c++ - 在另一个分配的矩阵中为特定范围创建cv::Mat header

我希望将给定ColumnVector的值右移并复制到比ColumnVector大一个元素的另一个RowVector上。我想通过将columnVector复制到RowVector内的某个范围来完成此操作。

// The given columnVector ( this code snippet is added to make the error reproducable)
int yDimLen = 26; // arbitrary value
cv::Mat columnVector(yDimLen, 1, CV_32SC1, cv::Scalar(13)); // one column AND yDimLen rows
for (int r = 0; r < yDimLen; r++)   // chech the values
{
    printf("%d\t", *(columnVector.ptr<int>(r)));
}

// Copy elemnets of columnVector to rowVector starting from column number 1
cv::Mat rowVector(1, yDimLen + 1, CV_32SC1, cv::Scalar(0));
cv::Mat rowVector_rangeHeader = rowVector.colRange(cv::Range(1, yDimLen + 1)); // header matrix, 1 in included AND (yDimLen + 1) is excluded
columnVector.copyTo(rowVector_rangeHeader);

// check the values
printf("\n---------------------------------\n");
for (int c = 0; c < yDimLen; c++)
{
    printf("%d\t", rowVector_rangeHeader.ptr<int>(0)[c]); // displays 13
}

printf("\n---------------------------------\n");
for (int c = 0; c < yDimLen + 1; c++)
{
    printf("%d\t", rowVector.ptr<int>(0)[c]); // displays 0 !!!
}

header 矩阵是否指向包含rowVector的相同内存地址？如果没有，如何将数据复制到行的某些部分？

最佳答案

从the docs of cv::Mat::copyTo :



只要目标的形状或数据类型与源不匹配，就需要重新分配。您的情况就是形状。为了演示这一点，让我们在copyTo调用周围添加一些跟踪:

std::cout << "Shapes involved in `copyTo`:\n";
std::cout << "Source: " << columnVector.size() << "\n";
std::cout << "Destination: " << rowVector_rangeHeader.size() << "\n";

columnVector.copyTo(rowVector_rangeHeader);

std::cout << "After the `copyTo`:\n";
std::cout << "Destination: " << rowVector_rangeHeader.size() << "\n";

输出:

Shapes involved in `copyTo`:
Source: [1 x 26]
Destination: [26 x 1]
After the `copyTo`:
Destination: [1 x 26]

解决方案很简单-reshape源匹配目标。

columnVector.reshape(1, 1).copyTo(rowVector_rangeHeader);

现在我们得到以下输出:

Shapes involved in `copyTo`:
Source: [1 x 26]
Destination: [26 x 1]
After the `copyTo`:
Destination: [26 x 1]

和rowVector的内容如下:

[0, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13]

作为替代方案，您可以只将std::copy与MatIterator一起使用。

std::copy(columnVector.begin<int>()
    , columnVector.end<int>()
    , rowVector_rangeHeader.begin<int>());

rowVector的内容与以前相同:

[0, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13]