opencv - OpenCV，我们如何将 Mat min 标准化为 max 并将 max 标准化为 min？

Question

我想将 a 标准化Mat为最小值为 255 和最大值为 0 （标准化0~255Mat之间的值）。

例如，如果我们有一个像[0.02, 0.002, 0.0002]标准化后的数组，我想得到这样的结果：[3, 26, 255]，但现在当我使用时，NORM_MINMAX我得到了[255, 26, 3]。

但是我没有找到任何函数来做NORM_MINMAX.

使用的代码：

cv::Mat mat(10, 10, CV_64F);
mat.setTo(0);
mat.row(0) = 0.02;
mat.row(1) = 0.002;
mat.row(2) = 0.0002;
cv::normalize(mat, mat, 255, 0, cv::NORM_MINMAX);
mat.convertTo(mat, CV_8UC1);
std::cout << mat << std::endl;

结果是：

[255, 255, 255, 255, 255, 255, 255, 255, 255, 255;
  26,  26,  26,  26,  26,  26,  26,  26,  26,  26;
   3,   3,   3,   3,   3,   3,   3,   3,   3,   3;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0]

但我想要上述结果的倒数。

更新：当我从垫子中减去 255 时：

cv::subtract(255, mat, mat, mat); // the last mat acts as mask
std::cout << mat << std::endl;

结果是：

[  0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
 229, 229, 229, 229, 229, 229, 229, 229, 229, 229;
 252, 252, 252, 252, 252, 252, 252, 252, 252, 252;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0]

标签： opencv

Answer 1

我终于找到了如何计算的方法，以下是步骤：

通过使用反比例公式，我们可以很容易地计算出NORM_MINMAX

x = a*b/c

其中a = mat 元素的最小值，b =255（最大值），c = 我们要计算它的元素。

cv::Mat mat(10, 10, CV_64F);

mat.setTo(0);
mat.row(0) = 0.02;
mat.row(1) = 0.002;
mat.row(2) = 0.0002;
std::cout << mat<< std::endl;

// craete a mask
cv::Mat mask(mat.size(), CV_8U);
mask.setTo(0);
mask.row(0) = 255;
mask.row(1) = 255;
mask.row(2) = 255;

// find the min value
double min;
cv::minMaxLoc(mat, &min, nullptr, nullptr, nullptr, mask);
std::cout << "min=" << min << std::endl;

// unfortunately opencv divide operation does not support mask, so we need some extra steps to perform.
cv::Mat result, maskNeg;
cv::divide(min*255, mat, result); // this is the magic line
cv::bitwise_not(mask, maskNeg);
mat.copyTo(result, maskNeg);
std::cout << result << std::endl;

// convert to 8bit
result .convertTo(result , CV_8UC1);
std::cout << "the final result:" << std::endl;
std::cout << temp << std::endl;

和输出：

original mat
[0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02;
 0.002, 0.002, 0.002, 0.002, 0.002, 0.002, 0.002, 0.002, 0.002, 0.002;
 0.0002, 0.0002, 0.0002, 0.0002, 0.0002, 0.0002, 0.0002, 0.0002, 0.0002, 0.0002;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

min=0.0002

the calculated min-max
[2.55, 2.55, 2.55, 2.55, 2.55, 2.55, 2.55, 2.55, 2.55, 2.55;
 25.5, 25.5, 25.5, 25.5, 25.5, 25.5, 25.5, 25.5, 25.5, 25.5;
 255, 255, 255, 255, 255, 255, 255, 255, 255, 255;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 the final result:
[  3,   3,   3,   3,   3,   3,   3,   3,   3,   3;
  26,  26,  26,  26,  26,  26,  26,  26,  26,  26;
 255, 255, 255, 255, 255, 255, 255, 255, 255, 255;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0;
   0,   0,   0,   0,   0,   0,   0,   0,   0,   0]

是的，这就是我想要的。