var maindata=[
{"date":"21-APR-16 04:19 AM","Delhi":30,"Mumbai":28},
{"date":"21-APR-16 05:19 AM","Delhi":32,"Mumbai":30},
{"date":"21-APR-16 06:19 AM","Delhi":34,"Mumbai":34},
{"date":"21-APR-16 10:19 AM","Kolkata":34},
{"date":"21-APR-16 11:19 AM","Delhi":48,"Chennai":40,"Kolkata":36}
];
var that = this;
var deepClonedCopy = jQuery.extend(true, {}, maindata);
var data;
data = $.map(deepClonedCopy, function(el) { return el });
// var passedheight = this.getHeight();
//var containerWidth = jQuery.sap.byId(this.oParent.sId).width() || 800; // gets super parent width
var margin = {top: 15, right: 30, bottom: 20, left: 30},
width = 960- margin.left - margin.right,
height = 500 - margin.top - margin.bottom;
if(data.length >= 1){
//alert(JSON.stringify(data));
var parseDate = d3.time.format("%d-%b-%y %H:%M %p").parse;
var maxDate = d3.time.hour.offset(parseDate(d3.max(data, function(d) { return d.date; })),+1);
var minDate = d3.time.hour.offset(parseDate(d3.min(data, function(d) { return d.date; })),-1);
var div = d3.select("#toolTip");
var x = d3.time.scale()
.domain([minDate, maxDate])
.range([0, width]);
var y = d3.scale.linear()
.range([Math.ceil(height), 0]);
var color = d3.scale.category10();
var xAxis = d3.svg.axis()
.scale(x)
.orient("bottom").ticks(4);
var yAxis = d3.svg.axis()
.scale(y)
.orient("left").ticks(4).tickSize(-width, 0, 0);
var line = d3.svg.line()
.x(function(d) { return x(d.date); })
.y(function(d) { return y(d.tonneValue); });
var svg = d3.select("body").append("svg")
.attr("width", width + margin.left + margin.right)
.attr("height", height + margin.top + margin.bottom)
.append("g")
.attr("transform", "translate(" + margin.left + "," + margin.top + ")");
color.domain(d3.keys(data[0]).filter(function(key) { return key !== "date" && key !== "maxLength"; }));
data.forEach(function(d) {
d.date = parseDate(d.date);
});
var wsfs = color.domain().map(function(name) {
return {
name: name,
values: data.map(function(d) {
return {date: d.date, tonneValue: +d[name]};
})
};
});
//x.domain(d3.extent(data, function(d) { return d.date; }));
y.domain([
//d3.min(wsfs, function(c) { return d3.min(c.values, function(v) { return v.tonneValue; }); }),
0,
Math.ceil(d3.max(wsfs, function(c) { return d3.max(c.values, function(v) { return v.tonneValue; }); }))
]);
svg.append("g")
.attr("class", "x axis")
.attr("transform", "translate(0," + height + ")")
.call(xAxis);
svg.append("g")
.attr("class", "y axis")
.call(yAxis);
var wsf = svg.selectAll(".wsf")
.data(wsfs)
.enter().append("g")
.attr("class", "wsf");
wsf.append("path")
.attr("class", "line")
.attr("d", function(d) { return line(d.values); })
.style("stroke", function(d) { return color(d.name); });
wsf.selectAll("dot")
.data(function(d) { return d.values;})
.enter().append("circle")
.attr("class", "dot")
.attr("r", 3)
.attr("cx", function(d) { return x(d.date); })
.attr("cy", function(d) { return y(d.tonneValue); })
.attr("stroke", function (d) {
return color(this.parentNode.__data__.name) })
.attr("fill", function (d) {
return color(this.parentNode.__data__.name) })
//.attr("fill-opacity", .5)
//.attr("stroke-width", 2)
.on("mouseover", function (d) {
formatDate = d3.time.format("%d-%m-%Y %H:%M %p");
div.transition().duration(100).style("opacity", .9);
div.html(/*this.parentNode.__data__.name + "<br/>" +*/ d.tonneValue /*+ "<br/>" + "<br/>"*/ +" Tonne handled at "+ formatDate(d.date))
.style("left", (d3.event.pageX) + "px").style("top", (d3.event.pageY - 28) + "px").attr('r', 8);
d3.select(this).attr('r', 8)
}).on("mouseout", function (d) {
div.transition().duration(600).style("opacity", 0)
d3.select(this).attr('r', 3);
});
var legendNames = d3.keys(data[0]).filter(function(key) { return key !== "date" });
data.forEach(function(d) {
d.ages = legendNames.map(function(name) { return {name: name, value: +d[name]}; });
});
var legend = svg.selectAll(".legend")
.data(legendNames.slice())
.enter().append("g")
.attr("class", "legend")
.attr("transform", function(d, i) { return "translate(0," + i * 20 + ")"; });
legend.append("rect")
.attr("x", width - 18)
.attr("width", 18)
.attr("height", 4)
.style("fill", function(d) {return color(d); });
legend.append("text")
.attr("x", width - 24)
.attr("y", 6)
.attr("dy", ".35em")
.style("text-anchor", "end")
.text(function(d) { return d; });
}
上面的代码给出了2行,直到前3个json元素,直到数据格式相似为止
但是从第4个元素开始,没有Delhi
/ Mumbai
,而且只有Kolkata
,但我找不到Kolkata
34行,{{1} } {36} Kolkata
&#39; s点,Delhi
&#39; s点应该从34延伸到48,而Chennai是一个简单的点,因为只有一个点是Chennai ie 40.
任何人都可以告诉我哪里做错了。这是fiddle。
答案 0 :(得分:1)
以下是更新后的fiddle
某些密钥(即Kolkata
)未包含在绑定数据中
原始代码:
color.domain(d3.keys(data[0]).filter(function(key) { return key !== "date" && key !== "maxLength"; }));
data.forEach(function(d) {
d.date = parseDate(d.date);
});
var wsfs = color.domain().map(function(name) {
return {
name: name,
values: data.map(function(d) {
return {date: d.date, tonneValue: +d[name]};
})
};
});
要(您需要更改['Delhi', 'Mumbai', 'Kolkata', 'Chennai']
以更正式地编码):
var wsfs = ['Delhi', 'Mumbai', 'Kolkata', 'Chennai'].map(function(name) {
return {
name: name,
values: data.map(function(d) {
return {date: d.date, tonneValue: +d[name]};
}).filter(function(d) {
return !isNaN(d.tonneValue)
})
};
});
过滤数据
原始代码:
var wsfs = color.domain().map(function(name) {
return {
name: name,
values: data.map(function(d) {
return {date: d.date, tonneValue: +d[name]};
})
};
});
要:
var wsfs = ['Delhi', 'Mumbai', 'Kolkata', 'Chennai'].map(function(name) {
return {
name: name,
values: data.map(function(d) {
return {date: d.date, tonneValue: +d[name]};
}).filter(function(d) {
return !isNaN(d.tonneValue)
})};
});
答案 1 :(得分:1)
代码中的主要问题(缺少行和点以及不完整的图例)是您的数据需要一些结构。
您的数据集有时会有可能存在或可能不存在的城市。
所以第一项任务是制作一个合适的数据集
由此:
var maindata = [{
"date": "21-APR-16 04:19 AM",
"Delhi": 30,
"Mumbai": 28
}, {
"date": "21-APR-16 05:19 AM",
"Delhi": 32,
"Mumbai": 30
}, {
"date": "21-APR-16 06:19 AM",
"Delhi": 34,
"Mumbai": 34
}, {
"date": "21-APR-16 10:19 AM",
"Kolkata": 34
}, {
"date": "21-APR-16 11:19 AM",
"Delhi": 48,
"Chennai": 40,
"Kolkata": 36
}];
对此:
[
{
"name":"Delhi",
"values":[
{
"date":"2016-04-20T22:49:00.000Z",
"value":30
},
{
"date":"2016-04-20T23:49:00.000Z",
"value":32
},
{
"date":"2016-04-21T00:49:00.000Z",
"value":34
},
{
"date":"2016-04-21T05:49:00.000Z",
"value":48
}
]
},
{
"name":"Mumbai",
"values":[
{
"date":"2016-04-20T22:49:00.000Z",
"value":28
},
{
"date":"2016-04-20T23:49:00.000Z",
"value":30
},
{
"date":"2016-04-21T00:49:00.000Z",
"value":34
}
]
},
{
"name":"Kolkata",
"values":[
{
"date":"2016-04-21T04:49:00.000Z",
"value":34
},
{
"date":"2016-04-21T05:49:00.000Z",
"value":36
}
]
},
{
"name":"Chennai",
"values":[
{
"date":"2016-04-21T05:49:00.000Z",
"value":40
}
]
}
]
所以现在所有城市都有各自的数据。 城市数量x因此行数为x。
要使数据集执行此操作:
var cities = []
var parseDate = d3.time.format("%d-%b-%y %H:%M %p").parse;
maindata.forEach(function(d) {
cities.push(d3.keys(d).filter(function(key) {
return key !== "date" && key !== "maxLength";
}));
});
cities = d3.set(d3.merge(cities)).values();//get all cites and make it unique
console.log(cities)
var myData = [];
var allValues = [];
var allDates =[];
//generate cities and its values as shown in my dataset above.
cities.forEach(function(city){
var cityData = {};
cityData.name = city;
cityData.values = [];
maindata.forEach(function(md){
if (md[city]){
allValues.push(md[city])
allDates.push(parseDate(md.date))
cityData.values.push({date: parseDate(md.date), value: md[city]})
}
})
myData.push(cityData)
});
接下来制作这样的行:
var wsf = svg.selectAll(".wsf")
.data(myData)
.enter().append("g")
.attr("class", "wsf");
wsf.append("path")
.attr("class", "line")
.attr("d", function(d) {
return line(d.values);
})
.style("stroke", function(d) {
return color(d.name);
});
传说就是这样:
var legend = svg.selectAll(".legend")
.data(cities)
.enter().append("g")
.attr("class", "legend")
.attr("transform", function(d, i) {
return "translate(0," + i * 20 + ")";
});
legend.append("rect")
.attr("x", width - 18)
.attr("width", 18)
.attr("height", 4)
.style("fill", function(d) {
return color(d);
});
legend.append("text")
.attr("x", width - 24)
.attr("y", 6)
.attr("dy", ".35em")
.style("text-anchor", "end")
.text(function(d) {
return d;
});
}
工作代码here