How to draw a pie chart using pdfbox?

Question

I have to draw a pie-chart using pdfbox.

Let the data be:

Subject Mark in Percentage Mark in Degrees Cumulative Degrees
Sub-1 80 80 80
Sub-2 70 70 150
Sub-3 65 65 215
Sub-4 90 90 305
Sub-5 55 55 360

Let the radius and centre be 100 pixels and ( 250, 400).

Let us take initial line parallel to x axis.
Drawing initial line statement will be:
contentStream.drawLine(250, 400, 350, 400);

I stuck up with:
a) finding x, y co-ordinates of point on the circle that is some degrees away from the initial line to draw radius
b) drawing arc between the two points using Bezier curve.

Any help on resolving the issues would be highly appreciated!

Answer 1

If you only want to draw some charts into a PDF using PDFBox and don't want to do all the math etc by yourself you can just use my PDFBox Graphics2D adapter. This allows you to use any Graphics2D based java library to draw your chart (e.g. JFreeChart). It will create a XForm which you can then place freely in your PDF.

If you really want to do this yourself you can still use the Java2D API to help with the shapes.

Arc2D.Float arc = new Arc2D.Float(x,y,w,h,start,extend, Arch2D.OPEN);
AffineTransform tf = new AffineTransform();
// You may need to setup tf to correctly position the drawing.
float[] coords = new float[6];
PathIterator pi = arc.getPathIterator(tf);
while (!pi.isDone()) {
    int segment = pi.currentSegment(coords);
    switch (segment) {
    case PathIterator.SEG_MOVETO:
        if (isFinite(coords, 2))
            contentStream.moveTo(coords[0], coords[1]);
        break;
    case PathIterator.SEG_LINETO:
        if (isFinite(coords, 2))
            contentStream.lineTo(coords[0], coords[1]);
        break;
    case PathIterator.SEG_QUADTO:
        if (isFinite(coords, 4))
            contentStream.curveTo1(coords[0], coords[1], coords[2], coors[3]);
            break;
    case PathIterator.SEG_CUBICTO:
        if (isFinite(coords, 6))
            contentStream.curveTo(coords[0], coords[1], coords[2], coords[3], coords[4], coords[5]);
        break;
    case PathIterator.SEG_CLOSE:
        contentStream.closePath();
        break;
    }
    pi.next();
}
contentStream.fill();

See also my graphics adapter shape walking code.

Answer 2

Finding the x, y coordinates on the circle depending on an angle is school math, i.e. sin() and cos(), the tricky part is to draw an arc with Bézier curves.

Here's some code that draws the pie chart you asked for. Note that createSmallArc() can only work with angles up to 90°. If you want more, you'd either have to modify the code by drawing several arcs until you go back to (0,0), or just draw several slices.

(createSmallArc() is by Hans Muller, license: Creative Commons Attribution 3.0. Changes made: implemented original AS code into java. Algorithm is by Aleksas Riškus)

public class PieChart
{
    public static void main(String[] args) throws IOException
    {
        PDDocument doc = new PDDocument();
        PDPage page = new PDPage();
        doc.addPage(page);
        PDPageContentStream cs = new PDPageContentStream(doc, page);

        cs.transform(Matrix.getTranslateInstance(250, 400));

        cs.setNonStrokingColor(Color.yellow);
        drawSlice(cs, 100, 0, 80);
        cs.fill();
        cs.setNonStrokingColor(Color.red);
        drawSlice(cs, 100, 80, 150);
        cs.fill();
        cs.setNonStrokingColor(Color.green);
        drawSlice(cs, 100, 150, 215);
        cs.fill();
        cs.setNonStrokingColor(Color.blue);
        drawSlice(cs, 100, 215, 305);
        cs.fill();
        cs.setNonStrokingColor(Color.ORANGE);
        drawSlice(cs, 100, 305, 360);
        cs.fill();

        cs.close();
        doc.save("piechart.pdf");
        doc.close();
    }

    private static void drawSlice(PDPageContentStream cs, float rad, float startDeg, float endDeg) throws IOException
    {
        cs.moveTo(0, 0);
        List<Float> smallArc = createSmallArc(rad, Math.toRadians(startDeg), Math.toRadians(endDeg));
        cs.lineTo(smallArc.get(0), smallArc.get(1));
        cs.curveTo(smallArc.get(2), smallArc.get(3), smallArc.get(4), smallArc.get(5), smallArc.get(6), smallArc.get(7));
        cs.closePath();
    }

    /**
     *  From https://hansmuller-flex.blogspot.com/2011/10/more-about-approximating-circular-arcs.html
     * 
     *  Cubic bezier approximation of a circular arc centered at the origin, 
     *  from (radians) a1 to a2, where a2-a1 &lt; pi/2.  The arc's radius is r.
     * 
     *  Returns a list with 4 points, where x1,y1 and x4,y4 are the arc's end points
     *  and x2,y2 and x3,y3 are the cubic bezier's control points.
     * 
     *  This algorithm is based on the approach described in:
     *  Aleksas Riškus, "Approximation of a Cubic Bezier Curve by Circular Arcs and Vice Versa," 
     *  Information Technology and Control, 35(4), 2006 pp. 371-378.
     */
    private static List<Float> createSmallArc(double r, double a1, double a2)
    {
        // Compute all four points for an arc that subtends the same total angle
        // but is centered on the X-axis
        double a = (a2 - a1) / 2;
        double x4 = r * Math.cos(a);
        double y4 = r * Math.sin(a);
        double x1 = x4;
        double y1 = -y4;
        double q1 = x1*x1 + y1*y1;

        double q2 = q1 + x1*x4 + y1*y4;
        double k2 = 4/3d * (Math.sqrt(2 * q1 * q2) - q2) / (x1 * y4 - y1 * x4);
        double x2 = x1 - k2 * y1;
        double y2 = y1 + k2 * x1;
        double x3 = x2; 
        double y3 = -y2;

        // Find the arc points' actual locations by computing x1,y1 and x4,y4 
        // and rotating the control points by a + a1

        double ar = a + a1;
        double cos_ar = Math.cos(ar);
        double sin_ar = Math.sin(ar);

        List<Float> list = new ArrayList<Float>();
        list.add((float) (r * Math.cos(a1)));
        list.add((float) (r * Math.sin(a1))); 
        list.add((float) (x2 * cos_ar - y2 * sin_ar)); 
        list.add((float) (x2 * sin_ar + y2 * cos_ar)); 
        list.add((float) (x3 * cos_ar - y3 * sin_ar)); 
        list.add((float) (x3 * sin_ar + y3 * cos_ar)); 
        list.add((float) (r * Math.cos(a2))); 
        list.add((float) (r * Math.sin(a2)));
        return list;
    }
}