在我的论文中,我使用opencv在c ++中编写了一个块匹配算法。 它正在制作灰度图片,并通过他的绝对像素来处理IPLImage。
我必须在相同大小的块(8x8 pxls)中分配IPLImage。为了访问块内的像素值,我计算像素值并以这种方式访问像素值:
for (int yBlock = 0; yBlock < maxYBlocks; yBlock++){
for (int xBlock = 0; yxlock < maxXBlocks; xBlock++){
for (int yPixel = 0; yPixel < 8; yPixel++){
for (int xPixel = 0; xPixel < 8; xPixel++){
pixelAdress = yBlock*imageWidth*8 + xBlock*8 + yPixel*imageWidth + xPixel;
unsigned char* imagePointer = (unsigned char*)(img->imageData);
pixelValue = imagePointer[pixelAdress];
}
}
}
}
我真的没有对行和列进行迭代,而且效果很好!
现在我有一个彩色的IPLImage(没有灰度),也不知道如何访问r,g,b像素值。
我在这个论坛上发现了这个
for( row = 0; row < img->height; row++ ){
for ( col = 0; col < img->width; col++ ){
b = (int)img->imageData[img->widthStep * row + col * 3];
g = (int)img->imageData[img->widthStep * row + col * 3 + 1];
r = (int)img->imageData[img->widthStep * row + col * 3 + 2];
}
}
但我不确定如何在我的计算像素上使用它。将它乘以3是否正确(因为我不迭代行并添加0,1或2?例如:
pixelValueR = imagePointer[pixelAdress*3 + 2];
pixelValueG = imagePointer[pixelAdress*3 + 1];
pixelValueB = imagePointer[pixelAdress*3 + 0];
或者我必须使用之前使用过imageWidth的widthStep,如下所示:
pixelAdressR = pixelAdress = yBlock*img->widthStep*8 + xBlock*8*3 + yPixel*img->widthStep + xPixel*3 + 2;
pixelAdressG = pixelAdress = yBlock*img->widthStep*8 + xBlock*8*3 + yPixel*img->widthStep + xPixel*3 + 1;
pixelAdressB = pixelAdress = yBlock*img->widthStep*8 + xBlock*8*3 + yPixel*img->widthStep + xPixel*3;
所以访问
pixelValueR = imagePointer[pixelAdressR];
pixelValueG = imagePointer[pixelAdressG];
pixelValueB = imagePointer[pixelAdressB];
答案 0 :(得分:1)
如果是多频道Mat
(本例中为BGR),您可以使用,如here所述
Vec3b intensity = img.at<Vec3b>(y, x);
uchar blue = intensity.val[0];
uchar green = intensity.val[1];
uchar red = intensity.val[2];
答案 1 :(得分:0)
适用于Mat
(例如Mat img
)
灰度(8UC1
):
uchar intensity = img.at<uchar>(y, x);
彩色图像(BGR颜色排序,imread
返回的默认格式):
Vec3b intensity = img.at<Vec3b>(y, x);
uchar blue = intensity.val[0];
uchar green = intensity.val[1];
uchar red = intensity.val[2];
适用于IplImage
(例如IplImage* img
)
灰度:
uchar intensity = CV_IMAGE_ELEM(img, uchar, h, w);
彩色图片:
uchar blue = CV_IMAGE_ELEM(img, uchar, y, x*3);
uchar green = CV_IMAGE_ELEM(img, uchar, y, x*3+1);
uchar red = CV_IMAGE_ELEM(img, uchar, y, x*3+2);
答案 2 :(得分:0)
不确定您的整个算法,目前无法对其进行测试,但对于IplImages,内存对齐如下:
1. row
baseadress + 0 = b of [0]
baseadress + 1 = g of [0]
baseadress + 2 = r of [0]
baseadress + 3 = b of [1]
etc
2. row
baseadress + widthStep + 0 = b
baseadress + widthStep + 1 = g
baseadress + widthStep + 2 = r
所以如果你有n*m
个大小为8x8
的无符号字符bgr数据块,并且想要在块[x,y]
中循环变量[bx,by]
,你就可以这样做:
baseadress + (by*8+ y_in_block)*widthStep + (bx*8+x)*3 +0 = b
baseadress + (by*8+ y_in_block)*widthStep + (bx*8+x)*3 +1 = g
baseadress + (by*8+ y_in_block)*widthStep + (bx*8+x)*3 +2 = r
因为行by*8+y is adress
baseadress +(* * 8 + y_in_block)* widthStep`
和列bx*8+x
是地址偏移(bx*8+x)*3