I wish to set values on a line whose endpoints are returned by the hough transforms to zero. I have written the following code snippet
imshow(img);
hold on
img_black = img;
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2]; %line end points
[x, y] = bresenham(xy(1,1),xy(1,2),xy(2,1),xy(2,2)); %returns all points on the line
for i = 1:length(x)
plot(x(i),y(i),'*'); %to plot individual pixels on line itself
img_black(x(i),y(i),:) = [0,0,0]; %set rgb values to zero
end
end
Although the points plotted on the image below are as expected
The image where the corresponding pixel values are being set to zero is not as expected.
What is happening here?
It looks like you have mixed up x and y with rows and columns.
img_black(x(i), y(i),:)
Should be
img_black(y(i), x(i),:);
This is because the first dimension of img_black is rows (y) and the second dimension is columns (x).
The resulting image looks like it does because your lines go the wrong way and (sometimes) go outside the bounds of the original image, but MATLAB gladly expands your image (with zeros) and sets the values that you request, hence all the black pixels on the right.
NOTE: This switching back and forth between row, column and x,y is common throughout MATLAB's built-in functions and you should always be careful to note what the output is. A class example is meshgrid vs ndgrid outputs.
Related
I have a matrix named figmat from which I obtain the following pcolor plot (Matlab-Version R 2016b).
Basically I only want to extract the bottom red high intensity line from this plot.
I thought of doing it in some way of extracting the maximum values from the matrix and creating some sort of mask on the main matrix. But I'm not understanding a possible way to achieve this. Can it be accomplished with the help of any edge/image detection algorithms?
I was trying something like this with the following code to create a mask
A=max(figmat);
figmat(figmat~=A)=0;
imagesc(figmat);
But this gives only the boundary of maximum values. I also need the entire red color band.
Okay, I assume that the red line is linear and its values can uniquely be separated from the rest of the picture. Let's generate some test data...
[x,y] = meshgrid(-5:.2:5, -5:.2:5);
n = size(x,1)*size(x,2);
z = -0.2*(y-(0.2*x+1)).^2 + 5 + randn(size(x))*0.1;
figure
surf(x,y,z);
This script generates a surface function. Its set of maximum values (x,y) can be described by a linear function y = 0.2*x+1. I added a bit of noise to it to make it a bit more realistic.
We now select all points where z is smaller than, let's say, 95 % of the maximum value. Therefore find can be used. Later, we want to use one-dimensional data, so we reshape everything.
thresh = min(min(z)) + (max(max(z))-min(min(z)))*0.95;
mask = reshape(z > thresh,1,n);
idx = find(mask>0);
xvec = reshape(x,1,n);
yvec = reshape(y,1,n);
xvec and yvec now contain the coordinates of all values > thresh.
The last step is to do some linear polynomial over all points.
pp = polyfit(xvec(idx),yvec(idx),1)
pp =
0.1946 1.0134
Obviously these are roughly the coefficients of y = 0.2*x+1 as it should be.
I do not know, if this also works with your data, since I made some assumptions. The threshold level must be chosen carefully. Maybe some preprocessing must be done to dynamically detect this level if you really want to process your images automatically. There might also be a simpler way to do it... but for me this one was straight forward without the need of any toolboxes.
By assuming:
There is only one band to extract.
It always has the maximum values.
It is linear.
I can adopt my previous answer to this case as well, with few minor changes:
First, we get the distribution of the values in the matrix and look for a population in the top values, that can be distinguished from the smaller values. This is done by finding the maximum value x(i) on the histogram that:
Is a local maximum (its bin is higher than that of x(i+1) and x(i-1))
Has more values above it than within it (the sum of the height of bins x(i+1) to x(end) < the height of bin x):
This is how it is done:
[h,x] = histcounts(figmat); % get the distribution of intesities
d = diff(fliplr(h)); % The diffrence in bin height from large x to small x
band_min_ind = find(cumsum(d)>size(figmat,2) & d<0, 1); % 1st bin that fit the conditions
flp_val = fliplr(x); % the value of x from large to small
band_min = flp_val(band_min_ind); % the value of x that fit the conditions
Now we continue as before. Mask all the unwanted values, interpolate the linear line:
mA = figmat>band_min; % mask all values below the top value mode
[y1,x1] = find(mA,1); % find the first nonzero row
[y2,x2] = find(mA,1,'last'); % find the last nonzero row
m = (y1-y2)/(x1-x2); % the line slope
n = y1-m*x1; % the intercept
f_line = #(x) m.*x+n; % the line function
And if we plot it we can see the red line where the band for detection was:
Next, we can make this line thicker for a better representation of this line:
thick = max(sum(mA)); % mode thickness of the line
tmp = (1:thick)-ceil(thick/2); % helper vector for expanding
rows = bsxfun(#plus,tmp.',floor(f_line(1:size(A,2)))); % all the rows for each column
rows(rows<1) = 1; % make sure to not get out of range
rows(rows>size(A,1)) = size(A,1); % make sure to not get out of range
inds = sub2ind(size(A),rows,repmat(1:size(A,2),thick,1)); % convert to linear indecies
mA(inds) = true; % add the interpolation to the mask
result = figmat.*mA; % apply the mask on figmat
Finally, we can plot that result after masking, excluding the unwanted areas:
imagesc(result(any(result,2),:))
I have a simple pcolor plot in Matlab (Version R 2016b) which I have uploaded as shown in the image below. I need to get only the blue sloped line which extends from the middle of the leftmost corner to the rightmost corner without hard-coding the matrix values.
For instance: One can see that the desired slope line has values somewhere approximately between 20 to 45 from the pcolor plot. (From a rough guess just by looking at the graph)
I'm applying the following code on the matrix named Slant which contains the plotted values.
load('Slant.mat');
Slant(Slant<20|Slant>50)=0;
pcolor(Slant); colormap(jet); shading interp; colorbar;
As one can see I hard-coded the values which I don't want to. Is there any method of detecting certain matrix values while making the rest equal to zero?
I used an other small algorithm of taking half the maximum value from the matrix and setting it to zero. But this doesn't work for other images.
[maxvalue, row] = max(Slant);
max_m=max(maxvalue);
Slant(Slant>max_m/2)=0;
pcolor(Slant); colormap(jet); shading interp; colorbar;
Here is another suggestion:
Remove all the background.
Assuming this "line" results in a Bimodal distribution of the data (after removing the zeros), find the lower mode.
Assuming the values of the line are always lower than the background, apply a logic mask that set to zeros all values above the minimum + 2nd_mode, as demonstrated in the figure below (in red circle):
Here is how it works:
A = Slant(any(Slant,2),:); % save in A only the nonzero data
Now we have A that looks like this:
[y,x] = findpeaks(histcounts(A)); % find all the mode in the histogram of A
sorted_x = sortrows([x.' y.'],-2); % sort them by their hight in decendet order
mA = A<min(A(:))+sorted_x(2,1); % mask all values above the second mode
result = A.*mA; % apply the mask on A
And we get the result:
The resulted line has some holes within it, so you might want to interpolate the whole line from the result. This can be done with simple math on the indices:
[y1,x1] = find(mA,1); % find the first nonzero row
[y2,x2] = find(mA,1,'last'); % find the last nonzero row
m = (y1-y2)/(x1-x2); % the line slope
n = y1-m*x1; % the intercept
f_line = #(x) m.*x+n; % the line function
So we get a line function f_line like this (in red below):
Now we want to make this line thicker, like the line in the data, so we take the mode of the thickness (by counting the values in each column, you might want to take max instead), and 'expand' the line by half of this factor to both sides:
thick = mode(sum(mA)); % mode thickness of the line
tmp = (1:thick)-ceil(thick/2); % helper vector for expanding
rows = bsxfun(#plus,tmp.',floor(f_line(1:size(A,2)))); % all the rows for each coloumn
rows(rows<1) = 1; % make sure to not get out of range
rows(rows>size(A,1)) = size(A,1); % make sure to not get out of range
inds = sub2ind(size(A),rows,repmat(1:size(A,2),thick,1)); % convert to linear indecies
mA(inds) = 1; % add the interpolation to the mask
result = A.*mA; % apply the mask on A
And now result looks like this:
Idea: Use the Hough transform:
First of all it is best to create a new matrix with only the rows and columns we are interested in.
In order to apply matlab's built in hough we have to create a binary image: As the line always has lower values than the rest, we could e.g. determine the lowest quartile of the brightnesses present in the picture (using quantile, and set these to white, everything else to black.
Then to find the line, we can use hough directly on that BW image.
I would like to resize a 512X512 image into 363X762 image which will be larger than the original image(of size 512X512). Those extra pixel values must be different values in the range of 0-255.
I tried the following code:
I=imread('photo.jpg'); %photo.jpg is a 512X512 image
B=zeros(363,726);
sizeOfMatrixB=size(B);
display(sizeOfMatrixB);
B(1:262144)=I(1:262144);
imshow(B);
B(262155:263538)=0;
But I think this is a lengthy one and the output is also not as desired. Could anyone suggest me with a better piece of code to perform this. Thank you in advance.
I think that the code you have is actually pretty close to ideal except that you have a lot of hard-coded values in there. Those should really be computed on the fly. We can do that using numel to count the number of elements in B.
B = zeros(363, 726);
%// Assign the first 262144 elements of B to the values in I
%// all of the rest will remain as 0
B(1:numel(I)) = I;
This flexibility is important and the importance is actually demonstrated via the typo in your last line:
B(262155:263538)=0;
%// Should be
B(262144:263538)=0;
Also, you don't need these extra assignments to zero at the end because you initialize the matrix to be all zeros in the first place.
A Side Note
It looks like you are spreading the original image data for each column across multiple columns. I'm guessing this isn't what you want. You probably only want to grab the first 363 rows of I to be placed into B. You can do that this way:
B = zeros(363, 726);
B(1:size(B, 1), 1:size(I, 2)) = I(1:size(B, 1), :);
Update
If you want the other values to be something other than zero, you can initialize your matrix to be that value instead.
value = 2;
B = zeros(363, 726) + value;
B(1:numel(I)) = I;
If you want them to be random integers between 0 and 255, use randi to initialize the matrix.
B = randi([0 255], 363, 726);
B(1:numel(I)) = I;
this is my situation: I have a 30x30 image and I want to calculate the radial and tangent component of the gradient of each point (pixel) along the straight line passing through the centre of the image (15,15) and the same (i,j) point.
[dx, dy] = gradient(img);
for i=1:30
for j=1:30
pt = [dx(i, j), dy(i,j)];
line = [i-15, j-15];
costh = dot(line, pt)/(norm(line)*norm(pt));
par(i,j) = norm(costh*line);
tang(i,j) = norm(sin(acos(costh))*line);
end
end
is this code correct?
I think there is a conceptual error in your code, I tried to get your results with a different approach, see how it compares to yours.
[dy, dx] = gradient(img);
I inverted x and y because the usual convention in matlab is to have the first dimension along the rows of a matrix while gradient does the opposite.
I created an array of the same size as img but with each pixel containing the angle of the vector from the center of the image to this point:
[I,J] = ind2sub(size(img), 1:numel(img));
theta=reshape(atan2d(I-ceil(size(img,1)/2), J-ceil(size(img,2)/2)), size(img))+180;
The function atan2d ensures that the 4 quadrants give distinct angle values.
Now the projection of the x and y components can be obtained with trigonometry:
par=dx.*sind(theta)+dy.*cosd(theta);
tang=dx.*cosd(theta)+dy.*sind(theta);
Note the use of the .* to achieve point-by-point multiplication, this is a big advantage of Matlab's matrix computations which saves you a loop.
Here's an example with a well-defined input image (no gradient along the rows and a constant gradient along the columns):
img=repmat(1:30, [30 1]);
The results:
subplot(1,2,1)
imagesc(par)
subplot(1,2,2)
imagesc(tang)
colorbar
A biologist friend of mine asked me if I could help him make a program to count the squama (is this the right translation?) of lizards.
He sent me some images and I tried some things on Matlab. For some images it's much harder than other, for example when there are darker(black) regions. At least with my method. I'm sure I can get some useful help here. How should I improve this? Have I taken the right approach?
These are some of the images.
I got the best results by following Image Processing and Counting using MATLAB. It's basically turning the image into Black and white and then threshold it. But I did add a bit of erosion.
Here's the code:
img0=imread('C:...\pic.png');
img1=rgb2gray(img0);
%The output image BW replaces all pixels in the input image with luminance greater than level with the value 1 (white) and replaces all other pixels with the value 0 (black). Specify level in the range [0,1].
img2=im2bw(img1,0.65);%(img1,graythresh(img1));
imshow(img2)
figure;
%erode
se = strel('line',6,0);
img2 = imerode(img2,se);
se = strel('line',6,90);
img2 = imerode(img2,se);
imshow(img2)
figure;
imshow(img1, 'InitialMag', 'fit')
% Make a truecolor all-green image. I use this later to overlay it on top of the original image to show which elements were counted (with green)
green = cat(3, zeros(size(img1)),ones(size(img1)), zeros(size(img1)));
hold on
h = imshow(green);
hold off
%counts the elements now defined by black spots on the image
[B,L,N,A] = bwboundaries(img2);
%imshow(img2); hold on;
set(h, 'AlphaData', img2)
text(10,10,strcat('\color{green}Objects Found:',num2str(length(B))))
figure;
%this produces a new image showing each counted element and its count id on top of it.
imshow(img2); hold on;
colors=['b' 'g' 'r' 'c' 'm' 'y'];
for k=1:length(B),
boundary = B{k};
cidx = mod(k,length(colors))+1;
plot(boundary(:,2), boundary(:,1), colors(cidx),'LineWidth',2);
%randomize text position for better visibility
rndRow = ceil(length(boundary)/(mod(rand*k,7)+1));
col = boundary(rndRow,2); row = boundary(rndRow,1);
h = text(col+1, row-1, num2str(L(row,col)));
set(h,'Color',colors(cidx),'FontSize',14,'FontWeight','bold');
end
figure;
spy(A);
And these are some of the results. One the top-left corner you can see how many were counted.
Also, I think it's useful to have the counted elements marked in green so at least the user can know which ones have to be counted manually.
There is one route you should consider: watershed segmentation. Here is a quick and dirty example with your first image (it assumes you have the IP toolbox):
raw=rgb2gray(imread('lCeL8.jpg'));
Icomp = imcomplement(raw);
I3 = imhmin(Icomp,20);
L = watershed(I3);
%%
imagesc(L);
axis image
Result shown with a colormap:
You can then count the cells as follows:
count = numel(unique(L));
One of the advantages is that it can be directly fed to regionprops and give you all the nice details about the individual 'squama':
r=regionprops(L, 'All');
imshow(raw);
for k=2:numel(r)
if r(k).Area>100 % I chose 100 to filter out the objects with a small are.
rectangle('Position',r(k).BoundingBox, 'LineWidth',1, 'EdgeColor','b', 'Curvature', [1 1]);
end
end
Which you could use to monitor over/under segmentation:
Note: special thanks to #jucestain for helping with the proper access to the fields in the r structure here