我正试图找到二进制图像的非零(x,y)坐标。
我找到了一些函数countNonZero()
的引用,它只计算非零坐标和findNonZero()
,我不确定如何访问或使用它,因为它似乎已被删除完整的文档。
This是我找到的最接近的参考,但仍然没有任何帮助。我将不胜感激。
编辑: - 要指定,这是使用C ++
答案 0 :(得分:32)
Here解释了findNonZero()
如何保存非零元素。以下代码对于访问二进制图像的非零坐标非常有用。方法1在OpenCV中使用findNonZero()
,方法2检查每个像素以找到非零(正)的。
方法1:
#include <iostream>
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
using namespace std;
using namespace cv;
int main(int argc, char** argv) {
Mat img = imread("binary image");
Mat nonZeroCoordinates;
findNonZero(img, nonZeroCoordinates);
for (int i = 0; i < nonZeroCoordinates.total(); i++ ) {
cout << "Zero#" << i << ": " << nonZeroCoordinates.at<Point>(i).x << ", " << nonZeroCoordinates.at<Point>(i).y << endl;
}
return 0;
}
方法2:
#include <iostream>
#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
using namespace std;
using namespace cv;
int main(int argc, char** argv) {
Mat img = imread("binary image");
for (int i = 0; i < img.cols; i++ ) {
for (int j = 0; j < img.rows; j++) {
if (img.at<uchar>(j, i) > 0) {
cout << i << ", " << j << endl; // Do your operations
}
}
}
return 0;
}
答案 1 :(得分:5)
以下源代码为supplied for OpenCV 2.4.3,可能会有所帮助:
#include <opencv2/core/core.hpp>
#include <vector>
/*! @brief find non-zero elements in a Matrix
*
* Given a binary matrix (likely returned from a comparison
* operation such as compare(), >, ==, etc, return all of
* the non-zero indices as a std::vector<cv::Point> (x,y)
*
* This function aims to replicate the functionality of
* Matlab's command of the same name
*
* Example:
* \code
* // find the edges in an image
* Mat edges, thresh;
* sobel(image, edges);
* // theshold the edges
* thresh = edges > 0.1;
* // find the non-zero components so we can do something useful with them later
* vector<Point> idx;
* find(thresh, idx);
* \endcode
*
* @param binary the input image (type CV_8UC1)
* @param idx the output vector of Points corresponding to non-zero indices in the input
*/
void find(const cv::Mat& binary, std::vector<cv::Point> &idx) {
assert(binary.cols > 0 && binary.rows > 0 && binary.channels() == 1 && binary.depth() == CV_8U);
const int M = binary.rows;
const int N = binary.cols;
for (int m = 0; m < M; ++m) {
const char* bin_ptr = binary.ptr<char>(m);
for (int n = 0; n < N; ++n) {
if (bin_ptr[n] > 0) idx.push_back(cv::Point(n,m));
}
}
}
注意 - 看起来函数签名错误,所以我将输出vector
更改为传递引用。
答案 2 :(得分:-3)
你可以在不使用findNonZero()这个opencv方法的情况下找到它。相反,你可以通过简单地使用2 for循环来获得它。这是片段。希望它可以帮到你。
**
for(int i = 0 ;i <image.rows() ; i++){// image : the binary image
for(int j = 0; j< image.cols() ; j++){
double[] returned = image.get(i,j);
int value = (int) returned[0];
if(value==255){
System.out.println("x: " +i + "\ty: "+j);// returned the (x,y) //co ordinates of all white pixels.
}
}
}
**