opencv – 获取深度图

2023年9月30日 387次阅读

我无法从差异中获得正常的深度图.

这是我的代码：

#include "opencv2/core/core.hpp"
#include "opencv2/calib3d/calib3d.hpp"
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include "opencv2/contrib/contrib.hpp"
#include <cstdio>
#include <iostream>
#include <fstream>

using namespace cv;
using namespace std;

ofstream out("points.txt");

int main()
{
    Mat img1, img2;
    img1 = imread("images/im7rect.bmp");
    img2 = imread("images/im8rect.bmp");

    //resize(img1, img1, Size(320, 280));
    //resize(img2, img2, Size(320, 280));

    Mat g1,g2, disp, disp8;
    cvtColor(img1, g1, CV_BGR2GRAY);
    cvtColor(img2, g2, CV_BGR2GRAY);

    int sadSize = 3;
    StereoSGBM sbm;
    sbm.SADWindowSize = sadSize;
    sbm.numberOfDisparities = 144;//144; 128
    sbm.preFilterCap = 10; //63
    sbm.minDisparity = 0; //-39; 0
    sbm.uniquenessRatio = 10;
    sbm.speckleWindowSize = 100;
    sbm.speckleRange = 32;
    sbm.disp12MaxDiff = 1;
    sbm.fullDP = true;
    sbm.P1 = sadSize*sadSize*4;
    sbm.P2 = sadSize*sadSize*32;
    sbm(g1, g2, disp);

    normalize(disp, disp8, 0, 255, CV_MINMAX, CV_8U);

    Mat dispSGBMscale; 
    disp.convertTo(dispSGBMscale,CV_32F, 1./16); 

    imshow("image", img1);

    imshow("disparity", disp8);

    Mat Q;
    FileStorage fs("Q.txt", FileStorage::READ);
    fs["Q"] >> Q;
    fs.release();

    Mat points, points1;
    //reprojectImageTo3D(disp, points, Q, true);
    reprojectImageTo3D(disp, points, Q, false, CV_32F);
    imshow("points", points);

    ofstream point_cloud_file;
    point_cloud_file.open ("point_cloud.xyz");
    for(int i = 0; i < points.rows; i++) {
        for(int j = 0; j < points.cols; j++) {
            Vec3f point = points.at<Vec3f>(i,j);
            if(point[2] < 10) {
                point_cloud_file << point[0] << " " << point[1] << " " << point[2]
                    << " " << static_cast<unsigned>(img1.at<uchar>(i,j)) << " " << static_cast<unsigned>(img1.at<uchar>(i,j)) << " " << static_cast<unsigned>(img1.at<uchar>(i,j)) << endl;
            }
        }
    }
    point_cloud_file.close(); 

    waitKey(0);

    return 0;
}

我的图片是：

视差图：

我得到像这点云的smth：

Q等于：
[1.,0.,0.,-3.2883545303344727e 02,0.1,1.,0,
       -2.3697290992736816e 02,0.0,0.0,0.4497170185417110e 02,0.,
       0,-1.4446083962336606e-02,0.]

我尝试了很多其他的东西.我尝试了不同的图像,但没有人能够获得正常的深度图.

我究竟做错了什么？我应该使用reprojectImageTo3D还是使用其他方法代替它？虚拟化深度图的最佳方法是什么？ (我试过point_cloud库)
或者你能给我提供数据集和校准信息的工作示例,我可以运行它并获得深度图.或者如何从middlebury立体数据库(http://vision.middlebury.edu/stereo/data/)获取depth_map,我认为没有足够的校准信息.

编辑：
现在我得到像：

它当然更好,但仍然不正常.

Edited2：
我试过你说的话：

Mat disp;
disp = imread("disparity-image.pgm", CV_LOAD_IMAGE_GRAYSCALE);

Mat disp64; 
disp.convertTo(disp64,CV_64F, 1.0/16.0); 
imshow("disp", disp);

我用行cv :: minMaxIdx(…)得到这个结果：

当我评论这一行时：

Ps：另外请你能告诉我如何计算深度,只知道基线和焦距(以像素为单位).

最佳答案 我在OpenCV的
reprojectImageTo3D()和我自己的(见下文)之间进行了简单的比较,并且还对正确的差异和Q矩阵进行了测试.

// Reproject image to 3D
void customReproject(const cv::Mat& disparity, const cv::Mat& Q, cv::Mat& out3D)
{
    CV_Assert(disparity.type() == CV_32F && !disparity.empty());
    CV_Assert(Q.type() == CV_32F && Q.cols == 4 && Q.rows == 4);

    // 3-channel matrix for containing the reprojected 3D world coordinates
    out3D = cv::Mat::zeros(disparity.size(), CV_32FC3);

    // Getting the interesting parameters from Q, everything else is zero or one
    float Q03 = Q.at<float>(0, 3);
    float Q13 = Q.at<float>(1, 3);
    float Q23 = Q.at<float>(2, 3);
    float Q32 = Q.at<float>(3, 2);
    float Q33 = Q.at<float>(3, 3);

    // Transforming a single-channel disparity map to a 3-channel image representing a 3D surface
    for (int i = 0; i < disparity.rows; i++)
    {
        const float* disp_ptr = disparity.ptr<float>(i);
        cv::Vec3f* out3D_ptr = out3D.ptr<cv::Vec3f>(i);

        for (int j = 0; j < disparity.cols; j++)
        {
            const float pw = 1.0f / (disp_ptr[j] * Q32 + Q33);

            cv::Vec3f& point = out3D_ptr[j];
            point[0] = (static_cast<float>(j)+Q03) * pw;
            point[1] = (static_cast<float>(i)+Q13) * pw;
            point[2] = Q23 * pw;
        }
    }
}

这两种方法产生了几乎相同的结果,它们对我来说都是正确的.你能在你的视差图和Q矩阵上试试吗？您可以在我的GitHub上拥有我的测试环境.

注1：还要注意不要缩放两倍的差异(注释掉行disparity.convertTo(disparity,CV_32F,1.0 / 16.0);如果你的差异也是缩放的.)

注意2：它是使用OpenCV 3.0构建的,您可能需要更改包含.

点赞