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I am using MATLAB. I want to use canny method for edge detection. But I need the edges that are diagonal or the edges that are only on 40 to 50 degree angle. how can i do that?
You need write canny edge detector's code by your own (you would get lots of implementation )in the internet. You would then be calculating the gradient magnitudes and gradient directions in the second step. There you need to filter out the angles and corresponding magnitudes.
Hope this helps you.
I've answered a similar question about how to use Matlab's edge function to find oriented edges with Canny ( Orientational Canny Edge Detection ), but I also wanted to try out a custom implementation as suggested by Avijit.
Canny Edge Detection steps:
Start with an image, I'll use a built in demo image.
A = im2double(rgb2gray(imread('peppers.png')));
Gaussian Filter
A_filter = imgaussfilt(A);
Sobel Edge Detection -- We can't use the built in implementation (edge(A_filter, 'Sobel')), because we want the edge angles, not just the edge locations, so we implement our own operator.
a. Convolution to find oriented gradients
%These filters measure the difference in values between vertically or horizontally adjacent pixels.
%Effectively, this finds vertical and horizontal gradients.
vertical_filter = [-1 0 1; -2 0 2; -1 0 1];
horizontal_filter = [-1 -2 -1; 0 0 0; 1 2 1];
A_vertical = conv2(A_filter, vertical_filter, 'same');
A_horizontal = conv2(A_filter, horizontal_filter, 'same');
b. Calculate the angles
A_angle = arctan(A_vertical./A_horizontal);
At this step, we traditionally bin edges by orientation (0°, 45°, 90°, 135°), but since you only want diagonal edges between 40 and 50 degrees, we will retain those edges and discard the rest.
% I lowered the thresholds to include more pixels
% But for your original post, you would use 40 and 50
lower_angle_threshold = 22.5;
upper_angle_threshold = 67.5;
diagonal_map = zeros(size(A), 'logical');
diagonal_map (A_angle>(lower_angle_threshold*pi/180) & A_angle<(upper_angle_threshold*pi/180)) = 1;
Perform non-max suppression on the remaining edges -- This is the most difficult portion to adapt to different angles. To find the exact edge location, you compare two adjacent pixels: for 0° edges, compare east-west, for 45° south-west pixel to north-east pixel, for 90° compare north-south, and for 135° north-west pixel to south-east pixel.
Since your desired angle is close to 45°, I just used south-west, but if you wanted 10° to 20°, for example, you'd have to put some more thought into these comparisons.
non_max = A_sobel;
[n_rows, n_col] = size(A);
%For every pixel
for row = 2:n_rows-1
for col = 2:n_col-1
%If we are at a diagonal edge
if(diagonal_map(row, col))
%Compare north east and south west pixels
if(A_sobel(row, col)<A_sobel(row-1, col-1) || ...
A_sobel(row, col)<A_sobel(row+1, col+1))
non_max(row, col) = 0;
end
else
non_max(row, col) = 0;
end
end
end
Edge tracking with hysteresis -- Decide whether weak edge pixels are close enough (I use a 3x3 window) to strong edge pixels. If they are, include them in the edge. If not, they are noise; remove them.
high_threshold = 0.5; %These thresholds are tunable parameters
low_threshold = 0.01;
weak_edge_pixels = non_max > low_threshold & non_max < high_threshold;
strong_edge_pixels = non_max > high_threshold;
final = strong_edge_pixels;
for row = 2:n_rows-1
for col = 2:n_col-1
window = strong_edge_pixels(row-1:row+1, col-1:col+1);
if(weak_edge_pixels(row, col) && any(window(:)))
final(row, col) = 1;
end
end
end
Here are my results.
As you can see, discarding the other edge orientations has a very negative effect on the hysteresis step because fewer strong pixels are detected. Adjusting the high_threshold would help somewhat. Another option would be to do Steps 5 and 6 using all edge orientations, and then use the diagonal_map to extract the diagonal edges.
BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.
I'm currently trying to write a function in MatLab which loops over each pixel, takes the mean intensity of the pixels within a radius around it and then applies that intensity to the central pixel, effectively blurring the image.
I start by declaring the function and finding the maximum width and height of the image, nx and ny:
function [] = immean(IMAGE, r)
[nx, ny] = size(IMAGE);
I then create a completely black image of the same size as the image variable IMAGE. This is so that I can store the value of each pixel, once the mean intensity of its neighbourhood has been found.
average = zeros(size(IMAGE));
I then loop through the image:
for x = 1:nx
for y = 1:ny
and apply a series of if-statements to deal with cases where the radius of the circle around the pixel does not fit the image. (For example, a pixel at (1,1) with a radius of 5 would have a starting point of -4, which would cause an error):
if x-r <= 0
startx = 1;
else
startx = x-r;
end
if x+r > nx
endx = nx;
else
endx = x+r;
end
if y-r <= 0
starty = 1;
else
starty = y-r;
end
if y+r > ny
endy = ny;
else
endy = y+r;
end
This effectively creates a square of values that may fall under the domain of the circular sample, which speeds up the program dramatically. After that, I iterate through the values within this square and find any pixels which fall within the radius of the central pixel. The intensities of these pixels are then added to a variable called total and the count pixelcount increments:
total = 0;
pixelcount = 0;
for xp = startx : endx
for yp = starty : endy
if (x-xp)^2 + (y-yp)^2 <= r^2
total = total + uint64(IMAGE(xp, yp));
pixelcount = pixelcount + 1;
end
end
end
I then find the mean intensity of the circular sample of pixels, by dividing total by pixelcount and then plug that value into the appropriate pixel of the completely black image average:
mean = total / pixelcount;
average(x,y) = mean;
The trouble is: this isn't working. Instead of a blurred version of the original image, I get an entirely white image instead. I'm not sure why - when I take the ; from the last line, it shows me that mean constitutes many values - it's not like they're all 255. So I figure that there must be something wrong with the assignment line average(x,y) = mean;, but I can't find out what that is.
Can anyone see why this is going wrong?
I am new in openCV, I already detect edge of paper sheet but my result image is blurred after draw lines on edge, How I can draw lines on edges of paper sheet so my image quality remain unaffected.
what I am Missing..
My code is below.
Many thanks.
-(void)forOpenCV
{
if( imageView.image != nil )
{
cv::Mat greyMat=[self cvMatFromUIImage:imageView.image];
vector<vector<cv::Point> > squares;
cv::Mat img= [self debugSquares: squares: greyMat ];
imageView.image =[self UIImageFromCVMat: img];
}
}
- (cv::Mat) debugSquares: (std::vector<std::vector<cv::Point> >) squares : (cv::Mat &)image
{
NSLog(#"%lu",squares.size());
// blur will enhance edge detection
Mat blurred(image);
medianBlur(image, blurred, 9);
Mat gray0(image.size(), CV_8U), gray;
vector<vector<cv::Point> > contours;
// find squares in every color plane of the image
for (int c = 0; c < 3; c++)
{
int ch[] = {c, 0};
mixChannels(&image, 1, &gray0, 1, ch, 1);
// try several threshold levels
const int threshold_level = 2;
for (int l = 0; l < threshold_level; l++)
{
// Use Canny instead of zero threshold level!
// Canny helps to catch squares with gradient shading
if (l == 0)
{
Canny(gray0, gray, 10, 20, 3); //
// Dilate helps to remove potential holes between edge segments
dilate(gray, gray, Mat(), cv::Point(-1,-1));
}
else
{
gray = gray0 >= (l+1) * 255 / threshold_level;
}
// Find contours and store them in a list
findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
// Test contours
vector<cv::Point> approx;
for (size_t i = 0; i < contours.size(); i++)
{
// approximate contour with accuracy proportional
// to the contour perimeter
approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);
// Note: absolute value of an area is used because
// area may be positive or negative - in accordance with the
// contour orientation
if (approx.size() == 4 &&
fabs(contourArea(Mat(approx))) > 1000 &&
isContourConvex(Mat(approx)))
{
double maxCosine = 0;
for (int j = 2; j < 5; j++)
{
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
if (maxCosine < 0.3)
squares.push_back(approx);
}
}
}
}
NSLog(#"%lu",squares.size());
for( size_t i = 0; i < squares.size(); i++ )
{
cv:: Rect rectangle = boundingRect(Mat(squares[i]));
if(i==squares.size()-1)////Detecting Rectangle here
{
const cv::Point* p = &squares[i][0];
int n = (int)squares[i].size();
NSLog(#"%d",n);
line(image, cv::Point(507,418), cv::Point(507+1776,418+1372), Scalar(255,0,0),2,8);
polylines(image, &p, &n, 1, true, Scalar(255,255,0), 5, CV_AA);
fx1=rectangle.x;
fy1=rectangle.y;
fx2=rectangle.x+rectangle.width;
fy2=rectangle.y+rectangle.height;
line(image, cv::Point(fx1,fy1), cv::Point(fx2,fy2), Scalar(0,0,255),2,8);
}
}
return image;
}
Instead of
Mat blurred(image);
you need to do
Mat blurred = image.clone();
Because the first line does not copy the image, but just creates a second pointer to the same data.
When you blurr the image, you are also changing the original.
What you need to do instead is, to create a real copy of the actual data and operate on this copy.
The OpenCV reference states:
by using a copy constructor or assignment operator, where on the right side it can
be a matrix or expression, see below. Again, as noted in the introduction, matrix assignment is O(1) operation because it only copies the header and increases the reference counter.
Mat::clone() method can be used to get a full (a.k.a. deep) copy of the matrix when you need it.
The first problem is easily solved by doing the entire processing on a copy of the original image. That way, after you get all the points of the square you can draw the lines on the original image and it will not be blurred.
The second problem, which is cropping, can be solved by defining a ROI (region of interested) in the original image and then copying it to a new Mat. I've demonstrated that in this answer:
// Setup a Region Of Interest
cv::Rect roi;
roi.x = 50
roi.y = 10
roi.width = 400;
roi.height = 450;
// Crop the original image to the area defined by ROI
cv::Mat crop = original_image(roi);
cv::imwrite("cropped.png", crop);
Sounds like I got the concept but cant seems to get the implementation correct. eI have a cluster (an ArrayList) with multiple points, and I want to calculate avg distance. Ex: Points in cluster (A, B, C, D, E, F, ... , n), Distance A-B, Distance A-C, Distance A-D, ... Distance A,N, Distance (B,C) Distance (B,D)... Distance (B,N)...
Thanks in advance.
You don't want to double count any segment, so your algorithm should be a double for loop. The outer loop goes from A to M (you don't need to check N, because there'll be nothing left for it to connect to), each time looping from curPoint to N, calculating each distance. You add all the distances, and divide by the number of points (n-1)^2/2. Should be pretty simple.
There aren't any standard algorithms for improving on this that I'm aware of, and this isn't a widely studied problem. I'd guess that you could get a pretty reasonable estimate (if an estimate is useful) by sampling distances from each point to a handful of others. But that's a guess.
(After seeing your code example) Here's another try:
public double avgDistanceInCluster() {
double totDistance = 0.0;
for (int i = 0; i < bigCluster.length - 1; i++) {
for (int j = i+1; j < bigCluster.length; j++) {
totDistance += distance(bigCluster[i], bigCluster[j]);
}
}
return totDistance / (bigCluster.length * (bigCluster.length - 1)) / 2;
}
Notice that the limit for the first loop is different.
Distance between two points is probably sqrt((x1 - x2)^2 + (y1 -y2)^2).
THanks for all the help, Sometimes after explaining the question on forum answer just popup to your mind. This is what I end up doing.
I have a cluster of point, and I need to calculate the avg distance of points (pairs) in the cluster. So, this is what I did. I am sure someone will come with a better answer if so please drop a note. Thanks in advance.
/**
* Calculate avg distance between points in cluster
* #return
*/
public double avgDistanceInCluster() {
double avgDistance = 0.0;
Stack<Double> holder = new Stack<Double>();
for (int i = 0; i < cluster.size(); i++) {
System.out.println(cluster.get(i));
for (int j = i+1; j < cluster.size(); j++) {
avgDistance = (cluster.get(i) + cluster.get(j))/2;
holder.push(avgDistance);
}
}
Iterator<Double> iter = holder.iterator();
double avgClusterDist = 0;
while (iter.hasNext()) {
avgClusterDist =+ holder.pop();
System.out.println(avgClusterDist);
}
return avgClusterDist/cluster.size();
}