Question

我这里有一个尺寸为2156x1728的黑/白png文件，我希望使用AffineTransform旋转90度。生成的图像没有正确的比例。这里有一些示例代码（假设我已成功将png文件加载到BufferedImage中）：

public BufferedImage transform(BufferedImage image){

    System.out.println("Input width: "+ image.getWidth());
    System.out.println("Input height: "+ image.getHeight());

    AffineTransform affineTransform = new AffineTransform();
    affineTransform.setToQuadrantRotation(1, image.getWidth() / 2, image.getHeight() / 2);

    AffineTransformOp opRotated = new AffineTransformOp(affineTransform, AffineTransformOp.TYPE_BILINEAR);
    BufferedImage transformedImage = opRotated.createCompatibleDestImage(image, image.getColorModel());
    System.out.println("Resulting width: "+ transformedImage.getWidth());
    System.out.println("Resulting height: "+ transformedImage.getHeight());

    transformedImage = opRotated.filter(image, transformedImage);
    return transformedImage;
}

输出相应：

输入宽度：2156

输入高度：1728

结果宽度：1942

结果身高：1942年

旋转如何返回完全不相关的尺寸？

Answer 1

我不是这方面的专家，但为什么不创建一个正确大小的BufferedImage？另请注意，您的旋转中心不正确。您将需要在[w / 2，w / 2]或[h / 2，h / 2]的中心上旋转（宽度为w，高度为h），具体取决于您要旋转的象限，1或3 ，以及图像的相对高度和宽度。例如：

import java.awt.geom.AffineTransform;
import java.awt.image.AffineTransformOp;
import java.awt.image.BufferedImage;
import java.io.IOException;
import java.net.MalformedURLException;
import java.net.URL;

import javax.imageio.ImageIO;
import javax.swing.ImageIcon;
import javax.swing.JLabel;
import javax.swing.JOptionPane;

public class RotateImage {
   public static final String IMAGE_PATH = "http://duke.kenai.com/"
         + "models/Duke3DprogressionSmall.jpg";

   public static void main(String[] args) {
      try {
         URL imageUrl = new URL(IMAGE_PATH);
         BufferedImage img0 = ImageIO.read(imageUrl);
         ImageIcon icon0 = new ImageIcon(img0);

         int numquadrants = 1;
         BufferedImage img1 = transform(img0, numquadrants );
         ImageIcon icon1 = new ImageIcon(img1);

         JOptionPane.showMessageDialog(null, new JLabel(icon0));
         JOptionPane.showMessageDialog(null, new JLabel(icon1));

      } catch (MalformedURLException e) {
         e.printStackTrace();
      } catch (IOException e) {
         e.printStackTrace();
      }
   }

   public static BufferedImage transform(BufferedImage image, int numquadrants) {
      int w0 = image.getWidth();
      int h0 = image.getHeight();
      int w1 = w0;
      int h1 = h0;

      int centerX = w0 / 2;
      int centerY = h0 / 2;

      if (numquadrants % 2 == 1) {
         w1 = h0;
         h1 = w0;
      }

      if (numquadrants % 4 == 1) {
         if (w0 > h0) {
            centerX = h0 / 2;
            centerY = h0 / 2;
         } else if (h0 > w0) {
            centerX = w0 / 2;
            centerY = w0 / 2;
         }
         // if h0 == w0, then use default
      } else if (numquadrants % 4 == 3) {
         if (w0 > h0) {
            centerX = w0 / 2;
            centerY = w0 / 2;
         } else if (h0 > w0) {
            centerX = h0 / 2;
            centerY = h0 / 2;
         }
         // if h0 == w0, then use default
      }

      AffineTransform affineTransform = new AffineTransform();
      affineTransform.setToQuadrantRotation(numquadrants, centerX, centerY);

      AffineTransformOp opRotated = new AffineTransformOp(affineTransform,
            AffineTransformOp.TYPE_BILINEAR);

      BufferedImage transformedImage = new BufferedImage(w1, h1,
            image.getType());

      transformedImage = opRotated.filter(image, transformedImage);
      return transformedImage;
   }
}

编辑1
你问：

你能解释一下为什么它必须是[w / 2，w / 2]或[h / 2，h / 2]吗？

为了解释这一点，最好对矩形进行可视化和物理操作：

切出一张长方形纸，将其放在一张纸上，使其左上角位于纸张的左上角 - 这就是您在屏幕上的图像。现在检查一下你需要旋转那个矩形1或3个象限的位置，使它的新左上角覆盖在纸张的上方，你就会明白为什么需要使用[w / 2，w / 2]或[h / 2，h / 2]。

Answer 2

上述解决方案存在宽度问题;图像的高度以下代码独立于w＆gt; h || h>瓦特

public static BufferedImage rotateImage(BufferedImage image, int quadrants) {

    int w0 = image.getWidth();
    int h0 = image.getHeight();
    int w1 = w0;
    int h1 = h0;
    int centerX = w0 / 2;
    int centerY = h0 / 2;

    if (quadrants % 2 == 1) {
        w1 = h0;
        h1 = w0;
    }

    if (quadrants % 4 == 1) {
        centerX = h0 / 2;
        centerY = h0 / 2;
    } else if (quadrants % 4 == 3) {
        centerX = w0 / 2;
        centerY = w0 / 2;
    }

    AffineTransform affineTransform = new AffineTransform();
    affineTransform.setToQuadrantRotation(quadrants, centerX, centerY);
    AffineTransformOp opRotated = new AffineTransformOp(affineTransform,
            AffineTransformOp.TYPE_BILINEAR);
    BufferedImage transformedImage = new BufferedImage(w1, h1,
            image.getType());
    transformedImage = opRotated.filter(image, transformedImage);

    return transformedImage;

}

Answer 3

furykid's回答很棒并帮了我很多忙。但它并不是那么完美。如果图像是矩形，则生成的旋转图像可能在一侧包含一些额外的黑色像素。

我尝试了一张Marty Feldman照片，原版和结果可在此链接中查看： Marty Feldman rotation tests

在黑色背景上很难看到，但在任何图像编辑软件上，很容易看到生成图像右侧和底部的黑色小边框。对于某些人来说这可能不是问题，但是如果它适合您，那么这里是固定代码（我保留了原文作为评论以便于比较）：

public BufferedImage rotateImage(BufferedImage image, int quadrants) {

    int w0 = image.getWidth();
    int h0 = image.getHeight();
    /* These are not necessary anymore
    * int w1 = w0;
    * int h1 = h0;
    */
    int centerX = w0 / 2;
    int centerY = h0 / 2;

    /* This is not necessary anymore
    * if (quadrants % 2 == 1) {
    *     w1 = h0;
    *     h1 = w0;
    * }
    */

    //System.out.println("Original dimensions: "+w0+", "+h0);
    //System.out.println("Rotated dimensions: "+w1+", "+h1);

    if (quadrants % 4 == 1) {
        centerX = h0 / 2;
        centerY = h0 / 2;
    } else if (quadrants % 4 == 3) {
        centerX = w0 / 2;
        centerY = w0 / 2;
    }

    //System.out.println("CenterX: "+centerX);
    //System.out.println("CenterY: "+centerY);

    AffineTransform affineTransform = new AffineTransform();
    affineTransform.setToQuadrantRotation(quadrants, centerX, centerY);
    AffineTransformOp opRotated = new AffineTransformOp(affineTransform,
            AffineTransformOp.TYPE_BILINEAR);

    /*Old code for comparison
    //BufferedImage transformedImage = new BufferedImage(w1, h1,image.getType());
    //transformedImage = opRotated.filter(image, transformedImage);
    */
    BufferedImage transformedImage = opRotated.filter(image, null);
    return transformedImage;

}

警告：提前意见。我不确定为什么会发生这种情况，但我有猜测。 如果你能更好地解释，请编辑。

我相信这个＆＃34;故障的原因＆＃34;是因为奇怪的尺寸。在计算新BufferedImage的尺寸时，如果正确值为136.5，则高度273将生成136的centerY。这可能会导致旋转发生在略微偏离中心的位置。但是，通过将null发送到filter作为目标图片，＆＃34;使用源BufferedImage＆＃34;创建ColorModel，这似乎有效最好的。

AffineTransform会截断图片，我有什么不对？

3 个答案: