OpenCV:拍摄3通道RGB图像,分割通道并仅使用R + G查看图像

Sri*_*sha 24 c++ opencv

我想只查看RGB图像中的R + G通道,因为当蓝色通道被移除时,我可以更好地对比检测对象.我使用OpenCV来分割通道,但是在将蓝色通道设置为0之后合并它时,我的代码无法编译.

#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main( int argc, char** argv )
{
    if( argc != 2)
    {
     cout <<" Usage: display_image ImageToLoadAndDisplay" << endl;
     return -1;
    }

    Mat image,fin_img;
    image = imread(argv[1], CV_LOAD_IMAGE_COLOR);   // Read the file

    if(! image.data )                              // Check for invalid input
    {
        cout <<  "Could not open or find the image" << std::endl ;
        return -1;
    }

   namedWindow( "Display window", CV_WINDOW_AUTOSIZE );// Create a window for display.
                      // Show our image inside it.

    // Create Windows
    namedWindow("Red",1);
    namedWindow("Green",1);
    namedWindow("Blue",1);

    // Create Matrices (make sure there is an image in input!)

    Mat channel[3];
    imshow( "Original Image", image ); 


    // The actual splitting.
    split(image, channel);


   channel[0]=Mat::zeros(Size(image.rows, image.cols), CV_8UC1);//Set blue channel to 0

    //Merging red and green channels

    merge(channel,image);
    imshow("R+G", image);

    waitKey(0);//Wait for a keystroke in the window
    return 0;
}
Run Code Online (Sandbox Code Playgroud)

我可以对我出错的地方有任何反馈意见吗?我怀疑它是将蓝色通道设置为0.有没有更好的方法将其设置为0?有没有办法使用cvMixChannels()来做到这一点?

Har*_*ris 32

您需要更改这些行

    channel[0]=Mat::zeros(Size(image.rows, image.cols), CV_8UC1);//Set blue channel to 0

    //Merging red and green channels
    merge(channel,image);
Run Code Online (Sandbox Code Playgroud)

    channel[0]=Mat::zeros(image.rows, image.cols, CV_8UC1);//Set blue channel to 0

    //Merging red and green channels
    merge(channel,3,image);
Run Code Online (Sandbox Code Playgroud)

编辑

根据您的评论,这里是完整的代码和结果.

#include <iostream>
#include "opencv2/opencv.hpp"
#include <stdio.h>    

using namespace cv;
using namespace std;

int main( int argc, char** argv )
{
    if( argc != 2)
    {
     cout <<" Usage: display_image ImageToLoadAndDisplay" << endl;
     return -1;
    }

    Mat image,fin_img;
    image = imread("bgr.png", CV_LOAD_IMAGE_COLOR);   // Read the file

    if(! image.data )                              // Check for invalid input
    {
        cout <<  "Could not open or find the image" << std::endl ;
        return -1;
    }

   namedWindow( "Display window", CV_WINDOW_AUTOSIZE );// Create a window for display.
                      // Show our image inside it.

    // Create Windows
    namedWindow("Red",1);
    namedWindow("Green",1);
    namedWindow("Blue",1);

    // Create Matrices (make sure there is an image in input!)

    Mat channel[3];
    imshow( "Original Image", image );


    // The actual splitting.
    split(image, channel);


   channel[0]=Mat::zeros(image.rows, image.cols, CV_8UC1);//Set blue channel to 0

    //Merging red and green channels

    merge(channel,3,image);
    imshow("R+G", image);
    imwrite("dest.jpg",image);

    waitKey(0);//Wait for a keystroke in the window
    return 0;
}
Run Code Online (Sandbox Code Playgroud)

来源图片

在此输入图像描述

没有蓝色成分的结果

在此输入图像描述


Sri*_*sha 6

好吧,我开始使用mixChannels():我在上面的代码片段中添加了一个附加内容:

Mat gr( image.rows, image.cols, CV_8UC3);

// forming an array of matrices is a quite efficient operation,
// because the matrix data is not copied, only the headers
   Mat out[] = {gr};
// bgr[1] -> gr[1],
// bgr[2] -> gr[2], 
int from_to[] = {1,1, 2,2 };
mixChannels( &image, 1, out, 2, from_to, 2 );

imshow("R+G",gr);
Run Code Online (Sandbox Code Playgroud)

谢谢哈尔莎


Adi*_*vit 6

最有效的方法是不进行任何拆分和合并.这节省了时间和内存.

只需按位和 - 你的图像,cv::Scalar(0,255,255)这将把你的蓝色通道设置为零.

如:imshow("R+G", src & cv::Scalar(0,255,255));.


stu*_*men 5

Scalar(255,0,0)另一种方法是从源图像中减去

#include <opencv2/opencv.hpp>
using namespace cv;

int main(int argc, char **argv)
{
    Mat src = imread(argv[1], CV_LOAD_IMAGE_COLOR);
    imshow("src", src );
    src -= Scalar(255,0,0);
    imshow("Green and Red channels", src );
    waitKey();
    return 0;
}
Run Code Online (Sandbox Code Playgroud)