有什么方法可以在Java中弯曲BufferedImage
吗?
我认为如果我将图像裁剪成更小的部分并旋转它们,那么我基本上会弯曲图像,但它似乎不起作用。
这是我创建的方法:
/**
* This is a recursive method that will accept an image the point where the bending will start and the point where the bending will end, as well as the angle of bending
*
* @param original:the original image
* @param startingPoint: the point where the bending should start
* @param endingPoint: the point where the bending should end
* @param radiands: the angle
* @return the bent image
*/
public static BufferedImage getBentImage(BufferedImage original, int startingPoint, int endingPoint, double radians) {
if (startingPoint >= endingPoint)
return original;
int type = BufferedImage.TYPE_INT_ARGB;
int width = original.getWidth();
int height = original.getHeight();
BufferedImage crop = original.getSubimage(0, 0, startingPoint, height);
BufferedImage crop0 = original.getSubimage(startingPoint, 0, width - startingPoint, height);
BufferedImage bendCrop = new BufferedImage(width, height, type);
BufferedImage image = new BufferedImage(width, height, type);
AffineTransform rotation = new AffineTransform();
rotation.translate(0, 0);
rotation.rotate(radians);
Graphics2D g = bendCrop.createGraphics();
g.drawImage(crop0, rotation, null);
g.dispose();
g = image.createGraphics();
g.drawImage(crop, 0, 0, null);
g.drawImage(bendCrop, startingPoint, 0, null);
g.dispose();
return getBentImage(image, startingPoint + 1, endingPoint, radians);
}
这是原始图片:
这是 getBentImage(image, 200, 220, Math.toRadians(1))
的结果:
我期待更接近的东西:
关于如何实际实现 getBentImage()
方法有什么想法吗?
最佳答案
正如评论中所建议的,一个简单的方法是将图像分为三部分:
- 与原件相同。
- 根据弯曲变换进行弯曲。
- 恒定对角延续。
这是一个快速但有点困惑的示例,显示了原始形状及其下方的结果形状。我只是为图像使用了标签图标,而不是进行自定义绘画。 (此外,我没有遵守 Java 变量 final
命名约定,因为它是数学而不是典型的编码。)
由于计算代码中有相当多的变量,所以我在最后添加了一个草图来显示变量代表的含义。
public class Main extends JFrame {
static BufferedImage image;
public static void main(String[] args) {
try {
image = ImageIO.read(ClassLoader.getSystemResource("img.png"));
} catch (IOException e) {
e.printStackTrace();
}
new Main();
}
public Main() {
getContentPane().setLayout(new BorderLayout(5, 10));
BufferedImage img2 = transform(15, 100, 300);
JLabel label1 = new JLabel(new ImageIcon(image));
label1.setHorizontalAlignment(JLabel.LEFT);
label1.setOpaque(true);
label1.setBackground(Color.YELLOW);
add(label1, BorderLayout.NORTH);
JLabel label2 = new JLabel(new ImageIcon(img2));
label2.setHorizontalAlignment(JLabel.LEFT);
label2.setOpaque(true);
label2.setBackground(Color.CYAN);
add(label2);
pack();
setDefaultCloseOperation(EXIT_ON_CLOSE);
setVisible(true);
}
static BufferedImage transform(int t, int x1, int x2) {
final double TH = Math.toRadians(t);
final int D = x2 - x1;
final int W = image.getWidth();
final int H = image.getHeight();
final int dD = (int) (D / (2 * TH) * Math.sin(2 * TH));
final int dH = (int) (D / TH * Math.pow(Math.sin(TH), 2));
final int pH = (int) ((W - x2) * Math.tan(2 * TH));
final int width = W - (D - dD);
final int height = (int) (H + dH + pH);
System.out.println(W + " " + H + " -> " + width + " " + height);
BufferedImage img2 = new BufferedImage(width, height, image.getType());
for (int x = 0; x < x1; x++) {
for (int y = 0; y < H; y++) {
int rgb = image.getRGB(x, y);
img2.setRGB(x, y, rgb);
}
}
for (int x = x1; x < x2; x++) {
for (int y = 0; y < H; y++) {
int rgb = image.getRGB(x, y);
int dx = (int) (D / (2 * TH) * Math.sin(2 * (x-x1) * TH / D));
int dy = (int) (D / TH * Math.pow(Math.sin((x-x1) * TH / D), 2));
img2.setRGB(x1 + dx, y + dy, rgb);
}
}
for (int x = x2; x < W; x++) {
for (int y = 0; y < H; y++) {
int rgb = image.getRGB(x, y);
int dp = (int) ((x - x2) * Math.tan(2 * TH));
img2.setRGB(x - (D - dD), y + dH + dp, rgb);
}
}
return img2;
}
}
至于计算,我留给你做作业;它只是几何/三角学,更属于Math.SE而不是SO。如果你不明白,我会给你指导。
请注意,此方法可能根本不快,但肯定可以优化,我也将其留给您。哦,而且不小心将 double
舍入为 int
,因此结果不是像素完美的。
关于java - 如何在java中弯曲图像,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38502480/