我有四个饼图,它们代表使用d3JS库制作的不同数据。它们的制作正确,但是当我包含其图例时,上一个图表的图例中的元素也将包含在下一个图表中。因此,在最后一个图表的图例中,所有先前图表的图例的元素都存在。
代码是这样的:
/*var divSkills, divKnows, divEnvs, divTools;
var dataSkills = [
{ label: 'Web Designing', count: 86 },
{ label: 'Web Development / Web Programming', count: 90 },
{ label: 'Graphic Designing / Logo Designing', count: 65 },
{ label: 'Programming', count: 73 },
{ label: 'Networking', count: 63 },
{ label: 'Content Writing', count: 56 }
];
var dataKnows = [
{ label: 'PHP', count: 75 },
{ label: 'SQL', count: 59 },
{ label: 'HTML 5', count: 90 },
{ label: 'CSS 3', count: 93 },
{ label: 'JavaScript', count: 80 },
{ label: 'C', count: 75 }
];
var dataEnvs = [
{ label: 'Adobe Dreamweaver', count: 75 },
{ label: 'SublimeText', count: 80 },
{ label: 'Notepad++', count: 80 },
{ label: 'Cloud9 (Online IDE)', count: 80 }
];
var dataTools = [
{ label: 'Adobe Photoshop', count: 65 },
{ label: 'Adobe Illustrator', count: 30 },
{ label: 'Adobe Fireworks', count: 50 },
{ label: 'AutoDesk Maya', count: 40 },
{ label: 'Git / GitHub / GitLab (Control Version System)', count: 70 }
];
var dataset = [dataSkills, dataKnows, dataEnvs, dataTools];
var width = 360;
var height = 360;
var radius = Math.min(width, height) / 2;
var color = d3.scale.category20b();
var legendRectSize = 18;
var legendSpacing = 4;
$(document).ready(function () {
divSkills = d3.select(".view-graph section:first-child .graph"),
divKnows = d3.select(".view-graph section:nth-child(2) .graph"),
divEnvs = d3.select(".view-graph section:nth-child(3) .graph"),
divTools = d3.select(".view-graph section:last-child .graph");
var divs = [divSkills, divKnows, divEnvs, divTools];
var arc = d3.svg.arc()
.outerRadius(radius);
var pie = d3.layout.pie()
.value(function(d) { return d.count; })
.sort(null);*/
for (var i = 0; i < divs.length; i++) {
var svg = divs[i]
.append('svg')
.attr('width', width)
.attr('height', height)
.append('g')
.attr('transform', 'translate(' + (width / 2) + ',' + (height / 2) + ')');
var path = svg.selectAll('path')
.data(pie( dataset[i]) )
.enter()
.append('path')
.attr('d', arc)
.attr('fill', function(d, i) {
return color(d.data.label);
});
var legend = svg.selectAll('.legend')
.data(color.domain())
.enter()
.append('g')
.attr('class', 'legend')
.attr('transform', function(d, j) {
var height = legendRectSize + legendSpacing;
var offset = height * color.domain().length / 2;
var horz = 15 * legendRectSize;
var vert = j * height - offset;
return 'translate(' + horz + ',' + vert + ')';
});
legend.append('rect')
.attr('width', legendRectSize)
.attr('height', legendRectSize)
.style('fill', color)
.style('stroke', color);
legend.append('text')
.attr('x', legendRectSize + legendSpacing)
.attr('y', legendRectSize - legendSpacing)
.text(function(d) { return d; });
}
});
如您所见,我正在使用for
循环来避免重复。我试图通过在代码中进行几个变体来修复它,但是没有一个起作用。这是笔。现在,我有两个问题:
大多数代码都来自本教程。
编辑:注释掉不相关的代码。
您的问题是对进行数据绑定color.domain()
。正如您color(d.data.label)
在每次迭代中所做的那样,域会不断增长。扔一个
console.log(color.domain());
在构建每个图例之前。你会看到的:
["Web Designing", "Web Development / Web Programming", "Graphic Designing / Logo Designing", "Programming", "Networking", "Content Writing"]
["Web Designing", "Web Development / Web Programming", "Graphic Designing / Logo Designing", "Programming", "Networking", "Content Writing", "PHP", "SQL", "HTML 5", "CSS 3", "JavaScript", "C"]
["Web Designing", "Web Development / Web Programming", "Graphic Designing / Logo Designing", "Programming", "Networking", "Content Writing", "PHP", "SQL", "HTML 5", "CSS 3", "JavaScript", "C", "Adobe Dreamweaver", "SublimeText", "Notepad++", "Cloud9 (Online IDE)"]
["Web Designing", "Web Development / Web Programming", "Graphic Designing / Logo Designing", "Programming", "Networking", "Content Writing", "PHP", "SQL", "HTML 5", "CSS 3", "JavaScript", "C", "Adobe Dreamweaver", "SublimeText", "Notepad++", "Cloud9 (Online IDE)", "Adobe Photoshop", "Adobe Illustrator", "Adobe Fireworks", "AutoDesk Maya", "Git / GitHub / GitLab (Control Version System)"]
一个简单的解决方法是将数据绑定到:
var pieLabels = dataset[i].map(function(d){
return d.label;
});
var legend = svg.selectAll('.legend')
.data(pieLabels);
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