Question

所以我有一个有火山的图像文件。其他一切都是0xFFFF00FF（不透明洋红色）。我想用0（透明）替换包含该颜色的每个像素。到目前为止，我的方法看起来像这样：

public static BufferedImage replace(BufferedImage image, int target, int preferred) {
    int width = image.getWidth();
    int height = image.getHeight();
    BufferedImage newImage = new BufferedImage(width, height, image.getType());
    int color;

    for (int i = 0; i < width; i++) {
        for (int j = 0; j < height; j++) {
            color = image.getRGB(i, j);
            if (color == target) {
                newImage.setRGB(i, j, preferred);
            }
            else {
                newImage.setRGB(i, j, color);
            }
        }
    }

    return newImage;
}

这很好但看起来很慢。我见过有人这样做，但我不知道发生了什么。如果有人知道更好的方法，我非常希望听到它。

Answer 1

虽然我还没有机会对此进行彻底测试，但使用LookupOp可能会从加速中受益：

public class ColorMapper
extends LookupTable {

    private final int[] from;
    private final int[] to;

    public ColorMapper(Color from,
                       Color to) {
        super(0, 4);

        this.from = new int[] {
            from.getRed(),
            from.getGreen(),
            from.getBlue(),
            from.getAlpha(),
        };
        this.to = new int[] {
            to.getRed(),
            to.getGreen(),
            to.getBlue(),
            to.getAlpha(),
        };
    }

    @Override
    public int[] lookupPixel(int[] src,
                             int[] dest) {
        if (dest == null) {
            dest = new int[src.length];
        }

        int[] newColor = (Arrays.equals(src, from) ? to : src);
        System.arraycopy(newColor, 0, dest, 0, newColor.length);

        return dest;
    }
}

使用它就像创建LookupOp一样简单：

Color from = Color.decode("#ff00ff");
Color to = new Color(0, true);
BufferedImageOp lookup = new LookupOp(new ColorMapper(from, to), null);
BufferedImage convertedImage = lookup.filter(image, null);

Answer 2

为避免迭代像素，请更改基础ColorModel。 Here就是一个例子。下面是作者采用原始BufferedImage并应用新颜色模型的片段。

 private static BufferedImage createImage() {
    int width = 200;
    int height = 200;
    // Generate the source pixels for our image
    // Lets just keep it to a simple blank image for now

    byte[] pixels = new byte[width * height];
    DataBuffer dataBuffer = new DataBufferByte(pixels, width*height, 0);
    SampleModel sampleModel = new SinglePixelPackedSampleModel(
    DataBuffer.TYPE_BYTE, width, height, new int[] {(byte)0xf});
    WritableRaster raster = Raster.createWritableRaster(
    sampleModel, dataBuffer, null);
    return new BufferedImage(createColorModel(0), raster, false, null);
}

private static ColorModel createColorModel(int n) {
    // Create a simple color model with all values mapping to
    // a single shade of gray
    // nb. this could be improved by reusing the byte arrays

    byte[] r = new byte[16];
    byte[] g = new byte[16];
    byte[] b = new byte[16];
    for (int i = 0; i < r.length; i++) {
        r[i] = (byte) n;
        g[i] = (byte) n;
        b[i] = (byte) n;
    }
    return new IndexColorModel(4, 16, r, g, b);
}

private BufferedImage image = createImage();
image = new BufferedImage(createColorModel(e.getX()), image.getRaster(), false, null);

Answer 3

您可以像这样获取缓冲图像的pixels[]数组

int[] pixels = ((DataBufferInt) newImg().getDataBuffer()).getData();

然后设置你的颜色

for (int i = 0; i < width; i++) {
    for (int j = 0; j < height; j++) {
        color = pixels[y * width + x];
        if (color == target) {
            pixels[y * width + x] = preferred;
        }
        else {
            pixels[y * width + x] = color;
        }
    }
}

这比使用setRGB()

略微加快

如何在BufferedImage中用另一种颜色替换颜色

3 个答案: