functionoid functionoid - 7 months ago 20
Javascript Question

D3.js stacked bar chart

I've spent enough time modifying the stacked bar chart sample that it would be wise to get a second pair of eyes to look at what I'm doing wrong. The modified sample is in a js fiddle.

My main changes are these:

1: On line 10 reversed the array for range

y = d3.scale.linear().range([height, 0])


2: On lines 17 and 22 added axis

var xAxis = d3.svg.axis()
.scale(x)
.orient("bottom")
.tickFormat(format);
var yAxis = d3.svg.axis()
.scale(y)
.orient("left")


3: Line 62 changed the addition of y0 and y (this is the change that might be creating problem)

.attr("y", function (d) { return y(d.y) - y(d.y0); })


As you can see bars are going way above the specified height.

If someone could tell me what I'm doing wrong I'd be very thankful. I spent too much time looking just by myself. I do think I'm very close as chart is displaying other than the height problem.

Answer

There is one more change, which you haven't mentioned: that you the chart containing g is now translated to the top-left point instead of the bottom-left point, as was in the original plot.

Changing the following two lines around the line you mentioned will solve the problem:

            .attr("y", function (d) { return y(d.y + d.y0); })
            .attr("height", function (d) { return y(d.y0) - y(d.y +d.y0); })

Demo

This is how I generally do these calculations with an inverted scale such as y = d3.scale.linear().range([height, 0]):

calculations with inverted scale

Here, the point S(y + y0) will be the value of the point closer to the top, or with the lower y value, and the height of the bar will be = S(y0) - S(y + y0) or = abs(S(y0+y) - S(y0)), whichever you find more palatable.

However, to be honest, there is plenty of trial and error involved before I actually draw this down. Then I do more trial and error to find the solution and convince myself that this drawing helped.

So ... your mileage might vary. :)