位图转换:创建从透明位图中排除透明边的位图

Emi*_*nin 8 java android bitmap image-processing

我有一组位图.它们在某种程度上都是透明的,我事先并不知道哪些部分是透明的.我想从原始位图创建一个新的位图,它排除透明部分,但是在正方形中.我认为这张图片解释了它:

在此输入图像描述

我知道如何从现有位图创建位图,但我不知道如何找出哪个部分是透明的以及如何使用它来实现我的目标.

这就是我计划这样做的方式:

public Bitmap cutImage(Bitmap image) {
        Bitmap newBitmap = null; 

        int width = image.getWidth(); 
        int height = image.getHeight(); 

        newBitmap = Bitmap.createBitmap(width, height, Bitmap.Config.ARGB_8888); 

        Canvas canvas = new Canvas(newBitmap); 

        //This is where I need to find out correct values of r1 and r1.

        Rect r1 = new Rect(?, ?, ?, ?);
        Rect r2 = new Rect(?, ?, ?, ?);

        canvas.drawBitmap(image, r1, r2, null);

        return newBitmap; 
     }
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有谁知道如何实现这一目标?

编辑:

我使用以下算法来查找left,right,top和bottom值:

private int x1;
private int x2;
private int y1;
private int y2;

private void findRectValues(Bitmap image)
{
    for(int x = 0; x < image.getWidth(); x++)
    {
        for(int y = 0; y < image.getHeight(); y++)
        {
            if(image.getPixel(x, y) != Color.TRANSPARENT)
            {
                System.out.println("X1 is: " + x);
                x1 = x;
                break;
            }
        }

        if(x1 != 0)
            break;

    }

    for(int x = image.getWidth()-1; x > 0; x--)
    {
        for(int y = 0; y < image.getHeight(); y++)
        {
            if(image.getPixel(x, y) != Color.TRANSPARENT)
            {
                System.out.println("X2 is: " + x);
                x2 = x;
                break;
            }
        }

        if(x2 != 0)
            break;

    }

    for(int y = 0; y < image.getHeight(); y++)
    {
        for(int x = 0; x < image.getWidth(); x++)
        {
            if(image.getPixel(x, y) != Color.TRANSPARENT)
            {
                System.out.println("Y1 is: " + y);
                y1 = y;
                break;
            }
        }

        if(y1 != 0)
            break;

    }

    for(int y = image.getHeight()-1; y > 0; y--)
    {
        for(int x = 0; x < image.getWidth(); x++)
        {
            if(image.getPixel(x, y) != Color.TRANSPARENT)
            {
                System.out.println("Y2 is: " + y);
                y2 = y;
                break;
            }
        }

        if(y2 != 0)
            break;

    }
}
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gru*_*bFX 6

我觉得这样效率更高,对我来说效果很好

public Bitmap cropBitmapToBoundingBox(Bitmap picToCrop, int unusedSpaceColor) {
    int[] pixels = new int[picToCrop.getHeight() * picToCrop.getWidth()];
    int marginTop = 0, marginBottom = 0, marginLeft = 0, marginRight = 0, i;
    picToCrop.getPixels(pixels, 0, picToCrop.getWidth(), 0, 0,
            picToCrop.getWidth(), picToCrop.getHeight());

    for (i = 0; i < pixels.length; i++) {
        if (pixels[i] != unusedSpaceColor) {
            marginTop = i / picToCrop.getWidth();
            break;
        }
    }

    outerLoop1: for (i = 0; i < picToCrop.getWidth(); i++) {
        for (int j = i; j < pixels.length; j += picToCrop.getWidth()) {
            if (pixels[j] != unusedSpaceColor) {
                marginLeft = j % picToCrop.getWidth();
                break outerLoop1;
            }
        }
    }

    for (i = pixels.length - 1; i >= 0; i--) {
        if (pixels[i] != unusedSpaceColor) {
            marginBottom = (pixels.length - i) / picToCrop.getWidth();
            break;
        }
    }

    outerLoop2: for (i = pixels.length - 1; i >= 0; i--) {
        for (int j = i; j >= 0; j -= picToCrop.getWidth()) {
            if (pixels[j] != unusedSpaceColor) {
                marginRight = picToCrop.getWidth()
                        - (j % picToCrop.getWidth());
                break outerLoop2;
            }
        }
    }

    return Bitmap.createBitmap(picToCrop, marginLeft, marginTop,
            picToCrop.getWidth() - marginLeft - marginRight,
            picToCrop.getHeight() - marginTop - marginBottom);
}
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Pau*_*tos 5

如果您要裁剪的所有图像都或多或少位于原始画布的中心,我想您可以这样:

  1. 从每个边框开始向内工作,搜索非透明像素的图像
  2. 找到左上角像素和右下角后,您将获得所需的目标.
  3. 随意复制图像

现在,问题仍然是您认为透明像素.alpha trasparency是否重要?如果是这样,那么在你决定透明度足以从图像中剪切之前,有多少alpha?