#include <opencv2\opencv.hpp>
#include <vector>
#include <algorithm>
using namespace std;
using namespace cv;
int main()
{
// Load the image
Mat3b img = imread("path_to_image", IMREAD_COLOR);
// Convert to grayscale
Mat1b gray;
cvtColor(img, gray, COLOR_BGR2GRAY);
// Get binary mask (remove jpeg artifacts)
gray = gray > 200;
// Get all non black points
vector<Point> pts;
findNonZero(gray, pts);
// Define the radius tolerance
int th_distance = 50; // radius tolerance
// Apply partition
// All pixels within the radius tolerance distance will belong to the same class (same label)
vector<int> labels;
// With lambda function (require C++11)
int th2 = th_distance * th_distance;
int n_labels = partition(pts, labels, [th2](const Point& lhs, const Point& rhs) {
return ((lhs.x - rhs.x)*(lhs.x - rhs.x) + (lhs.y - rhs.y)*(lhs.y - rhs.y)) < th2;
});
// You can save all points in the same class in a vector (one for each class), just like findContours
vector<vector<Point>> contours(n_labels);
for (int i = 0; i < pts.size(); ++i)
{
contours[labels[i]].push_back(pts[i]);
}
// Get bounding boxes
vector<Rect> boxes;
for (int i = 0; i < contours.size(); ++i)
{
Rect box = boundingRect(contours[i]);
boxes.push_back(box);
}
// Get largest bounding box
Rect largest_box = *max_element(boxes.begin(), boxes.end(), [](const Rect& lhs, const Rect& rhs) {
return lhs.area() < rhs.area();
});
// Draw largest bounding box in RED
Mat3b res = img.clone();
rectangle(res, largest_box, Scalar(0, 0, 255));
// Draw enlarged BOX in GREEN
Rect enlarged_box = largest_box + Size(20,20);
enlarged_box -= Point(10,10);
rectangle(res, enlarged_box, Scalar(0, 255, 0));
imshow("Result", res);
waitKey();
return 0;
}
2 回答
您可以使用partition根据给定谓词对白色像素进行分组 . 在这种情况下,您的谓词可以是:对给定欧几里德距离内的所有白色像素进行分组 .
然后,您可以计算每个组的边界框,保留最大的框(在下面的RED中),并最终放大它(在下面的绿色中):
码:
您可以计算每行和每列的积分 . 然后搜索这个积分不断增长的地方 . 在这里你也可以添加一些移动平均线来排除噪音等 . 然后这个地方意味着这里比其他部分更白 . 现在,您可以使用openCV中的矩形函数在此区域周围绘制矩形(http://docs.opencv.org/2.4/modules/core/doc/drawing_functions.html#rectangle) .