从 cv mat 中删除行或列的最佳方法是啥

Posted 2023-02-22

【中文标题】从 cv mat 中删除行或列的最佳方法是啥

【英文标题】：what is the best way to remove a row or col from a cv mat从 cv mat 中删除行或列的最佳方法是什么

【发布时间】：2015-04-17 10:32:44

【问题描述】：

假设我有一个 mat 对象，如下所示：

mat = 
   [75, 97, 66, 95, 15, 22;
    24, 21, 71, 72, 34, 66;
    21, 69, 88, 72, 64, 1;
    26, 47, 26, 40, 95, 24;
    70, 37, 9, 83, 16, 83];

我想从中删除一行，说第二行有这样的垫子：

 [75, 97, 66, 95, 15, 22;
 21, 69, 88, 72, 64, 1;
 26, 47, 26, 40, 95, 24;
 70, 37, 9, 83, 16, 83]

或删除一个 col 说 col 3：

[75, 97,  95, 15, 22;
 24, 21,  72, 34, 66;
 21, 69,  72, 64, 1;
 26, 47,  40, 95, 24;
 70, 37,  83, 16, 83]

最快的方法是什么？我可以将矩阵分解为 ROI，然后将它们相互合并，但是有没有更好的方法？

【参考方案1】：

我测试了两种方式：

使用cv::Rect 和cv::Mat::copyTo：

// Removing a row
cv::Mat matIn;    // Matrix of which a row will be deleted.
int row;          // Row to delete.
int col;          // Column to delete.
cv::Mat matOut;   // Result: matIn less that one row.

if ( row > 0 ) // Copy everything above that one row.

    cv::Rect rect( 0, 0, size.width, row );
    matIn( rect ).copyTo( matOut( rect ) );


if ( row < size.height - 1 ) // Copy everything below that one row.

    cv::Rect rect1( 0, row + 1, size.width, size.height - row - 1 );
    cv::Rect rect2( 0, row, size.width, size.height - row - 1 );
    matIn( rect1 ).copyTo( matOut( rect2 ) );


// Removing a column
if ( col > 0 ) // Copy everything left of that one column.

    cv::Rect rect( 0, 0, col, size.height );
    matIn( rect ).copyTo( matOut( rect ) );


if ( col < size.width - 1 ) // Copy everything right of that one column.

    cv::Rect rect1( col + 1, 0, size.width - col - 1, size.height );
    cv::Rect rect2( col,     0, size.width - col - 1, size.height );
    matIn( rect1 ).copyTo( matOut( rect2 ) );

使用std::memcpy 和cv::Mat::data：

// Removing a row
int rowSizeInBytes = size.width * sizeof( T );

if ( row > 0 )

    int numRows  = row;
    int numBytes = rowSizeInBytes * numRows;
    std::memcpy( matOut.data, matIn.data, numBytes );


if ( row < size.height - 1 )

    int matOutOffset = rowSizeInBytes * row;
    int matInOffset  = matOutOffset + rowSizeInBytes;

    int numRows  = size.height - ( row + 1 );
    int numBytes = rowSizeInBytes * numRows;
    std::memcpy( matOut.data + matOutOffset , matIn.data + matInOffset, numBytes );


// Removing a column
int rowInInBytes  = size.width * sizeof( T );
int rowOutInBytes = ( size.width - 1 ) * sizeof( T );

if ( col > 0 )

    int matInOffset = 0;
    int matOutOffset = 0;
    int numCols = col;
    int numBytes = numCols * sizeof( T );

    for ( int y = 0; y < size.height; ++y )
    
        std::memcpy( matOut.data + matOutOffset, matIn.data + matInOffset, numBytes );

        matInOffset  += rowInInBytes;
        matOutOffset += rowOutInBytes;
    


if ( col < size.width - 1 )

    int matInOffset = ( col + 1 ) * sizeof( T );
    int matOutOffset = col * sizeof( T );
    int numCols = size.width - ( col + 1 );
    int numBytes = numCols * sizeof( T );

    for ( int y = 0; y < size.height; ++y )
    
        std::memcpy( matOut.data + matOutOffset, matIn.data + matInOffset, numBytes );

        matInOffset  += rowInInBytes;
        matOutOffset += rowOutInBytes;
    

第一种方法的时序测试显示：

Removed:      row
Method:       cv::Rect + cv::Mat::copyTo()
Iterations:   10000
Size:         [500 x 500]
Best time:    67ms
Worst time:   526ms
Average time: 70.9061ms
Median time:  70ms

Removed:      column
Method:       cv::Rect + cv::Mat::copyTo()
Iterations:   10000
Size:         [500 x 500]
Best time:    64ms
Worst time:   284ms
Average time: 80.3893ms
Median time:  79ms

对于第二种方法：

Removed:      row
Method:       std::memcpy and/or for-loop
Iterations:   10000
Size:         [500 x 500]
Best time:    31ms
Worst time:   444ms
Average time: 68.9445ms
Median time:  68ms

Removed:      column
Method:       std::memcpy and/or for-loop
Iterations:   10000
Size:         [500 x 500]
Best time:    49ms
Worst time:   122ms
Average time: 79.3948ms
Median time:  78ms

因此，考虑到接近的时序结果和简短的实现，第一种方法似乎更合适。 我发布了minimal working example on github 以验证此测试的结果。

【参考方案2】：

删除第 N 行：

memmove(mat + N * x_size, 
        mat + (N + 1) * x_size, 
        x_size * sizeof(int) * (y_size - N - 1));

删除列 N：

for(int y = 0; y < y_size; y++)
  memmove(mat + N + y * (x_size - 1), 
  mat + N + y * x_size + 1, 
  (x_size - 1) * sizeof(int));

ATTN：第二个代码（删除列）读取矩阵后面的额外行。在大多数情况下，这是可以接受的，并且算法保持简单。如果需要，修改代码以将正确的大小传递到最后一个 memmove。