D3.js Choropleth 使用 JSON 中的数据
D3.js Choropleth using data in JSON
我正在尝试制作类似于 Mike Bostock 的 Choropleth (https://bl.ocks.org/mbostock/4060606) 的地图。
<!DOCTYPE html>
<meta charset="utf-8">
<style>
.counties {
fill: none;
}
.states {
fill: none;
stroke: #fff;
stroke-linejoin: round;
}
</style>
<svg width="960" height="600"></svg>
<script src="https://d3js.org/d3.v4.min.js"></script>
<script src="https://d3js.org/d3-scale-chromatic.v1.min.js"></script>
<script src="https://d3js.org/topojson.v2.min.js"></script>
<script>
var svg = d3.select("svg"),
width = +svg.attr("width"),
height = +svg.attr("height");
var unemployment = d3.map();
var path = d3.geoPath();
var x = d3.scaleLinear()
.domain([1, 10])
.rangeRound([600, 860]);
var color = d3.scaleThreshold()
.domain(d3.range(2, 10))
.range(d3.schemeBlues[9]);
var g = svg.append("g")
.attr("class", "key")
.attr("transform", "translate(0,40)");
g.selectAll("rect")
.data(color.range().map(function(d) {
d = color.invertExtent(d);
if (d[0] == null) d[0] = x.domain()[0];
if (d[1] == null) d[1] = x.domain()[1];
return d;
}))
.enter().append("rect")
.attr("height", 8)
.attr("x", function(d) { return x(d[0]); })
.attr("width", function(d) { return x(d[1]) - x(d[0]); })
.attr("fill", function(d) { return color(d[0]); });
g.append("text")
.attr("class", "caption")
.attr("x", x.range()[0])
.attr("y", -6)
.attr("fill", "#000")
.attr("text-anchor", "start")
.attr("font-weight", "bold")
.text("Unemployment rate");
g.call(d3.axisBottom(x)
.tickSize(13)
.tickFormat(function(x, i) { return i ? x : x + "%"; })
.tickValues(color.domain()))
.select(".domain")
.remove();
d3.queue()
.defer(d3.json, "https://d3js.org/us-10m.v1.json")
.defer(d3.tsv, "unemployment.tsv", function(d) {
console.log(d);
unemployment.set(d.id, +d.rate); })
.await(ready);
function ready(error, us) {
if (error) throw error;
svg.append("g")
.attr("class", "counties")
.selectAll("path")
.data(topojson.feature(us, us.objects.counties).features)
.enter().append("path")
.attr("fill", function(d) { return color(d.rate = unemployment.get(d.id)); })
.attr("d", path)
.append("title")
.text(function(d) { return d.rate + "%"; });
svg.append("path")
.datum(topojson.mesh(us, us.objects.states, function(a, b) { return a !== b; }))
.attr("class", "states")
.attr("d", path);
}
</script>
我不想使用 TSV 文件,而是想对我的数据使用 JSON。我已经尝试弄清楚如何将脚本转换为使用 JSON,但没有成功。目前,我正在使用 JSON 作为示例结构:
[{"id": "01001", "rate": 5.1},
{"id": "01003", "rate": 4.9},
{"id": "01005", "rate": 8.6},
{"id": "01007", "rate": 6.2},
{"id": "01009", "rate": 5.1},
{"id": "01011", "rate": 7.1}]
最终,我想下拉到 select 要显示的内容 "column",甚至弄清楚如何过滤年份范围。我的最终 JSON 结构如下:
[{
"ID": "02",
"Volume": "NULL",
"SubCategory": "Other Housing",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Ohio"
},
{
"ID": "02",
"Volume": 24800000,
"SubCategory": "Student Loans",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Ohio"
},
{
"ID": "02",
"DateExcelNo": "12/31/14",
"Volume": 440200000,
"SubCategory": "Mort Rev Bonds",
"Category": "Use",
"Type": "PAB",
"date": "2014-12-31",
"State": "Ohio"
},
{
"ID": "02",
"Volume": 1205000000,
"SubCategory": "Total Cap",
"Category": "VolumeCapacity",
"Type": "PAB",
"date": "2014-12-31",
"State": "Ohio"
},
{ "ID": "01",
"Volume": "NULL",
"SubCategory": "Other Housing",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 24800000,
"SubCategory": "Student Loans",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"DateExcelNo": "12/31/14",
"Volume": 440200000,
"SubCategory": "Mort Rev Bonds",
"Category": "Use",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 1205000000,
"SubCategory": "Total Cap",
"Category": "VolumeCapacity",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 663400000,
"SubCategory": "Total Carryforward to Next Year",
"Category": "Carryfwd",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
}]
这里的问题很简单:d3.json
不接受行函数。
我相信你认为,鉴于 JSON 你在你的问题中分享,你可以简单地改变这一行...
.defer(d3.tsv, "unemployment.tsv", function(d) { unemployment.set(d.id, +d.rate); })
...为此:
.defer(d3.json, "unemployment.json", function(d) { unemployment.set(d.id, +d.rate); })
但是,这将不起作用,因为您不能将行函数(第二个逗号后的所有内容都是行函数)与 d3.json
.
一起使用
解决方法:
创建您的 JSON 文件(此处命名为 "unemployment.json")并在队列中执行此操作:
.defer(d3.json, "unemployment.json")
然后,在ready
函数中,加载JSON作为第三个参数(这里我称之为un
):
function ready(error, us, un) {
最后,填充您的 d3.map()
:
un.forEach(function(d) {
unemployment.set(d.id, +d.rate);
});
这是更新后的 bl.ocks(当然,只有 6 个县有颜色,因为我使用的 JSON 只有 6 个数据点):https://bl.ocks.org/anonymous/211bf9775e61c5b07935536ed1926225
PS:在 S.O 上每个问题只问 一个问题 是一种很好的做法。因此,我建议你post另一个关于下拉问题的问题。
我正在尝试制作类似于 Mike Bostock 的 Choropleth (https://bl.ocks.org/mbostock/4060606) 的地图。
<!DOCTYPE html>
<meta charset="utf-8">
<style>
.counties {
fill: none;
}
.states {
fill: none;
stroke: #fff;
stroke-linejoin: round;
}
</style>
<svg width="960" height="600"></svg>
<script src="https://d3js.org/d3.v4.min.js"></script>
<script src="https://d3js.org/d3-scale-chromatic.v1.min.js"></script>
<script src="https://d3js.org/topojson.v2.min.js"></script>
<script>
var svg = d3.select("svg"),
width = +svg.attr("width"),
height = +svg.attr("height");
var unemployment = d3.map();
var path = d3.geoPath();
var x = d3.scaleLinear()
.domain([1, 10])
.rangeRound([600, 860]);
var color = d3.scaleThreshold()
.domain(d3.range(2, 10))
.range(d3.schemeBlues[9]);
var g = svg.append("g")
.attr("class", "key")
.attr("transform", "translate(0,40)");
g.selectAll("rect")
.data(color.range().map(function(d) {
d = color.invertExtent(d);
if (d[0] == null) d[0] = x.domain()[0];
if (d[1] == null) d[1] = x.domain()[1];
return d;
}))
.enter().append("rect")
.attr("height", 8)
.attr("x", function(d) { return x(d[0]); })
.attr("width", function(d) { return x(d[1]) - x(d[0]); })
.attr("fill", function(d) { return color(d[0]); });
g.append("text")
.attr("class", "caption")
.attr("x", x.range()[0])
.attr("y", -6)
.attr("fill", "#000")
.attr("text-anchor", "start")
.attr("font-weight", "bold")
.text("Unemployment rate");
g.call(d3.axisBottom(x)
.tickSize(13)
.tickFormat(function(x, i) { return i ? x : x + "%"; })
.tickValues(color.domain()))
.select(".domain")
.remove();
d3.queue()
.defer(d3.json, "https://d3js.org/us-10m.v1.json")
.defer(d3.tsv, "unemployment.tsv", function(d) {
console.log(d);
unemployment.set(d.id, +d.rate); })
.await(ready);
function ready(error, us) {
if (error) throw error;
svg.append("g")
.attr("class", "counties")
.selectAll("path")
.data(topojson.feature(us, us.objects.counties).features)
.enter().append("path")
.attr("fill", function(d) { return color(d.rate = unemployment.get(d.id)); })
.attr("d", path)
.append("title")
.text(function(d) { return d.rate + "%"; });
svg.append("path")
.datum(topojson.mesh(us, us.objects.states, function(a, b) { return a !== b; }))
.attr("class", "states")
.attr("d", path);
}
</script>
我不想使用 TSV 文件,而是想对我的数据使用 JSON。我已经尝试弄清楚如何将脚本转换为使用 JSON,但没有成功。目前,我正在使用 JSON 作为示例结构:
[{"id": "01001", "rate": 5.1},
{"id": "01003", "rate": 4.9},
{"id": "01005", "rate": 8.6},
{"id": "01007", "rate": 6.2},
{"id": "01009", "rate": 5.1},
{"id": "01011", "rate": 7.1}]
最终,我想下拉到 select 要显示的内容 "column",甚至弄清楚如何过滤年份范围。我的最终 JSON 结构如下:
[{
"ID": "02",
"Volume": "NULL",
"SubCategory": "Other Housing",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Ohio"
},
{
"ID": "02",
"Volume": 24800000,
"SubCategory": "Student Loans",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Ohio"
},
{
"ID": "02",
"DateExcelNo": "12/31/14",
"Volume": 440200000,
"SubCategory": "Mort Rev Bonds",
"Category": "Use",
"Type": "PAB",
"date": "2014-12-31",
"State": "Ohio"
},
{
"ID": "02",
"Volume": 1205000000,
"SubCategory": "Total Cap",
"Category": "VolumeCapacity",
"Type": "PAB",
"date": "2014-12-31",
"State": "Ohio"
},
{ "ID": "01",
"Volume": "NULL",
"SubCategory": "Other Housing",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 24800000,
"SubCategory": "Student Loans",
"Category": "Use",
"Type": "PAB",
"date": "2013-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"DateExcelNo": "12/31/14",
"Volume": 440200000,
"SubCategory": "Mort Rev Bonds",
"Category": "Use",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 1205000000,
"SubCategory": "Total Cap",
"Category": "VolumeCapacity",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
},
{
"ID": "01",
"Volume": 663400000,
"SubCategory": "Total Carryforward to Next Year",
"Category": "Carryfwd",
"Type": "PAB",
"date": "2014-12-31",
"State": "Connecticut"
}]
这里的问题很简单:d3.json
不接受行函数。
我相信你认为,鉴于 JSON 你在你的问题中分享,你可以简单地改变这一行...
.defer(d3.tsv, "unemployment.tsv", function(d) { unemployment.set(d.id, +d.rate); })
...为此:
.defer(d3.json, "unemployment.json", function(d) { unemployment.set(d.id, +d.rate); })
但是,这将不起作用,因为您不能将行函数(第二个逗号后的所有内容都是行函数)与 d3.json
.
解决方法:
创建您的 JSON 文件(此处命名为 "unemployment.json")并在队列中执行此操作:
.defer(d3.json, "unemployment.json")
然后,在ready
函数中,加载JSON作为第三个参数(这里我称之为un
):
function ready(error, us, un) {
最后,填充您的 d3.map()
:
un.forEach(function(d) {
unemployment.set(d.id, +d.rate);
});
这是更新后的 bl.ocks(当然,只有 6 个县有颜色,因为我使用的 JSON 只有 6 个数据点):https://bl.ocks.org/anonymous/211bf9775e61c5b07935536ed1926225
PS:在 S.O 上每个问题只问 一个问题 是一种很好的做法。因此,我建议你post另一个关于下拉问题的问题。