如何在不改变其他通道的情况下有效地将cv :: Mat的给定通道设置为给定值？

Question

如何在cv::Mat不改变其他通道的情况下有效地将a的给定通道设置为给定值？例如,我想将其第四个通道(alpha通道)值设置为120(即半透明),如:

cv::Mat mat; // with type CV_BGRA
...
mat.getChannel(3) = Scalar(120); // <- this is what I want to do

PS:我当前的解决方案首先分成mat多个通道并设置alpha通道,然后将它们合并.

PS2:我知道如果我也想通过以下方式更改其他频道,我可以快速完成此操作:

mat.setTo(Scalar(54, 154, 65, 120)); 

---

使用通用解决方案更新:

这两种方法都可以将给定通道的所有mat值设置为给定值.它们将适用于所有矩阵,无论它们是否连续.

方法-1 - 更有效

- >基于@Antonio的答案,并由@MichaelBurdinov进一步改进

// set all mat values at given channel to given value
void setChannel(Mat &mat, unsigned int channel, unsigned char value)
{
    // make sure have enough channels
    if (mat.channels() < channel + 1)
        return;

    const int cols = mat.cols;
    const int step = mat.channels();
    const int rows = mat.rows;
    for (int y = 0; y < rows; y++) {
        // get pointer to the first byte to be changed in this row
        unsigned char *p_row = mat.ptr(y) + channel; 
        unsigned char *row_end = p_row + cols*step;
        for (; p_row != row_end; p_row += step)
            *p_row = value;
    }
}

方法-2 - 更优雅

- >基于@ MichaelBurdinov的回答

// set all mat values at given channel to given value
void setChannel(Mat &mat, unsigned int channel, unsigned char value)
{
    // make sure have enough channels
    if (mat.channels() < channel+1)
        return;

    // check mat is continuous or not
    if (mat.isContinuous())
        mat.reshape(1, mat.rows*mat.cols).col(channel).setTo(Scalar(value));
    else{
        for (int i = 0; i < mat.rows; i++)
            mat.row(i).reshape(1, mat.cols).col(channel).setTo(Scalar(value));
    }
}

---

PS:值得注意的是,根据文档,创建的矩阵Mat::create()总是连续的.但是,如果使用Mat::col(),Mat::diag()等等提取矩阵的一部分,或者为外部分配的数据构造矩阵头,则此类矩阵可能不再具有此属性.

Answer 1

如果您的图像在内存中是连续的,您可以使用以下技巧:

mat.reshape(1,mat.rows*mat.cols).col(3).setTo(Scalar(120));

如果不连续:

for(int i=0; i<mat.rows; i++)
    mat.row(i).reshape(1,mat.cols).col(3).setTo(Scalar(120));

编辑(感谢Antonio的评论):

请注意,此代码可能是最短的,并且它没有分配新的内存,但它根本没有效率.它可能比分割/合并方法更慢.当OpenCV应该对连续1个像素的非连续矩阵执行操作时,效率非常低.如果时间表现很重要,你应该使用@Antonio提出的解决方案.

只是对他的解决方案的一个小改进:

const int cols = img.cols;
const int step = img.channels();
const int rows = img.rows;
for (int y = 0; y < rows; y++) {
    unsigned char* p_row = img.ptr(y) + SELECTED_CHANNEL_NUMBER; //gets pointer to the first byte to be changed in this row, SELECTED_CHANNEL_NUMBER is 3 for alpha
    unsigned char* row_end = p_row + cols*step;
    for(; p_row != row_end; p_row += step)
         *p_row = value;
    }
}

这样可以节省x的递增操作和寄存器中的少一个值.在资源有限的系统上,它可以提供约5%的加速.否则时间表现将是相同的.

Answer 2

Mat img;
[...]
const int cols = img.cols;
const int step = img.channels();
const int rows = img.rows;
for (int y = 0; y < rows; y++) {
    unsigned char* p_row = img.ptr(y) + SELECTED_CHANNEL_NUMBER; //gets pointer to the first byte to be changed in this row, SELECTED_CHANNEL_NUMBER is 3 for alpha
    for (int x = 0; x < cols; x++) {
         *p_row = value;
         p_row += step; //Goes to the next byte to be changed
    }
}

注意:根据opencv术语的使用,这适用于连续矩阵和非连续矩阵:http://docs.opencv.org/modules/core/doc/basic_structures.html#bool%20Mat:: isContinuous%28% 29%20const