我正在尝试使用this example这样的条形图来创建旋转地球仪。您可以看到我的示例here。一切顺利,直到酒吧超越了地平线。我不知道当行星在另一侧时,如何从底部切下条形。有人可以建议我怎么做吗?
/*
* Original code source
* http://codepen.io/teetteet/pen/Dgvfw
*/
var width = 400;
var height = 400;
var scrollSpeed = 50;
var current = 180;
var longitudeScale = d3.scale.linear()
.domain([0, width])
.range([-180, 180]);
var planetProjection = d3.geo.orthographic()
.scale(200)
.rotate([longitudeScale(current), 0])
.translate([width / 2, height / 2])
.clipAngle(90);
var barProjection = d3.geo.orthographic()
.scale(200)
.rotate([longitudeScale(current), 0])
.translate([width / 2, height / 2])
.clipAngle(90);
var path = d3.geo.path()
.projection(planetProjection);
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height);
d3.json("https://dl.dropboxusercontent.com/s/4hp49mvf7pa2cg2/world-110m.json?dl=1", function(error, world) {
if (error) throw error;
var planet = svg.append("path")
.datum(topojson.feature(world, world.objects.land))
.attr("class", "land")
.attr("d", path);
d3.csv("https://dl.dropboxusercontent.com/s/v4kn2hrnjlgx1np/data.csv?dl=1", function(error, data) {
if (error) throw error;
var max = d3.max(data, function(d) {
return parseInt(d.Value);
})
var lengthScale = d3.scale.linear()
.domain([0, max])
.range([200, 250])
var bars = svg.selectAll(".bar")
.data(data)
.enter()
.append("line")
.attr("class", "bar")
.attr("stroke", "red")
.attr("stroke-width", "2");
function bgscroll() {
current += 1;
planetProjection.rotate([longitudeScale(current), 0]);
barProjection.rotate([longitudeScale(current), 0]);
planet.attr("d", path);
bars.attr("x1", function(d) {
return planetProjection([d.Longitude, d.Latitude])[0];
}).attr("y1", function(d) {
return planetProjection([d.Longitude, d.Latitude])[1];
}).attr("x2", function(d) {
barProjection.scale(lengthScale(d.Value));
return barProjection([d.Longitude, d.Latitude])[0];
}).attr("y2", function(d) {
barProjection.scale(lengthScale(d.Value));
return barProjection([d.Longitude, d.Latitude])[1];
});
}
// bgscroll();
setInterval(bgscroll, scrollSpeed);
})
})
最佳答案
为了剪掉地平线上的条形图,我们添加了一个以Globe 2D中心为中心并具有其半径的蒙版。然后,仅当底边越过地平线时才应用此蒙版(通过跟踪经度)。
创建 mask
// get the center of the circle
var center = planetProjection.translate();
// edge point
var edge = planetProjection([-90, 90])
// radius
var r = Math.pow(Math.pow(center[0] - edge[0], 2) + Math.pow(center[1] - edge[1], 2), 0.5);
svg.append("defs")
.append("clipPath")
.append("circle")
.attr("id", "edgeCircle")
.attr("cx", center[0])
.attr("cy", center[1])
.attr("r", r)
var mask = svg.append("mask").attr("id", "edge")
mask.append("rect")
.attr("x", 0)
.attr("y", 0)
.attr("width", "100%")
.attr("height", "100%")
.attr("fill", "white");
mask.append("use")
.attr("xlink:href", "#edgeCircle")
.attr("fill", "black");
涂上口罩
.... bars ....
.attr("mask", function (d) {
// make the range from 0 to 360, so that it's easier to compare
var longitude = Number(d.Longitude) + 180;
// +270 => -90 => the position of the left edge when the center is at 0
// -value because a rotation to the right => left edge longitude is reducing
// 360 because we want the range from 0 to 360
var startLongitude = 360 - ((longitudeScale(current) + 270) % 360);
// the right edge is start edge + 180
var endLongitude = (startLongitude + 180) % 360;
if ((startLongitude < endLongitude && longitude > startLongitude && longitude < endLongitude) ||
// wrap around
(startLongitude > endLongitude && (longitude > startLongitude || longitude < endLongitude)))
return null;
else
return "url(#edge)";
});
我们也可以通过测量距离来做到这一点。
fiddle -http://jsfiddle.net/gp3wvm8o/
关于d3.js - d3.js。旋转地球仪酒吧,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/31479802/