Stacked Bar Graph in d3js - d3.js

I am new to D3Js.i am following this http://codepen.io/benlister/pres/bNeLQy for stacked bar graph.I am not sure how to make the y axis absolute.Here it is shown in % .I tried adding
y.domain([0,500]);
It didnt work.Please help.

First, if you are talking about the % symbol, you have to change the tick format.
The code is this:
var yAxis = d3.svg.axis()
.scale(y)
.orient("left")
.tickFormat(d3.format(".0%"));
Just remove the tickFormat, or change it to the format you want.
But your problem is bigger than that: the y axis shows percent because the data is being transformed to percentages. This is the code that does the math:
data.forEach(function (d) {
var y0 = 0;
d.rates = color.domain().map(function (name) {
console.log();;
return {
name: name,
y0: y0,
y1: y0 += +d[name],
amount: d[name]
};
});
d.rates.forEach(function (d) {
d.y0 /= y0;
d.y1 /= y0;
});
Check this example to see how to do it with absolute values:
https://bl.ocks.org/mbostock/3886208

Try changing this line
var yAxis = d3.svg.axis().scale(y).orient("left").tickFormat(d3.format(".0%"));
to
var yAxis = d3.svg.axis().scale(y).orient("left").tickFormat(d3.format("d"));
for integer values on your y axis.
See here for more details.

Related

d3.v4: How to set ticks every Math.PI/2

In the d3.v4 documentation the following is stated:
To generate ticks every fifteen minutes with a time scale, say:
axis.tickArguments([d3.timeMinute.every(15)]);
Is there a similar approach that can be used with values other than time? I am plotting sine and cosine curves, so I'd like the ticks to begin at -2*Math.PI, end at 2*Math.PI, and between these values I'd like a tick to occur every Math.PI/2. I could, of course, explicitly compute the tick values and supply them to the tickValue method; however, if there is a simpler way to accomplish this, as in the time-related example quoted above, I'd prefer to use that.
Setting the end ticks and specifying the precise space of the ticks in a linear scale is a pain in the neck. The reason is that D3 axis generator was created in such a way that the ticks are automatically generated and spaced. So, what is handy for someone who doesn't care too much for customisation can be a nuisance for those that want a precise customisation.
My solution here is a hack: create two scales, one linear scale that you'll use to plot your data, and a second scale, that you'll use only to make the axis and whose values you can set at your will. Here, I choose a scalePoint() for the ordinal scale.
Something like this:
var realScale = d3.scaleLinear()
.range([10,width-10])
.domain([-2*Math.PI, 2*Math.PI]);
var axisScale = d3.scalePoint()
.range([10,width-10])
.domain(["-2 \u03c0", "-1.5 \u03c0", "-\u03c0", "-0.5 \u03c0", "0",
"0.5 \u03c0", "\u03c0", "1.5 \u03c0", "2 \u03c0"]);
Don't mind the \u03c0, that's just π (pi) in Unicode.
Check this demo, hover over the circles to see their positions:
var width = 500,
height = 150;
var data = [-2, -1, 0, 0.5, 1.5];
var realScale = d3.scaleLinear()
.range([10, width - 10])
.domain([-2 * Math.PI, 2 * Math.PI]);
var axisScale = d3.scalePoint()
.range([10, width - 10])
.domain(["-2 \u03c0", "-1.5 \u03c0", "-\u03c0", "-0.5 \u03c0", "0", "0.5 \u03c0", "\u03c0", "1.5 \u03c0", "2 \u03c0"]);
var svg = d3.select("body").append("svg")
.attr("width", width)
.attr("height", height);
var circles = svg.selectAll("circle").data(data)
.enter()
.append("circle")
.attr("r", 8)
.attr("fill", "teal")
.attr("cy", 50)
.attr("cx", function(d) {
return realScale(d * Math.PI)
})
.append("title")
.text(function(d) {
return "this circle is at " + d + " \u03c0"
});
var axis = d3.axisBottom(axisScale);
var gX = svg.append("g")
.attr("transform", "translate(0,100)")
.call(axis);
<script src="https://d3js.org/d3.v4.min.js"></script>
I was able to implement an x axis in units of PI/2, under program control (not manually laid out), by targetting the D3 tickValues and tickFormat methods. The call to tickValues sets the ticks at intervals of PI/2. The call to tickFormat generates appropriate tick labels. You can view the complete code on GitHub:
https://github.com/quantbo/sine_cosine
My solution is to customise tickValues and tickFormat. Only 1 scale is needed, and delegate d3.ticks function to give me the new tickValues that are proportional to Math.PI.
const piChar = String.fromCharCode(960);
const tickFormat = val => {
const piVal = val / Math.PI;
return piVal + piChar;
};
const convertSIToTrig = siDomain => {
const trigMin = siDomain[0] / Math.PI;
const trigMax = siDomain[1] / Math.PI;
return d3.ticks(trigMin, trigMax, 10).map(v => v * Math.PI);
};
const xScale = d3.scaleLinear().domain([-Math.PI * 2, Math.PI * 2]).range([0, 600]);
const xAxis = d3.axisBottom(xScale)
.tickValues(convertSIToTrig(xScale.domain()))
.tickFormat(tickFormat);
This way if your xScale's domain were changed via zoom/pan, the new tickValues are nicely generated with smaller/bigger interval

sine wave not going to the amplitude height in d3.js

Here is a jsbin of what I have so far.
My sine wave is not going to the y value of 1 or -1, i.e the amplitude.
My yScale is defined like this:
const yScaleAxis = d3.scale.linear()
.domain([-1, 1])
.range([radius, -radius]);
And I am creating the values like this:
const xValues = [0, 1.57, 3.14, 4.71, 6.28]; // 0 to 2PI
const sineData = xValues.map((x) => {
console.log(Math.sin(x));
return {x: x, y: Math.sin(x)};
});
The values for y are logged as:
0
0.9999996829318346
0.0015926529164868282
-0.999997146387718
-0.0031853017931379904
I then use the scale to set the values:
const sine = d3.svg.line()
.interpolate('basis')
.x( (d) => {return xScaleAxis(d.x);})
.y( (d) => {return yScaleAxis(d.y);});
circleGroup.append('path')
.datum(sineData)
.attr('class', 'sine-curve')
.attr('d', sine);
But as you can see in the jsbin the amplitude of the sine wave is not reaching 1 or -1 and I am not sure why.
Change the line interpolation method to monotone, basis corresponds to a B-spline
More info about the interpolation options provided by d3

Logarithmic time scale

How to make a logarithmic datetime scale in D3?
a simple time scale is like this:
d3.time.scale()
.domain([new Date(2014, 0, 1), new Date()])
.range([0, 500])
and a simple log scale is like:
d3.scale.log()
.domain([new Date(2014, 0, 1), new Date()])
.rangeRound([0, 500])
.base(10)
Tried to chain their syntax in a various ways with no effect.
Chart will position users by last login date. Range will be about one year. If we space data linearly, most users will collide during last days/hours. With logarithm we can zoom last hours.
Solution could be by interactive zoom or several charts. But goal here is to make single static chart with nonlinear overview of year.
One alternative could be to convert datetime to "days from now", a number. It would work for data. But then I wouldn't know how to label axis ticks like "01-01-2014"...
Something like the below seems to fool d3js into thinking it has a real scale object. It should make a good starting point:
var xt = d3.scaleUtc()
.domain([start, now])
.range([1, width])
var xp = d3.scalePow()
.exponent(2)
.domain([1, width])
.range([0, width])
// Fool d3js into thinking that it is looking at a scale object.
function x_copy() {
var x = function(t) { return xp(xt(t)) }
x.domain = xt.domain
x.range = xp.range
x.copy = x_copy
x.tickFormat = xt.tickFormat
x.ticks = xt.ticks
return x
}
x = x_copy()
var xAxis = d3.axisBottom(x)
Create two scales and use one after the other. First use the time scale and than the log or pow scale.
var parseDate = d3.time.format("%Y-%m-%d").parse;
var x = d3.time.scale()
.range([0,width]);
var xLog = d3.scale.pow().exponent(4)
.domain([1,width])
.range([0,width]);
than I'm using .forEach to get the linear points:
x.domain(d3.extent(data, function(d) { return parseDate(d.start); }));
data.forEach(function(d) {
d.start = x(parseDate(d.start));
});
when I'm drawing the objects I add the log scale:
.attr('cx', function (d) { return xLog(d.start)})

Don't show negative sign d3 axis labels

I have a scale in which I don't want the negative signs to appear before the numbers - how can this be done in the d3 formatter? The scale is built as follows:
var formater = d3.format("0");
self.x = d3.scale.linear().domain([self.xmin, self.xmax]).range([0, self.settings.width])
self.axis = d3.svg.axis()
.scale(self.x)
.orient("bottom")
.tickFormat(formater);
self.axisLabels = self.svg.append("g")
.attr("class", "axis")
.attr("id", "axis")
.call(self.axis)
I see an option to add a "+" sign but not remove a "-" sign https://github.com/mbostock/d3/wiki/Formatting#wiki-d3_format
Also, is it possible to remove one label? I'm labeling from -5 to 5 on the scale, and don't want the negative signs to appear, and I don't want to label 0. Thanks.
You are using a formatter already and you do not need to rely on D3 to remove the '-' sign, you can do it yourself:
var formatter = d3.format("0");
// ...
self.axis = d3.svg.axis()
.scale(self.x)
.orient("bottom")
.tickFormat(function (d) {
if (d === 0) return ''; // No label for '0'
else if (d < 0) d = -d; // No nagative labels
return formatter(d);
});

how do you draw linear line in scatter plot with d3.js

I am looking to implement ggplot2 type of graphs using d3.js library for interactivey purpose. I love ggplot2 but users are interested in interactive graphs. I've been exploring d3.js library and there seems to be lots of different graph capability but I really did not see any statistical graphs like linear line, forecast etc. Given a scatter plot, is it possible to also add linear line to the graph.
I have this sample script that draws scatter plot. How would I add linear line to this graph in d3.js?
// data that you want to plot, I've used separate arrays for x and y values
var xdata = [5, 10, 15, 20],
ydata = [3, 17, 4, 6];
// size and margins for the chart
var margin = {top: 20, right: 15, bottom: 60, left: 60}
, width = 960 - margin.left - margin.right
, height = 500 - margin.top - margin.bottom;
// x and y scales, I've used linear here but there are other options
// the scales translate data values to pixel values for you
var x = d3.scale.linear()
.domain([0, d3.max(xdata)]) // the range of the values to plot
.range([ 0, width ]); // the pixel range of the x-axis
var y = d3.scale.linear()
.domain([0, d3.max(ydata)])
.range([ height, 0 ]);
// the chart object, includes all margins
var chart = d3.select('body')
.append('svg:svg')
.attr('width', width + margin.right + margin.left)
.attr('height', height + margin.top + margin.bottom)
.attr('class', 'chart')
// the main object where the chart and axis will be drawn
var main = chart.append('g')
.attr('transform', 'translate(' + margin.left + ',' + margin.top + ')')
.attr('width', width)
.attr('height', height)
.attr('class', 'main')
// draw the x axis
var xAxis = d3.svg.axis()
.scale(x)
.orient('bottom');
main.append('g')
.attr('transform', 'translate(0,' + height + ')')
.attr('class', 'main axis date')
.call(xAxis);
// draw the y axis
var yAxis = d3.svg.axis()
.scale(y)
.orient('left');
main.append('g')
.attr('transform', 'translate(0,0)')
.attr('class', 'main axis date')
.call(yAxis);
// draw the graph object
var g = main.append("svg:g");
g.selectAll("scatter-dots")
.data(ydata) // using the values in the ydata array
.enter().append("svg:circle") // create a new circle for each value
.attr("cy", function (d) { return y(d); } ) // translate y value to a pixel
.attr("cx", function (d,i) { return x(xdata[i]); } ) // translate x value
.attr("r", 10) // radius of circle
.style("opacity", 0.6); // opacity of circle
To add a line to your plot, all that you need to do is to append some line SVGs to your main SVG (chart) or to the group that contains your SVG elements (main).
Your code would look something like the following:
chart.append('line')
.attr('x1',x(10))
.attr('x2',x(20))
.attr('y1',y(5))
.attr('y2',y(10))
This would draw a line from (10,5) to (20,10). You could similarly create a data set for your lines and append a whole bunch of them.
One thing you might be interested in is the SVG path element. This is more common for lines than drawing one straight segment at a time. The documentation is here.
On another note you may find it easier to work with data in d3 if you create it all as one object. For example, if your data was in the following form:
data = [{x: 5, y:3}, {x: 10, y:17}, {x: 15, y:4}, {x: 20, y:6}]
You could use:
g.selectAll("scatter-dots")
.data(ydata) // using the values in the ydata array
.enter().append("svg:circle") // create a new circle for each value
.attr("cy", function (d) { return y(d.y); } ) //set y
.attr("cx", function (d,i) { return x(d.x); } ) //set x
This would eliminate potentially messy indexing if your data gets more complex.
UPDATE: Here is the relevant block: https://bl.ocks.org/HarryStevens/be559bed98d662f69e68fc8a7e0ad097
I wrote this function to calculate a linear regression from data, formatted as JSON.
It takes 5 parameters:
1) Your data
2) The column name of the data plotted on your x-axis
3) The column name of the data plotted on your y-axis
4) The minimum value of your x-axis
5) The minimum value of your y-axis
I got the formula for calculating a linear regression from http://classroom.synonym.com/calculate-trendline-2709.html
function calcLinear(data, x, y, minX, minY){
/////////
//SLOPE//
/////////
// Let n = the number of data points
var n = data.length;
var pts = [];
data.forEach(function(d,i){
var obj = {};
obj.x = d[x];
obj.y = d[y];
obj.mult = obj.x*obj.y;
pts.push(obj);
});
// Let a equal n times the summation of all x-values multiplied by their corresponding y-values
// Let b equal the sum of all x-values times the sum of all y-values
// Let c equal n times the sum of all squared x-values
// Let d equal the squared sum of all x-values
var sum = 0;
var xSum = 0;
var ySum = 0;
var sumSq = 0;
pts.forEach(function(pt){
sum = sum + pt.mult;
xSum = xSum + pt.x;
ySum = ySum + pt.y;
sumSq = sumSq + (pt.x * pt.x);
});
var a = sum * n;
var b = xSum * ySum;
var c = sumSq * n;
var d = xSum * xSum;
// Plug the values that you calculated for a, b, c, and d into the following equation to calculate the slope
// m = (a - b) / (c - d)
var m = (a - b) / (c - d);
/////////////
//INTERCEPT//
/////////////
// Let e equal the sum of all y-values
var e = ySum;
// Let f equal the slope times the sum of all x-values
var f = m * xSum;
// Plug the values you have calculated for e and f into the following equation for the y-intercept
// y-intercept = b = (e - f) / n = (14.5 - 10.5) / 3 = 1.3
var b = (e - f) / n;
// return an object of two points
// each point is an object with an x and y coordinate
return {
ptA : {
x: minX,
y: m * minX + b
},
ptB : {
y: minY,
x: (minY - b) / m
}
}
}

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