位图转换:创建从透明位图中排除透明边的位图

时间:2011-07-16 12:44:08

标签: java android image-processing bitmap

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:

enter image description here

我知道如何从现有位图创建位图,但我不知道如何找出透明的部分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; 
     }

有谁知道如何实现这个目标?

修改

我使用以下算法来查找左,右,上和下值:

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;

    }
}

3 个答案:

答案 0 :(得分:6)

我认为这样效率更高一些,对我来说非常有用

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);
}

答案 1 :(得分:5)

如果你要裁剪的所有图像都或多或少地位于原始画布的中心,我想你可以这样:

  1. 从每个边框开始向前搜索图像,搜索非透明像素
  2. 找到左上角像素和右下角后,您将获得所需的目标。
  3. 随意复制图片
  4. 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,那么在确定透明度足以从图像中剪切之前,有多少alpha?

答案 2 :(得分:0)

要查找位图的非透明区域,请迭代x和y中的位图,找到非透明区域的最小值和最大值。然后将位图裁剪为这些坐标。

Bitmap CropBitmapTransparency(Bitmap sourceBitmap)
{
    int minX = sourceBitmap.getWidth();
    int minY = sourceBitmap.getHeight();
    int maxX = -1;
    int maxY = -1;
    for(int y = 0; y < sourceBitmap.getHeight(); y++)
    {
        for(int x = 0; x < sourceBitmap.getWidth(); x++)
        {
            int alpha = (sourceBitmap.getPixel(x, y) >> 24) & 255;
            if(alpha > 0)   // pixel is not 100% transparent
            {
                if(x < minX)
                    minX = x;
                if(x > maxX)
                    maxX = x;
                if(y < minY)
                    minY = y;
                if(y > maxY)
                    maxY = y;
            }
        }
    }
    if((maxX < minX) || (maxY < minY))
        return null; // Bitmap is entirely transparent

    // crop bitmap to non-transparent area and return:
    return Bitmap.createBitmap(sourceBitmap, minX, minY, (maxX - minX) + 1, (maxY - minY) + 1);
}