I have a binary image (BW_roi.mat), displayed below. And I wanted to measure the length of every line segment.
I tried the solution given in this link hough-transform. But it does not work out for me. It just measured the length of some lines as shown below.
I tried the other code
clc; clear all; close all;
load BW_ROI.mat
boundaries = bwboundaries(BW_roi_nodot);
patchno=1 %Let select patch 1
b = boundaries{patchno}; % Extract N by 2 matrix of (x,y) locations.
x = b(:, 1);
y = b(:, 2);
It though gives me points (x,y) that make up these polygons. But how can I get the line segment length of specific patch?
I'd propose using convexhull in combination with the algorithm here to reduce the number of vertices in the detected polygons. Seems to work pretty well. I'll leave it to you to use the adjacency matrix produced by bwboundaries to throw away duplicates.
load BW_ROI.mat
[B,L,N,A] = bwboundaries(BW_roi_nodot);
for k = 1:N
boundary = fliplr(B{k});
% take a convex hull of the bounded polygon
idxn = convhull(boundary, 'Simplify', true);
% use this algorithm to further simplify the polygon
% https://ww2.mathworks.cn/matlabcentral/fileexchange/41986-ramer-douglas-peucker-algorithm
point_reduced = DouglasPeucker(boundary(idxn,:), 5);
% make the shape, get the side lengths
shp = polyshape(point_reduced);
vertex_deltas = diff([shp.Vertices; shp.Vertices(1,:)], 1);
edge_lengths = sqrt(sum(vertex_deltas.^2, 2));
% plot animation
imshow(BW_roi_nodot), hold on
plot(shp), drawnow, pause(0.5)
end
Related
I have a polygon which intersects itself multiple times. I try to create a mask from this polygon, i.e., to find all points/pixels location within the polygon. I use the Matlab function poly2mask for this. However, due to the multiple self-intersections this is the results I obtain:
Resulting mask from poly2mask for multi-self-intersecting polygon
So, some areas remain unmasked, because of the intersections. I think Matlab sees this as some sort of inclusions. The Matlab help for poly2mask doesn't mention anything about this. Does anyone have an idea how to also include these regions in the mask?
I obtain good results combining a small erosion/dilation step and imfill as follows:
data = load('polygon_edge.mat');
x = data.polygon_edge(:, 1);
y = data.polygon_edge(:, 2);
bw1 = poly2mask(x,y,ceil(max(y)),ceil(max(x)));
se = strel('sphere',1);
bw2 = imerode(imdilate(bw1,se), se);
bw3 = imfill(bw2, 'holes');
figure
imshow(bw3)
hold on
plot(x(:, 1),y(:, 1),'g','LineWidth',2)
The small erosion and dilation step is needed to be sure that all the regions are connected even at places where the polygon is only connected through a single point, otherwise imfill may see some non-existing holes.
you can use inpolygon:
bw1 = poly2mask(x,y,1000,1000);
subplot(131)
imshow(bw1)
hold on
plot(x([1:end 1]),y([1:end 1]),'g','LineWidth',2)
title('using poly2mask')
[xq,yq] = meshgrid(1:1000);
[IN,ON] = inpolygon(xq,yq,x,y);
bw2 = IN | ON;
subplot(132)
imshow(bw2)
hold on
plot(x([1:end 1]),y([1:end 1]),'g','LineWidth',2)
title('using inpolygon')
% boundary - seggested by another answer
k = boundary(x, y, 1); % 1 == tightest single-region boundary
bw3 = poly2mask(x(k), y(k), 1000, 1000);
subplot(133)
imshow(bw3)
hold on
plot(x([1:end 1]),y([1:end 1]),'g','LineWidth',2)
title('using boundary')
Update - I updated my answer to include boundary - it not seems to work well in my case.
You should first calculate the boundary of your polygon and use this to create your mask.
k = boundary(x, y, 0.99); % 1 == tightest single-region boundary
BW = poly2mask(x(k), y(k), m, n)
Using a shrink factor of 0.99 instead of 1 avoids undercutting, but sharp non-convex corners are still not fitted correctly.
I segmented a mouse and get its image-properties using bwlabel. Thereby I have access to the position of the centroid and the orientation of the mouse. I also get the perimeter of the mouse using bwperim.
I want to find the two points of the straight line passing through the centroid and having the same direction than the orientation of the mouse cutting the perimeter.
I find the equation of the straight line using that code :
% E is a 2*2 matrix containing the coordinates of the centroid and the
% coordinates of the point which belong to the straight line and making
% the right angle given by the orientation
coeffs = polyfit(E(:,1),E(:,2),1);
% Create the equation of the straight line
x = 1:width;
yfit = coeffs(1)*x+coeffs(2);
% Make sure there are only int values.
yfit = uint16(yfit);
I convert my values to uint16 because i want to fill a new matrix that I will compare with the matrix containing the perimeter. Here is what I do then :
% Create a matrix of zeros and set to 1 all the pixels which belong to the
% straight line
k = 1;
temp = false;
m = false(size(iPerim));
while temp~=true
temp = false;
if yfit(k) > 0
m(yfit(k),k)=1;
temp = true;
end
k = k+1;
end
[t,p] = ind2sub(size(m), find(m==1));
minM = [min(p),min(t)];
% complete the straight line to don't have little holes
x = linspace(minM(1),D(1),width);
y = coeffs(1)*x+coeffs(2);
idx = sub2ind(size(m),round(y),round(x));
m(idx) = 1;
Then I compare m with iPerim which is the matrix containing my perimeter:
% Compare the matrix of the perimeter and the matrix of the straight line
% and find the two points in common. It is the points where the straight
% line cut the perimeter
p = m & iPerim;
% Extract thoses coordinates
[coordsY,coordsX] = ind2sub(size(p), find(p==1));
Well I am a new user of Matlab so I think this is not a elegant solution but there is the result:
Matrix m
Perimeter in which I plot yfit
As you can see the algorithm detects only one point and not the second one (the yellow spot)... I figure why but I can't find the solution. It is because the line straight is cutting the perimeter through a diagonal but there are not coordinates in common...
Somebody has a solution to my problem ? And of course I am taking any advises conerning my code :)
Thank you very much !
Edit: If there is a easier solution I take it obviously
When the coordinate of the point where the mouse-perimeter and the line cross are E(2,:), then the position of this point in the line is where the distance is minimal. E.g. like:
[xLine, yLine] = find(m); % x,y positions of the line
dX = abs(xline-E(2,1)) % x-distance to x-coordinate of direction-point
dY = abs(yLine-E(2,2)) % y-distance to y-coordinate of direction-point
distP = sqrt(dX.^2+dY.^2) % distance of line-points to directon-point
[~,indMin] = min(distP); % index of line-point which has the minimum distance
xPoint = xLine(indMin(1));
yPoint = yLine(indMin(1));
The abs and sqrtfunctions are not necessary here for finding the right point, only for the correct intermediate values...
From the Matlab Documentation about ind2sub:
For matrices, [I,J] = ind2sub(size(A),find(A>5)) returns the same values as [I,J] = find(A>5).
I have a binary image of a human. In MATLAB, boundary points and the center of the image are also defined, and they are two column matrices. Now I want to draw lines from the center to the boundary points so that I can obtain all points of intersection between these lines and the boundary of the image. How can I do that? Here is the code I have so far:
The code that is written just to get the one intersection point if anyone can help please
clear all
close all
clc
BW = im2bw(imread('C:\fyc-90_1-100.png'));
BW = imfill(BW,'holes');
[Bw m n]=preprocess(BW);
[bord sk pr_sk]=border_skeleton(BW);
boundry=bord;
L = bwlabel(BW);
s = regionprops(L, 'centroid');
centroids = cat(1, s.Centroid);
Step #1 - Generating your line
The first thing you need to do is figure out how to draw your line. To make this simple, let's assume that the centre of the human body is stored as an array of cen = [x1 y1] as you have said. Now, supposing you click anywhere in your image, you get another point linePt = [x2 y2]. Let's assume that both the x and y co-ordinates are the horizontal and vertical components respectively. We can find the slope and intercept of this line, then create points between these two points parameterized by the slope and intercept to generate your line points. One thing I will point out is that if we draw a slope with a vertical line, by definition the slope would be infinity. As such, we need to place in a check to see if we have this situation. If we do, we assume that all of the x points are the same, while y varies. Once you have your slope and intercept, simply create points in between the line. You'll have to choose how many points you want along this line yourself as I have no idea about the resolution of your image, nor how big you want the line to be. We will then store this into a variable called linePoints where the first column consists of x values and the second column consists of y values. In other words:
In other words, do this:
%// Define number of points
numPoints = 1000;
%// Recall the equation of the line: y = mx + b, m = (y2-y1)/(x2-x1)
if abs(cen(1) - linePt(1)) < 0.00001 %// If x points are close
yPts = linspace(cen(2), linePt(2), numPoints); %// y points are the ones that vary
xPts = cen(1)*ones(numPoints, 1); %//Make x points the same to make vertical line
else %// Normal case
slp = (cen(2) - linePt(2)) / cen(1) - linePt(1)); %// Solve for slope (m)
icept = cen(2) - slp*cen(1); %// Solve for intercept (b)
xPts = linspace(cen(1), linePt(1), numPoints); %// Vary the x points
yPts = slp*xPts + icept; %// Solve for the y points
end
linePoints = [xPts(:) yPts(:)]; %// Create point matrix
Step #2 - Finding points of intersection
Supposing you have a 2D array of points [x y] where x denotes the horizontal co-ordinates and y denotes the vertical co-ordinates of your line. We can simply find the distance between all of these points in your boundary with all of your points on the line. Should any of the points be under a certain threshold (like 0.0001 for example), then this indicates an intersection. Note that due to the crux of floating point data, we can't check to see if the distance is 0 due to the step size in between each discrete point in your data.
I'm also going to assume border_skeleton returns points of the same format. This method works without specifying what the centroid is. As such, I don't need to use the centroids in the method I'm proposing. Also, I'm going to assume that your line points are stored in a matrix called linePoints that is of the same type that I just talked about.
In other words, do this:
numBoundaryPoints = size(boundry, 1); %// boundary is misspelled in your code BTW
ptsIntersect = []; %// Store points of intersection here
for idx = 1 : numBoundaryPoints %// For each boundary point...
%//Obtain the i'th boundary point
pt = boundry(:,idx);
%//Get distances - This computes the Euclidean distance
%//between the i'th boundary point and all points along your line
dists = sqrt(sum(bsxfun(#minus, linePoints, pt).^2, 2));
%//Figure out which points intersect and store
ptsIntersect = [ptsIntersect; linePoints(dists < 0.0001, :)];
end
In the end, ptsIntersect will store all of the points along the boundary that intersect with this line. Take note that I have made a lot of assumptions here because you haven't (or seem reluctant to) give any more details than what you've specified in your comments.
Good luck.
I am having difficulty with calculating 2D area of contours produced from a Kernel Density Estimation (KDE) in Matlab. I have three variables:
X and Y = meshgrid which variable 'density' is computed over (256x256)
density = density computed from the KDE (256x256)
I run the code
contour(X,Y,density,10)
This produces the plot that is attached. For each of the 10 contour levels I would like to calculate the area. I have done this in some other platforms such as R but am having trouble figuring out the correct method / syntax in Matlab.
C = contourc(density)
I believe the above line would store all of the values of the contours allowing me to calculate the areas but I do not fully understand how these values are stored nor how to get them properly.
This little script will help you. Its general for contour. Probably working for contour3 and contourf as well, with adjustments of course.
[X,Y,Z] = peaks; %example data
% specify certain levels
clevels = [1 2 3];
C = contour(X,Y,Z,clevels);
xdata = C(1,:); %not really useful, in most cases delimters are not clear
ydata = C(2,:); %therefore further steps to determine the actual curves:
%find curves
n(1) = 1; %n: indices where the certain curves start
d(1) = ydata(1); %d: distance to the next index
ii = 1;
while true
n(ii+1) = n(ii)+d(ii)+1; %calculate index of next startpoint
if n(ii+1) > numel(xdata) %breaking condition
n(end) = []; %delete breaking point
break
end
d(ii+1) = ydata(n(ii+1)); %get next distance
ii = ii+1;
end
%which contourlevel to calculate?
value = 2; %must be member of clevels
sel = find(ismember(xdata(n),value));
idx = n(sel); %indices belonging to choice
L = ydata( n(sel) ); %length of curve array
% calculate area and plot all contours of the same level
for ii = 1:numel(idx)
x{ii} = xdata(idx(ii)+1:idx(ii)+L(ii));
y{ii} = ydata(idx(ii)+1:idx(ii)+L(ii));
figure(ii)
patch(x{ii},y{ii},'red'); %just for displaying purposes
%partial areas of all contours of the same plot
areas(ii) = polyarea(x{ii},y{ii});
end
% calculate total area of all contours of same level
totalarea = sum(areas)
Example: peaks (by Matlab)
Level value=2 are the green contours, the first loop gets all contour lines and the second loop calculates the area of all green polygons. Finally sum it up.
If you want to get all total areas of all levels I'd rather write some little functions, than using another loop. You could also consider, to plot just the level you want for each calculation. This way the contourmatrix would be much easier and you could simplify the process. If you don't have multiple shapes, I'd just specify the level with a scalar and use contour to get C for only this level, delete the first value of xdata and ydata and directly calculate the area with polyarea
Here is a similar question I posted regarding the usage of Matlab contour(...) function.
The main ideas is to properly manipulate the return variable. In your example
c = contour(X,Y,density,10)
the variable c can be returned and used for any calculation over the isolines, including area.
I want to plot a line from one well-defined point to another and then turn it into an image matrix to use a Gaussian filter on it for smoothing. For this I use the functions line and getframe to plot a line and capture the figure window in an image, but getframe is very slow and not very reliable. I noticed that it does not capture anything when the computer is locked and I got an out of memory error after 170 executions.
My questions are:
Is there a substitute to getframe that I can use?
Is there a way to create the image matrix and draw the line directly in it?
Here is a minimal code sample:
figure1=line([30 35] ,[200 60]);
F= getframe;
hsize=40; sigma=20;
h = fspecial('gaussian',hsize,sigma);
filteredImg = imfilter(double(F.cdata), h,256);
imshow(uint8(filteredImg));
[update]
High-Performance Mark's idea with linspace looks very promising, but how do I access the matrix coordinates calculated with linspace? I tried the following code, but it does not work as I think it should. I assume it is a very simple and basic MATLAB thing, but I just can not wrap my head around it:
matrix=zeros(200,60);
diagonal=round([linspace(30,200,numSteps); linspace(35,60,numSteps)]);
matrix(diagonal(1,:), diagonal(2,:))=1;
imshow(matrix);
Here's one example of drawing a line directly into a matrix. First, we'll create a matrix of zeros for an empty image:
mat = zeros(250, 250, 'uint8'); % A 250-by-250 matrix of type uint8
Then, let's say we want to draw a line running from (30, 35) to (200, 60). We'll first compute how many pixels long the line will have to be:
x = [30 200]; % x coordinates (running along matrix columns)
y = [35 60]; % y coordinates (running along matrix rows)
nPoints = max(abs(diff(x)), abs(diff(y)))+1; % Number of points in line
Next, we compute row and column indices for the line pixels using linspace, convert them from subscripted indices to linear indices using sub2ind, then use them to modify mat:
rIndex = round(linspace(y(1), y(2), nPoints)); % Row indices
cIndex = round(linspace(x(1), x(2), nPoints)); % Column indices
index = sub2ind(size(mat), rIndex, cIndex); % Linear indices
mat(index) = 255; % Set the line pixels to the max value of 255 for uint8 types
You can then visualize the line and the filtered version with the following:
subplot(1, 2, 1);
image(mat); % Show original line image
colormap(gray); % Change colormap
title('Line');
subplot(1, 2, 2);
h = fspecial('gaussian', 20, 10); % Create filter
filteredImg = imfilter(mat, h); % Filter image
image(filteredImg); % Show filtered line image
title('Filtered line');
If you have Computer Vision System toolbox there is a ShapeInserter object available. This can be used to draw lines, circles, rectangles and polygons on the image.
mat = zeros(250,250,'uint8');
shapeInserter = vision.ShapeInserter('Shape', 'Lines', 'BorderColor', 'White');
y = step(shapeInserter, mat, int32([30 60 180 210]));
imshow(y);
http://www.mathworks.com/help/vision/ref/vision.shapeinserterclass.html
You can check my answer here. It is robust way to achieve what you are asking for. Advantage of my approach is that it doesn't need additional parameters to control density of the line drawn.
Something like this:
[linspace(30,200,numSteps); linspace(35,60,numSteps)]
Does that work for you ?
Mark