我想在饼图切片的中间绘制一个带有三角形的饼图。 目前,我在切片中间绘制了切片和三角形的piechat,但三角形不是直角。我需要知道如何以正确的方式定位三角形。我的代码和结果:
import java.awt.*;
import java.awt.geom.Ellipse2D;
import javax.swing.*;
class Slice {
double value;
Color color;
public Slice(double value, Color color) {
this.value = value;
this.color = color;
}
}
class PieChart extends JPanel {
private Color a = Color.RED;
private Color b = Color.BLUE;
private Color c = Color.YELLOW;
Slice[] slices = {
new Slice(60, a),
new Slice(100, b),
new Slice(200, c)
};
public PieChart(){
}
@Override
protected void paintComponent(Graphics g) {
Graphics2D g2d = (Graphics2D)g;
super.paintComponent(g2d);
this.setBackground(new Color(255,255,255));
g2d.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
double total = 0.0D;
for (int i = 0; i < slices.length; i++) {
total += slices[i].value;
}
double curValue = 90.0D;
int startAngle = 0;
for (int i = 0; i < slices.length; i++) {
startAngle = (int) (curValue * 360 / total);
int arcAngle = (int) (slices[i].value * 360 / total);
g2d.setColor(slices[i].color);
g2d.fillArc(20, 20, 200, 200, startAngle, arcAngle);
g2d.setPaint(Color.BLACK);
int x = (int)(110+100*Math.cos(((-(startAngle+(arcAngle/2)))*Math.PI)/180));
int y = (int)(110+100*Math.sin(((-(startAngle+(arcAngle/2)))*Math.PI)/180));
Polygon p = new Polygon(new int[] {x, x+14, x+7}, new int[] {y, y, y-14}, 3); // this values seems to be important
g2d.draw(p);
g2d.fill(p);
curValue += slices[i].value;
}
}
}
编辑:应该如下所示:
答案 0 :(得分:2)
我从0点开始做第一个弧(我想你的意思是这样做)。
由于您使用的fillArc
需要int
秒,因此向下舍入的double
可能不会达到全部数量,并且切片之间会有间隙:
相反,使用Arc2D.Double
来获得更好的精确度:
class Slice {
double value;
Color color;
public Slice(double value, Color color) {
this.value = value;
this.color = color;
}
public Color getColor() {
return color;
}
public double getValue() {
return value;
}
}
class PieChart extends JPanel {
private final int SIZE = 500, START = 40, START_DEG = 90;
private final int TRIG_HBASE = 66, TRIG_HEIGHT = 36;
private final int x0 =(START + SIZE / 2), y0 = START;
private final Polygon poly;
private Color a = Color.RED;
private Color b = Color.BLUE;
private Color c = Color.YELLOW;
Slice[] slices = {new Slice(65, a), new Slice(123, b), new Slice(212, c)};
PieChart() {
setBackground(Color.WHITE);
int x1 = x0 + TRIG_HBASE, y1 = y0;
int x2 = x0 - TRIG_HBASE, y2 = y0;
int x3 = x0, y3 = y0 - TRIG_HEIGHT;
poly = new Polygon(new int[] {x1, x2, x3}, new int[] {y1, y2, y3}, 3);
}
@Override
protected void paintComponent(Graphics g) {
Graphics2D g2d = (Graphics2D) g;
super.paintComponent(g2d);
g2d.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
g2d.setColor(Color.LIGHT_GRAY);
g2d.fillRect(START, START, SIZE, SIZE);
double total = 0d;
for (Slice slice : slices) {
total += slice.getValue();
}
double startAngle = START_DEG;
double arcAngle, centerAngle;
double x, y;
for (Slice slice : slices) {
arcAngle = (slice.getValue() * 360 / total);
g2d.setColor(slice.getColor());
g2d.fill(new Arc2D.Double(START, START, SIZE, SIZE, startAngle, arcAngle, Arc2D.PIE));
centerAngle = Math.toRadians(((startAngle - START_DEG) + arcAngle / 2));
x = (START + SIZE / 2 * (1 - Math.sin(centerAngle)));
y = (START + SIZE / 2 * (1 - Math.cos(centerAngle)));
AffineTransform trans = AffineTransform.getTranslateInstance(x - x0, y - y0);
AffineTransform rot = AffineTransform.getRotateInstance(-centerAngle, x, y);
Shape s = trans.createTransformedShape(poly);
s = rot.createTransformedShape(s);
g2d.setColor(slice.getColor().darker());
g2d.fill(s);
startAngle += arcAngle;
}
}
@Override
public Dimension getPreferredSize() {
return new Dimension(START * 2 + SIZE, START * 2 + SIZE);
}
}
poly
用作基本三角形,朝上,其基点以0点钟为中心。每个弧都会对此多边形进行平移和变换(副本),使其基点位于弧长的中心,并使其向外指向。
备注:强>
setBackground
,在外面拨打电话。它会使绘制机制自动在每次重绘上绘制背景。如果你把它放在里面,你只需要重写每次重绘的指令。或者,您可以使用paintComponent
将背景设置为白色(或g.clearRect
设置不同的颜色。)fillRect
方法以与其内容兼容。getPreferredSize
)而不是内联数字。这样,您只需要在一个地方更改它们,并考虑所有依赖项。final
循环。for each
s并且仅在最新点转换为double
,否则您将失去精确度(将角度转换为int
,然后将其用作int
} argument。。double
和Math.toRadians
值得熟悉。Math.toDegrees
常数以使用它们的大小。我也给它们着色以了解哪个三角形属于哪个弧。以下是您的参数的结果(并没有特殊颜色):
TRIG
答案 1 :(得分:1)
这里我提出了一个解决方案(稍微改变算法):
public class Chart {
public static void main(String[] args) {
JFrame frame= new JFrame();
frame.add(new PieChart());
frame.setSize(new Dimension(400, 400));
frame.setDefaultCloseOperation(WindowConstants.DISPOSE_ON_CLOSE);
EventQueue.invokeLater( () ->{
frame.setVisible(true);
});
}
public static class Slice {
double value;
Color color;
public Slice(double value, Color color) {
this.value = value;
this.color = color;
}
}
public static class PieChart extends JPanel {
private static final long serialVersionUID = 1L;
private Color a = Color.RED;
private Color b = Color.BLUE;
private Color c = Color.YELLOW;
Slice[] slices = { new Slice(60, a), new Slice(100, b),
new Slice(200, c) };
public PieChart() {
}
@Override
protected void paintComponent(Graphics g) {
Graphics2D g2d = (Graphics2D) g;
super.paintComponent(g2d);
this.setBackground(Color.WHITE);
g2d.setRenderingHint(RenderingHints.KEY_ANTIALIASING,RenderingHints.VALUE_ANTIALIAS_ON);
int upperLeftX = 20;
int upperLeftY = 20;
int r = 100;
int startAngle = 0;
double curValue = 0.0;
for (int i = 0; i < slices.length; i++) {
startAngle = (int) curValue ;
int arcAngle = (int) slices[i].value ;
g2d.setColor(slices[i].color);
g2d.fillArc(upperLeftX, upperLeftY, r*2, r*2, startAngle, arcAngle);
g2d.setPaint(Color.BLACK);
double qi = curValue+slices[i].value/2;
int x = upperLeftX + r + (int)(Math.cos(qi*Math.PI/180)*r);
int y = upperLeftY + r - (int)(Math.sin(qi*Math.PI/180)*r);
//point touching the circle (x,y)->half point of the base
int x1 = x - (int)(7*Math.sin(qi*Math.PI/180));
int y1 = y - (int)(7*Math.cos(qi*Math.PI/180));
int x2 = x + (int)(7*Math.sin(qi*Math.PI/180));
int y2 = y + (int)(7*Math.cos(qi*Math.PI/180));
int x3 = upperLeftX + r + (int)(Math.cos(qi*Math.PI/180)*(r+12));
int y3 = upperLeftY + r - (int)(Math.sin(qi*Math.PI/180)*(r+12));
Polygon p = new Polygon(new int[] { x1, x2, x3 },
new int[] { y1, y2, y3 }, 3); // this values seems to
// be important
g2d.draw(p);
g2d.fill(p);
curValue += slices[i].value;
}
}
}
}