Question

我想将Mat标准化为最小值，将最大值设为255，将最大值设为0（将Mat标准化为 0〜255 之间）。

例如，如果规范化后我们有一个像[0.02, 0.002, 0.0002]这样的数组，我想得到这样的结果：[3, 26, 255]，但是现在当我使用NORM_MINMAX时，我得到了{{1 }}。

但是我没有找到任何执行[255, 26, 3]反向操作的函数。

使用的代码：

NORM_MINMAX

结果是：

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;

但是我想要以上结果的倒数。

更新：当我从垫子上减去255时：

[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]

结果是：

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

Answer 1

我终于找到了计算方法，步骤如下：

通过使用反比例公式，我们可以轻松计算NORM_MINMAX的反比例

x = a*b/c

其中 a =垫子元素的最小值， b = 255（最大值）和 c =我们要使用的元素要计算它。

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。

OpenCV，如何将Mat min最小化为max并将max最小化为min？

1 个答案: