opencv背景减法

Question

我有一个背景场景的图像和前面有物体的同一场景的图像.现在我想用背景减法创建前景中对象的蒙版.两个图像都是RGB.

我已经创建了以下代码:

cv::Mat diff;
diff.create(orgImage.dims, orgImage.size, CV_8UC3);
diff = abs(orgImage-refImage);

cv::Mat mask(diff.rows, diff.cols, CV_8U, cv::Scalar(0,0,0));
//mask = (diff > 10);

for (int j=0; j<diff.rows; j++) {
    // get the address of row j
    //uchar* dataIn= diff.ptr<uchar>(j);
    //uchar* dataOut= mask.ptr<uchar>(j);
    for (int i=0; i<diff.cols; i++) {
        if(diff.at<cv::Vec3b>(j,i)[0] > 30 || diff.at<cv::Vec3b>(j,i)[1] > 30 || diff.at<cv::Vec3b>(j,i)[2] > 30)
            mask.at<uchar>(j,i) = 255;
    }
}

我不知道我这样做是否正确？

Answer 1

看看OpenCV 的inRange函数.这将允许您为3通道图像同时设置多个阈值.

因此,要创建您正在寻找的面具,请执行以下操作:

inRange(diff, Scalar(30, 30, 30), Scalar(255, 255, 255), mask);

这也应该比尝试自己访问每个像素更快.

编辑:如果皮肤检测是你想要做的,我会先做皮肤检测,然后做背景扣除删除背景.否则,您的皮肤探测器将不得不考虑减法引起的强度偏移.

看看我的其他答案,关于皮肤检测的好技术.

编辑:

这更快吗？

int main(int argc, char* argv[])
{
    Mat fg = imread("fg.jpg");
    Mat bg = imread("bg.jpg");

    cvtColor(fg, fg, CV_RGB2YCrCb);
    cvtColor(bg, bg, CV_RGB2YCrCb);

    Mat distance = Mat::zeros(fg.size(), CV_32F);

    vector<Mat> fgChannels;
    split(fg, fgChannels);

    vector<Mat> bgChannels;
    split(bg, bgChannels);

    for(size_t i = 0; i < fgChannels.size(); i++)
    {
        Mat temp = abs(fgChannels[i] - bgChannels[i]);
        temp.convertTo(temp, CV_32F);

        distance = distance + temp;
    }


    Mat mask;
    threshold(distance, mask, 35, 255, THRESH_BINARY);

    Mat kernel5x5 = getStructuringElement(MORPH_RECT, Size(5, 5));
    morphologyEx(mask, mask, MORPH_OPEN, kernel5x5);

    imshow("fg", fg);
    imshow("bg", bg);
    imshow("mask", mask);

    waitKey();

    return 0;
}

此代码根据您的输入图像生成此蒙版:

最后,这是我使用简单的阈值方法得到的:

    Mat diff = fgYcc - bgYcc;
    vector<Mat> diffChannels;
    split(diff, diffChannels);

    // only operating on luminance for background subtraction...
    threshold(diffChannels[0], bgfgMask, 1, 255.0, THRESH_BINARY_INV);

    Mat kernel5x5 = getStructuringElement(MORPH_RECT, Size(5, 5));
    morphologyEx(bgfgMask, bgfgMask, MORPH_OPEN, kernel5x5);

这产生以下面具: