本文介绍了如何实现D3比例以让孩子从毕业的父母那里继承色彩?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有一棵D3.js树，该树具有应用于节点和链接的不同颜色.
I have a D3.js tree that have different colours applied to nodes and links.
着色是硬编码的:
nodeUpdate.select("circle")
.attr("r", 10)
.style("fill", function(d) {
if(d.name == "Top Level") {
return d._children ? "blue" : "#fff";
}
if(d.name == "Second A") {
return d._children ? "red" : "#fff";
}
if(d.name == "Second B") {
return d._children ? "green" : "#fff";
}
if(d.name == "Second C") {
return d._children ? "purple" : "#fff";
}
if(d.name == "Second D") {
return d._children ? "gold" : "#fff";
}
})
.style("stroke", function(d) {
if(d.name == "Top Level") {
return "blue";
}
if(d.name == "Second A") {
return "red";
}
if(d.name == "Second B") {
return "green";
}
if(d.name == "Second C") {
return "purple";
}
if(d.name == "Second D") {
return "gold";
}
});
和
link.enter().insert("path", "g")
.attr("class", "link")
.attr("stroke-width", function(d){
return 1;
})
.attr("d", function(d) {
var o = {x: source.x0, y: source.y0};
return diagonal({source: o, target: o});
})
.style("stroke", function(d) {
return linkColor(d.target.name);
});
和
function linkColor(node_name) {
switch (node_name)
{
case 'Second A': case 'Third A': case 'Third B':
return 'red';
break;
case 'Second B': case 'Third C': case 'Third D':
return 'green';
break;
case 'Second C': case 'Third E': case 'Third F':
return 'purple';
break;
case 'Second D': case 'Third G': case 'Third H':
return 'gold';
}
}
请参见小提琴
我不是要对每个节点和链接的颜色进行硬编码，而是要实现 d3-scale 为节点和线着色.该解决方案将让孩子沿着在给定x个孩子的情况下，自动对该颜色进行分级.
Rather than hardcoding the colour of each node and link, I want to implement d3-scale to colour the nodes and lines. The solution would have the children 'inherit' their parent's colour along a graduation of that colour automatically given x number of children.
https://www.d3indepth.com/scales/
我该如何实现?
一如既往，感谢您的帮助.
As always, thanks for your help.
推荐答案
您需要的标度是一个序数标度，如下所示:
The scale you need is an ordinal scale, like this:
var colourScale = d3.scale.ordinal()
.domain(["Top Level","Second A", "Second B", "Second C", "Second D"])
.range(["blue","red", "green", "purple", "gold"]);
然后，您可以更改整个内容...
Then, you can change this whole thing...
.style("fill", function(d) {
if (d.name == "Top Level") {
return d._children ? "blue" : "#fff";
}
if (d.name == "Second A") {
return d._children ? "red" : "#fff";
}
if (d.name == "Second B") {
return d._children ? "green" : "#fff";
}
if (d.name == "Second C") {
return d._children ? "purple" : "#fff";
}
if (d.name == "Second D") {
return d._children ? "gold" : "#fff";
}
})
...变成:
.style("fill", function(d) {
return colourScale(d.name)
})
并摆脱该linkColor函数.
请注意，尽管您已经链接了D3 v5文档，但是您正在使用D3 v3.
Pay attention to the fact that, despite you have linked D3 v5 documentation, you're using D3 v3.
这是您所做的更改的代码:
Here is your code with that change:
var treeData = [{
"name": "Top Level",
"children": [{
"name": "Second A",
"children": [{
"name": "Third A"
}, {
"name": "Third B"
}]
}, {
"name": "Second B",
"children": [{
"name": "Third C"
}, {
"name": "Third D"
}]
}, {
"name": "Second C",
"children": [{
"name": "Third E"
}, {
"name": "Third F"
}]
}, {
"name": "Second D",
"children": [{
"name": "Third G"
}, {
"name": "Third H"
}, ]
}, ]
}];
var colourScale = d3.scale.ordinal()
.domain(["Top Level", "Second A", "Second B", "Second C", "Second D"])
.range(["blue", "red", "green", "purple", "gold"]);
// ************** Generate the tree diagram *****************
var margin = {
top: 20,
right: 120,
bottom: 20,
left: 120
},
width = 960 - margin.right - margin.left,
height = 500 - margin.top - margin.bottom;
var i = 0,
duration = 750,
root;
var tree = d3.layout.tree()
.size([height, width]);
var diagonal = d3.svg.diagonal()
.projection(function(d) {
return [d.y, d.x];
});
var svg = d3.select("body").append("svg")
.attr("width", width + margin.right + margin.left)
.attr("height", height + margin.top + margin.bottom)
.append("g")
.attr("transform", "translate(" + margin.left + "," + margin.top + ")");
root = treeData[0];
root.x0 = height / 2;
root.y0 = 0;
update(root);
d3.select(self.frameElement).style("height", "500px");
// Collapse after the second level
root.children.forEach(collapse);
update(root);
// Collapse the node and all it's children
function collapse(d) {
if (d.children) {
d._children = d.children
d._children.forEach(collapse)
d.children = null
}
}
function update(source) {
// Compute the new tree layout.
var nodes = tree.nodes(root).reverse(),
links = tree.links(nodes);
// Normalize for fixed-depth.
nodes.forEach(function(d) {
d.y = d.depth * 180;
});
// Update the nodes…
var node = svg.selectAll("g.node")
.data(nodes, function(d) {
return d.id || (d.id = ++i);
});
// Enter any new nodes at the parent's previous position.
var nodeEnter = node.enter().append("g")
.attr("class", "node")
.attr("transform", function(d) {
return "translate(" + source.y0 + "," + source.x0 + ")";
})
.on("click", click);
nodeEnter.append("circle")
.attr("r", 1e-6)
.style("fill", function(d) {
return d._children ? "#C0C0C0" : "#fff";
});
nodeEnter.append("text")
.attr("x", function(d) {
return d.children || d._children ? -13 : 13;
})
.attr("dy", ".35em")
.attr("text-anchor", function(d) {
return d.children || d._children ? "end" : "start";
})
.text(function(d) {
return d.name;
})
.style("fill-opacity", 1e-6);
// Transition nodes to their new position.
var nodeUpdate = node.transition()
.duration(duration)
.attr("transform", function(d) {
return "translate(" + d.y + "," + d.x + ")";
});
nodeUpdate.select("circle")
.attr("r", 10)
.style("fill", function(d) {
return d.depth === 2 ? colourScale(d.parent.name) : colourScale(d.name);
})
.style("stroke", function(d) {
return d.depth === 2 ? colourScale(d.parent.name) : colourScale(d.name);
});
nodeUpdate.select("text")
.style("fill-opacity", 1);
// Transition exiting nodes to the parent's new position.
var nodeExit = node.exit().transition()
.duration(duration)
.attr("transform", function(d) {
return "translate(" + source.y + "," + source.x + ")";
})
.remove();
nodeExit.select("circle")
.attr("r", 1e-6);
nodeExit.select("text")
.style("fill-opacity", 1e-6);
// Update the links…
var link = svg.selectAll("path.link")
.data(links, function(d) {
return d.target.id;
});
// Enter any new links at the parent's previous position.
link.enter().insert("path", "g")
.attr("class", "link")
.attr("stroke-width", function(d) {
return 1;
})
.attr("d", function(d) {
var o = {
x: source.x0,
y: source.y0
};
return diagonal({
source: o,
target: o
});
})
.style("stroke", function(d) {
return d.target.depth === 2 ? colourScale(d.target.parent.name) : colourScale(d.target.name);
});
// Transition links to their new position.
link.transition()
.duration(duration)
.attr("d", diagonal);
// Transition exiting nodes to the parent's new position.
link.exit().transition()
.duration(duration)
.attr("d", function(d) {
var o = {
x: source.x,
y: source.y
};
return diagonal({
source: o,
target: o
});
})
.remove();
// Stash the old positions for transition.
nodes.forEach(function(d) {
d.x0 = d.x;
d.y0 = d.y;
});
}
// Toggle children on click.
function click(d) {
if (d.children) {
d._children = d.children;
d.children = null;
} else {
d.children = d._children;
d._children = null;
}
update(d);
}.node {
cursor: pointer;
}
.node circle {
fill: #fff;
stroke: #C0C0C0;
stroke-width: 1.5px;
}
.node text {
font: 10px sans-serif;
}
.link {
fill: none;
stroke: #C0C0C0;
stroke-width: 1.5px;
}<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.5.17/d3.min.js"></script>这篇关于如何实现D3比例以让孩子从毕业的父母那里继承色彩?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！