如何在不包含(或带有几个数字)背景像素的情况下调整或调整对象内部矩形的大小? [英] How to adapt or resize a rectangle inside an object without including (or with a few numbers) of background pixels?
问题描述
应用阈值并找到对象的轮廓后，我使用以下代码获取对象周围的笔直矩形(或输入其指令的旋转矩形):
After I applied thresholding and finding the contours of the object, I used the following code to get the straight rectangle around the object (or the rotated rectangle inputting its instruction):
img = cv2.imread('image.png')
imgray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
ret,thresh = cv2.threshold(imgray,127,255,cv2.THRESH_BINARY)
# find contours
contours, hierarchy = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
cnt = contours[0]
# straight rectangle
x,y,w,h = cv2.boundingRect(cnt)
img= cv2.rectangle(img,(x,y),(x+w,y+h),(0,255,0),2)
查看图片
然后我使用以下代码计算了直角矩形内的对象和背景像素的数量:
Then I have calculated the number of object and background pixels inside the straight rectangle using the following code:
# rectangle area (total number of object and background pixels inside the rectangle)
area_rect = w*h
# white or object pixels (inside the rectangle)
obj = cv2.countNonZero(imgray)
# background pixels (inside the rectangle)
bac = area_rect - obj
现在，我想根据背景像素和对象像素之间的关系来调整对象的矩形，例如，使一个矩形占据对象的较大部分，例如没有背景像素或具有较少背景像素:
Now I want to adapt the rectangle of the object as a function of the relationship between the background pixel and those of the object, ie to have a rectangle occupying the larger part of the object without or with less background pixel, for example:
如何创建它?
推荐答案
可以将此问题表示为找到非凸多边形上刻有的最大矩形.
可以在以下中找到近似解决方案链接.
这个问题也可以表达为:对于每个角度，找到矩阵中仅包含零的最大矩形，对此进行了探讨
This problem can be formulated also as: for each angle, find the largest rectangle containing only zeros in a matrix, explored in this SO question.
我的解决方案基于此答案.这将仅找到与轴对齐的矩形，因此您可以轻松地将图像旋转给定角度并将此解决方案应用于每个角度. 我的解决方案是C ++，但是您可以轻松地将其移植到Python，因为我主要使用的是OpenCV功能，或者可以考虑轮换来调整上述答案中的解决方案.
My solution is based on this answer. This will find only axis aligned rectangles, so you can easily rotate the image by a given angle and apply this solution for every angle. My solution is C++, but you can easily port it to Python, since I'm using mostly OpenCV function, or adjust the solution in the above mentioned answer accounting for rotation.
我们在这里:
#include <opencv2\opencv.hpp>
#include <iostream>
using namespace cv;
using namespace std;
// https://stackoverflow.com/a/30418912/5008845
Rect findMinRect(const Mat1b& src)
{
Mat1f W(src.rows, src.cols, float(0));
Mat1f H(src.rows, src.cols, float(0));
Rect maxRect(0,0,0,0);
float maxArea = 0.f;
for (int r = 0; r < src.rows; ++r)
{
for (int c = 0; c < src.cols; ++c)
{
if (src(r, c) == 0)
{
H(r, c) = 1.f + ((r>0) ? H(r-1, c) : 0);
W(r, c) = 1.f + ((c>0) ? W(r, c-1) : 0);
}
float minw = W(r,c);
for (int h = 0; h < H(r, c); ++h)
{
minw = min(minw, W(r-h, c));
float area = (h+1) * minw;
if (area > maxArea)
{
maxArea = area;
maxRect = Rect(Point(c - minw + 1, r - h), Point(c+1, r+1));
}
}
}
}
return maxRect;
}
RotatedRect largestRectInNonConvexPoly(const Mat1b& src)
{
// Create a matrix big enough to not lose points during rotation
vector<Point> ptz;
findNonZero(src, ptz);
Rect bbox = boundingRect(ptz);
int maxdim = max(bbox.width, bbox.height);
Mat1b work(2*maxdim, 2*maxdim, uchar(0));
src(bbox).copyTo(work(Rect(maxdim - bbox.width/2, maxdim - bbox.height / 2, bbox.width, bbox.height)));
// Store best data
Rect bestRect;
int bestAngle = 0;
// For each angle
for (int angle = 0; angle < 90; angle += 1)
{
cout << angle << endl;
// Rotate the image
Mat R = getRotationMatrix2D(Point(maxdim,maxdim), angle, 1);
Mat1b rotated;
warpAffine(work, rotated, R, work.size());
// Keep the crop with the polygon
vector<Point> pts;
findNonZero(rotated, pts);
Rect box = boundingRect(pts);
Mat1b crop = rotated(box).clone();
// Invert colors
crop = ~crop;
// Solve the problem: "Find largest rectangle containing only zeros in an binary matrix"
// https://stackoverflow.com/questions/2478447/find-largest-rectangle-containing-only-zeros-in-an-n%C3%97n-binary-matrix
Rect r = findMinRect(crop);
// If best, save result
if (r.area() > bestRect.area())
{
bestRect = r + box.tl(); // Correct the crop displacement
bestAngle = angle;
}
}
// Apply the inverse rotation
Mat Rinv = getRotationMatrix2D(Point(maxdim, maxdim), -bestAngle, 1);
vector<Point> rectPoints{bestRect.tl(), Point(bestRect.x + bestRect.width, bestRect.y), bestRect.br(), Point(bestRect.x, bestRect.y + bestRect.height)};
vector<Point> rotatedRectPoints;
transform(rectPoints, rotatedRectPoints, Rinv);
// Apply the reverse translations
for (int i = 0; i < rotatedRectPoints.size(); ++i)
{
rotatedRectPoints[i] += bbox.tl() - Point(maxdim - bbox.width / 2, maxdim - bbox.height / 2);
}
// Get the rotated rect
RotatedRect rrect = minAreaRect(rotatedRectPoints);
return rrect;
}
int main()
{
Mat1b img = imread("path_to_image", IMREAD_GRAYSCALE);
// Compute largest rect inside polygon
RotatedRect r = largestRectInNonConvexPoly(img);
// Show
Mat3b res;
cvtColor(img, res, COLOR_GRAY2BGR);
Point2f points[4];
r.points(points);
for (int i = 0; i < 4; ++i)
{
line(res, points[i], points[(i + 1) % 4], Scalar(0, 0, 255), 2);
}
imshow("Result", res);
waitKey();
return 0;
}
结果图像为:
注意
我想指出的是，此代码未经过优化，因此可能会更好地执行.有关近似解决方案的信息，请参见此处 ，并在那里报道了论文.
I'd like to point out that this code is not optimized, so it can probably perform better. For an approximized solution, see here, and the papers reported there.
对相关问题的答案使我朝着正确的方向前进.
This answer to a related question put me in the right direction.
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