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:
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;
}
}
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:
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);
}