D3力布图绘制--节点间的多条关系连接线的方法（转）
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
<!DOCTYPE html>
<head>
<meta charset="utf-8">
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.5.0/lodash.min.js"></script>
<script src="https://d3js.org/d3.v3.min.js"></script>
</head>
<body>
</body>
// 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
