dx = (d.target.x - d.source.x),
            dy = (d.target.y - d.source.y),
            dr = Math.sqrt(dx * dx + dy * dy),
            unevenCorrection = (d.sameUneven ? 0 : 0.5),
            arc = ((dr * d.maxSameHalf) / (d.sameIndexCorrected - unevenCorrection));

        if (d.sameMiddleLink) {
            arc = 0;
        }

        return "M" + d.source.x + "," + d.source.y + "A" + arc + "," + arc + " 0 0," + d.sameArcDirection + " " + d.target.x + "," + d.target.y;
    }

本文转自： http://bl.ocks.org/thomasdobber/9b78824119136778052f64a967c070e0