Question

我认为以下示例显示了warpAffine（带有预编译的Win64 dll的OpenCV 3.1）中的错误：

    Mat x(1,20, CV_32FC1);
    for (int iCol(0); iCol<x.cols; iCol++) { x.col(iCol).setTo(iCol); }
    Mat crop;
    Point2d c(10., 0.);
    double scale(1.3);
    int cropSz(11);
    double vals[6] = { scale, 0.0, c.x-(cropSz/2)*scale, 0.0, scale, c.y };
    Mat map(2, 3, CV_64FC1, vals);
    warpAffine(x, crop, map, Size(cropSz, 1), WARP_INVERSE_MAP |  INTER_LINEAR);
    float dx = (crop.at<float>(0, crop.cols-1) - crop.at<float>(0, 0))/(crop.cols-1);
    Mat constGrad = crop.clone().setTo(0);
    for (int iCol(0); iCol<constGrad.cols; iCol++) {
      constGrad.col(iCol) = c.x + (iCol-cropSz/2)*scale;
    }
    Mat diff = crop - constGrad;
    double err = norm(diff, NORM_INF);
    if (err>1e-4) {
      cout << "Problem:" << endl;
      cout << "computed output: " << crop << endl;
      cout << "expected output: " << constGrad << endl;
      cout << "difference: " << diff << endl;
      Mat dxImg;
      Mat dxFilt(1, 2, CV_32FC1);
      dxFilt.at<float>(0) = -1.0f;
      dxFilt.at<float>(1) = 1.0f;
      filter2D(crop, dxImg, crop.depth(), dxFilt);
      cout << "x-derivative in computed output: " << dxImg(Rect(1,0,10,1)) << endl;
      cout << "Note: We expect a constant difference of 1.3" << endl;
    }

这是程序输出：

Problem:
computed output: [3.5, 4.8125, 6.09375, 7.40625, 8.6875, 10, 11.3125, 12.59375, 13.90625, 15.1875, 16.5]
expected output: [3.5, 4.8000002, 6.0999999, 7.4000001, 8.6999998, 10, 11.3, 12.6, 13.9, 15.2, 16.5]
difference: [0, 0.012499809, -0.0062499046, 0.0062499046, -0.012499809, 0, 0.012499809, -0.0062503815, 0.0062503815, -0.012499809, 0]
x-derivative in computed output: [1.3125, 1.28125, 1.3125, 1.28125, 1.3125, 1.3125, 1.28125, 1.3125, 1.28125, 1.3125]
Note: We expect a constant difference of 1.3

我使用条目0,1,2，... n-1创建一个图像，并使用比例1.3切割（10,0）周围的区域。我还创建了一个预期的图像constGrad。但是，它们并不相同。更进一步，由于输入图像在x方向上具有恒定的导数并且映射是仿射的，因此在期望的图像中也期望恒定的梯度。问题不是边界问题，在图像内部也是如此。它也与WARP_INVERSE_MAP无关。

这是一个已知问题吗？对此有何评论？

cv :: warpAffine中的错误？

0 个答案: