OpenCV，我们如何将Mat min规范化为max和max规范化为min？

我想将一个Mat标准化为最小值为255，最大值为0。将Mat标准化为0~255之间的值

例如，如果我们有一个像[0.02,0.002,0.0002]这样的数组，经过归一化后，我希望得到这样的结果：[3,26,255]，但现在当我使用NORM_MINMAX时，我得到了[255,26,3]

但是我没有找到任何函数来做NORM_MINMAX的反向运算

使用的代码：

但是我想要上面结果的倒数

更新：当我从垫子上减去255时，如下所示：

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

---

我终于找到了计算的方法，以下是步骤：

通过使用逆比例公式，我们可以很容易地计算范数_MINMAX的逆

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;

---

是的，这就是我想要的。< /P> Python或C++ + @ MasStkeleC++可以做一些类似于结果＝255的结果normalization@api55问题更新了，但结果不是我想要的。@BahramdunAdil api55的答案是正确的，很高兴看到你做到了：D对你的努力投赞成票；

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

[  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]

x = a*b/c

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;

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]