我想在C ++接口(cv名称空间)中使用k-means和OpenCV对图像进行分色,我得到了奇怪的结果。我需要它来减少一些噪音。这是我的代码:
#include "cv.h"
#include "highgui.h"
using namespace cv;
int main() {
Mat imageBGR, imageHSV, planeH, planeS, planeV;
imageBGR = imread("fruits.jpg");
imshow("original", imageBGR);
cv::Mat labels, data;
cv::Mat centers(8, 1, CV_32FC1);
imageBGR.convertTo(data, CV_32F);
cv::kmeans(data, 8, labels,
cv::TermCriteria(CV_TERMCRIT_ITER, 10, 1.0),
3, cv::KMEANS_PP_CENTERS, ¢ers);
imshow("posterized hue", data);
data.convertTo(data, CV_32FC3);
waitKey();
return 0;
}
但是我得到了一个奇怪的结果
第一张图片:原创
第二张图片:k-means之后。
有什么建议吗?
#include "cv.h"
#include "highgui.h"
#include <iostream>
using namespace cv;
using namespace std;
int main() {
Mat src;
src = imread("fruits.jpg");
imshow("original", src);
blur(src, src, Size(15,15));
imshow("blurred", src);
Mat p = Mat::zeros(src.cols*src.rows, 5, CV_32F);
Mat bestLabels, centers, clustered;
vector<Mat> bgr;
cv::split(src, bgr);
// i think there is a better way to split pixel bgr color
for(int i=0; i<src.cols*src.rows; i++) {
p.at<float>(i,0) = (i/src.cols) / src.rows;
p.at<float>(i,1) = (i%src.cols) / src.cols;
p.at<float>(i,2) = bgr[0].data[i] / 255.0;
p.at<float>(i,3) = bgr[1].data[i] / 255.0;
p.at<float>(i,4) = bgr[2].data[i] / 255.0;
}
int K = 8;
cv::kmeans(p, K, bestLabels,
TermCriteria( CV_TERMCRIT_EPS+CV_TERMCRIT_ITER, 10, 1.0),
3, KMEANS_PP_CENTERS, centers);
int colors[K];
for(int i=0; i<K; i++) {
colors[i] = 255/(i+1);
}
// i think there is a better way to do this mayebe some Mat::reshape?
clustered = Mat(src.rows, src.cols, CV_32F);
for(int i=0; i<src.cols*src.rows; i++) {
clustered.at<float>(i/src.cols, i%src.cols) = (float)(colors[bestLabels.at<int>(0,i)]);
// cout << bestLabels.at<int>(0,i) << " " <<
// colors[bestLabels.at<int>(0,i)] << " " <<
// clustered.at<float>(i/src.cols, i%src.cols) << " " <<
// endl;
}
clustered.convertTo(clustered, CV_8U);
imshow("clustered", clustered);
waitKey();
return 0;
}
结果:
答案 0 :(得分:8)
我不是OpenCV的专家,因此我将提供与您的问题相关的一般性建议K-means采用基本上是矩阵的向量列表:
[x0, y0, r0, g0, b0]
[x1, y1, r1, g1, b1]
[x2, y2, r2, g2, b2]
.
.
.
你正在给它一个不起作用的图像。首先必须将图像转换为此k均值矩阵格式。对于源图像的每个像素,在结果矩阵中有一行。另请注意,您应该缩放值,使它们都具有相似的值。如果不这样做,x和y坐标通常会比颜色高得多“重力”,导致结果不令人满意。 C ++伪代码:
int pixel_index = 0;
for (int y = 0; y < image height; y++) {
for (int x = 0; x < image width; x++) {
matrix[pixel_index][0] = (float)x / image width;
matrix[pixel_index][1] = (float)y / image height;
matrix[pixel_index][2] = (float)pixel(x, y).r / 255.0f;
matrix[pixel_index][3] = (float)pixel(x, y).g / 255.0f;
matrix[pixel_index][4] = (float)pixel(x, y).b / 255.0f;
}
}
// Pass the matrix to kmeans...
因此,您将获得每个像素的标签,这些标签对应于已分配给它的群集。然后,您需要确定群集的颜色 - 这可以从获取中心像素颜色值到计算群集的平均/中值颜色。确定颜色后,只需移动图像并将像素设置为其簇颜色:
for (int y = 0; y < image height; y++) {
for (int x = 0; x < image width; x++) {
int index = y * image width + x; // This corresponds to pixel_index above
int cluster_index = labels[index]; // 0 to 7 in your case
Color color = colors[cluster_index]; // Colors is an array of 8 colors of the clusters
image.setpixel(x, y, color)
}
}
如果您更喜欢使用HSV而不是RGB,只需使用HSV值而不是RGB值。
OpenCV可能具有完全执行上述转换的功能,但我无法使用Google快速找到它们。
答案 1 :(得分:8)
如果您不需要k-means中的x,y坐标,则可以使用reshape命令更快地排列数据:
int origRows = img.rows;
notes << "original image is: " << img.rows << "x" << img.cols << endl;
Mat colVec = img.reshape(1, img.rows*img.cols); // change to a Nx3 column vector
cout << "colVec is of size: " << colVec.rows << "x" << colVec.cols << endl;
Mat colVecD, bestLabels, centers, clustered;
int attempts = 5;
int clusts = 8;
double eps = 0.001;
colVec.convertTo(colVecD, CV_32FC3, 1.0/255.0); // convert to floating point
double compactness = kmeans(colVecD, clusts, bestLabels,
TermCriteria(CV_TERMCRIT_EPS+CV_TERMCRIT_ITER, attempts, eps),
attempts, KMEANS_PP_CENTERS, centers);
Mat labelsImg = bestLabels.reshape(1, origRows); // single channel image of labels
cout << "Compactness = " << compactness << endl;