I have a 512*512 matrix which contains calculated probabilities. I am trying to plot the ocuucrences of 0s and 1s of these using a heatmap so that my final image is somewhat like an inverted flattened gaussian function;
the problem is my code returns an image where all values are along the 0 only; this is not possible as I have an almost equal probability of 0s and 1s. Is there some problem with my plotting of the values?
X = reshape(prob_to_1,512,512); % prob_to_1 is the matrix of probabilities which
% is reshaped to a 512*512 matrix
colormap('hot');
imagesc(X);
set(gca, 'XTick', [0:0.05:1]*512, 'XTickLabel',[0:0.05:1]) % 10 ticks
set(gca, 'YTick', [0:0.1:1]*512, 'YTickLabel',[0:0.1:1]*100) % 20
colorbar('YTickLabel',{'100%','90%','80%','70%','60%','50%','40%','30%','20%','10%','0%'})
I have attached an image showing my output. As you can see, the black line which shows the no.of 1s should have been at the extreme right on the x-axes (close to prob of 1).
Any suggestions/ideas?
Thanks!
Well, the second image is a histogram of the same data. hist(X,nbins) is sufficient for plotting it.
Related
I have the following code in MATLAB:
I=imread(image);
h=fspecial('gaussian',si,sigma);
I=im2double(I);
I=imfilter(I,h,'conv');
figure,imagesc(I),impixelinfo,title('Original Image after Convolving with gaussian'),colormap('gray');
How can I define and apply a Gaussian filter to an image without imfilter, fspecial and conv2?
It's really unfortunate that you can't use the some of the built-in methods from the Image Processing Toolbox to help you do this task. However, we can still do what you're asking, though it will be a bit more difficult. I'm still going to use some functions from the IPT to help us do what you're asking. Also, I'm going to assume that your image is grayscale. I'll leave it to you if you want to do this for colour images.
Create Gaussian Mask
What you can do is create a grid of 2D spatial co-ordinates using meshgrid that is the same size as the Gaussian filter mask you are creating. I'm going to assume that N is odd to make my life easier. This will allow for the spatial co-ordinates to be symmetric all around the mask.
If you recall, the 2D Gaussian can be defined as:
The scaling factor in front of the exponential is primarily concerned with ensuring that the area underneath the Gaussian is 1. We will deal with this normalization in another way, where we generate the Gaussian coefficients without the scaling factor, then simply sum up all of the coefficients in the mask and divide every element by this sum to ensure a unit area.
Assuming that you want to create a N x N filter, and with a given standard deviation sigma, the code would look something like this, with h representing your Gaussian filter.
%// Generate horizontal and vertical co-ordinates, where
%// the origin is in the middle
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
%// Create Gaussian Mask
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
%// Normalize so that total area (sum of all weights) is 1
h = h / sum(h(:));
If you check this with fspecial, for odd values of N, you'll see that the masks match.
Filter the image
The basics behind filtering an image is for each pixel in your input image, you take a pixel neighbourhood that surrounds this pixel that is the same size as your Gaussian mask. You perform an element-by-element multiplication with this pixel neighbourhood with the Gaussian mask and sum up all of the elements together. The resultant sum is what the output pixel would be at the corresponding spatial location in the output image. I'm going to use the im2col that will take pixel neighbourhoods and turn them into columns. im2col will take each of these columns and create a matrix where each column represents one pixel neighbourhood.
What we can do next is take our Gaussian mask and convert this into a column vector. Next, we would take this column vector, and replicate this for as many columns as we have from the result of im2col to create... let's call this a Gaussian matrix for a lack of a better term. With this Gaussian matrix, we will do an element-by-element multiplication with this matrix and with the output of im2col. Once we do this, we can sum over all of the rows for each column. The best way to do this element-by-element multiplication is through bsxfun, and I'll show you how to use it soon.
The result of this will be your filtered image, but it will be a single vector. You would need to reshape this vector back into matrix form with col2im to get our filtered image. However, a slight problem with this approach is that it doesn't filter pixels where the spatial mask extends beyond the dimensions of the image. As such, you'll actually need to pad the border of your image with zeroes so that we can properly do our filter. We can do this with padarray.
Therefore, our code will look something like this, going with your variables you have defined above:
N = 5; %// Define size of Gaussian mask
sigma = 2; %// Define sigma here
%// Generate Gaussian mask
ind = -floor(N/2) : floor(N/2);
[X Y] = meshgrid(ind, ind);
h = exp(-(X.^2 + Y.^2) / (2*sigma*sigma));
h = h / sum(h(:));
%// Convert filter into a column vector
h = h(:);
%// Filter our image
I = imread(image);
I = im2double(I);
I_pad = padarray(I, [floor(N/2) floor(N/2)]);
C = im2col(I_pad, [N N], 'sliding');
C_filter = sum(bsxfun(#times, C, h), 1);
out = col2im(C_filter, [N N], size(I_pad), 'sliding');
out contains the filtered image after applying a Gaussian filtering mask to your input image I. As an example, let's say N = 9, sigma = 4. Let's also use cameraman.tif that is an image that's part of the MATLAB system path. By using the above parameters, as well as the image, this is the input and output image we get:
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.
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
I have two black and white images and I need to calculate the mutual information.
Image 1 = X
Image 2 = Y
I know that the mutual information can be defined as:
MI = entropy(X) + entropy(Y) - JointEntropy(X,Y)
MATLAB already has built-in functions to calculate the entropy but not to calculate the joint entropy. I guess the true question is: How do I calculate the joint entropy of two images?
Here is an example of the images I'd like to find the joint entropy of:
X =
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
Y =
0 0 0 0 0 0
0 0 0.38 0.82 0.38 0.04
0 0 0.32 0.82 0.68 0.17
0 0 0.04 0.14 0.11 0
0 0 0 0 0 0
To calculate the joint entropy, you need to calculate the joint histogram between two images. The joint histogram is essentially the same as a normal 1D histogram but the first dimension logs intensities for the first image and the second dimension logs intensities for the second image. This is very similar to what is commonly referred to as a co-occurrence matrix. At location (i,j) in the joint histogram, it tells you how many intensity values we have encountered that have intensity i in the first image and intensity j in the second image.
What is important is that this logs how many times we have seen this pair of intensities at the same corresponding locations. For example, if we have a joint histogram count of (7,3) = 2, this means that when we were scanning both images, when we encountered the intensity of 7, at the same corresponding location in the second image, we encountered the intensity of 3 for a total of 2 times.
Constructing a joint histogram is very simple to do.
First, create a 256 x 256 matrix (assuming your image is unsigned 8-bit integer) and initialize them to all zeroes. Also, you need to make sure that both of your images are the same size (width and height).
Once you do that, take a look at the first pixel of each image, which we will denote as the top left corner. Specifically, take a look at the intensities for the first and second image at this location. The intensity of the first image will serve as the row while the intensity of the second image will serve as the column.
Find this location in the matrix and increment this spot in the matrix by 1.
Repeat this for the rest of the locations in your image.
After you're done, divide all entries by the total number of elements in either image (remember they should be the same size). This will give us the joint probability distribution between both images.
One would be inclined to do this with for loops, but as it is commonly known, for loops are notoriously slow and should be avoided if at all possible. However, you can easily do this in MATLAB in the following way without loops. Let's assume that im1 and im2 are the first and second images you want to compare to. What we can do is convert im1 and im2 into vectors. We can then use accumarray to help us compute the joint histogram. accumarray is one of the most powerful functions in MATLAB. You can think of it as a miniature MapReduce paradigm. Simply put, each data input has a key and an associated value. The goal of accumarray is to bin all of the values that belong to the same key and do some operation on all of these values. In our case, the "key" would be the intensity values, and the values themselves are the value of 1 for every intensity value. We would then want to add up all of the values of 1 that map to the same bin, which is exactly how we'd compute a histogram. The default behaviour for accumarray is to add all of these values. Specifically, the output of accumarray would be an array where each position computes the sum of all values that mapped to that key. For example, the first position would be the summation of all values that mapped to the key of 1, the second position would be the summation of all values that mapped to the key of 2 and so on.
However, for the joint histogram, you want to figure out which values map to the same intensity pair of (i,j), and so the keys here would be a pair of 2D coordinates. As such, any intensities that have an intensity of i in the first image and j in the second image in the same spatial location shared between the two images go to the same key. Therefore in the 2D case, the output of accumarray would be a 2D matrix where each element (i,j) contains the summation of all values that mapped to key (i,j), similar to the 1D case that was mentioned previously which is exactly what we are after.
In other words:
indrow = double(im1(:)) + 1;
indcol = double(im2(:)) + 1; %// Should be the same size as indrow
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
With accumarray, the first input are the keys and the second input are the values. A note with accumarray is that if each key has the same value, you can simply assign a constant to the second input, which is what I've done and it's 1. In general, this is an array with the same number of rows as the first input. Also, take special note of the first two lines. There will inevitably be an intensity of 0 in your image, but because MATLAB starts indexing at 1, we need to offset both arrays by 1.
Now that we have the joint histogram, it's really simple to calculate the joint entropy. It is similar to the entropy in 1D, except now we are just summing over the entire joint probability matrix. Bear in mind that it will be very likely that your joint histogram will have many 0 entries. We need to make sure that we skip those or the log2 operation will be undefined. Let's get rid of any zero entries now:
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
Take notice that I searched the joint histogram instead of the joint probability matrix. This is because the joint histogram consists of whole numbers while the joint probability matrix will lie between 0 and 1. Because of the division, I want to avoid comparing any entries in this matrix with 0 due to numerical roundoff and instability. The above will also convert our joint probability matrix into a stacked 1D vector, which is fine.
As such, the joint entropy can be calculated as:
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
If my understanding of calculating entropy for an image in MATLAB is correct, it should calculate the histogram / probability distribution over 256 bins, so you can certainly use that function here with the joint entropy that was just calculated.
What if we have floating-point data instead?
So far, we have assumed that the images that you have dealt with have intensities that are integer-valued. What if we have floating point data? accumarray assumes that you are trying to index into the output array using integers, but we can still certainly accomplish what we want with this small bump in the road. What you would do is simply assign each floating point value in both images to have a unique ID. You would thus use accumarray with these IDs instead. To facilitate this ID assigning, use unique - specifically the third output from the function. You would take each of the images, put them into unique and make these the indices to be input into accumarray. In other words, do this instead:
[~,~,indrow] = unique(im1(:)); %// Change here
[~,~,indcol] = unique(im2(:)); %// Change here
%// Same code
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
Note that with indrow and indcol, we are directly assigning the third output of unique to these variables and then using the same joint entropy code that we computed earlier. We also don't have to offset the variables by 1 as we did previously because unique will assign IDs starting at 1.
Aside
You can actually calculate the histograms or probability distributions for each image individually using the joint probability matrix. If you wanted to calculate the histograms / probability distributions for the first image, you would simply accumulate all of the columns for each row. To do it for the second image, you would simply accumulate all of the rows for each column. As such, you can do:
histogramImage1 = sum(jointHistogram, 1);
histogramImage2 = sum(jointHistogram, 2);
After, you can calculate the entropy of both of these by yourself. To double check, make sure you turn both of these into PDFs, then compute the entropy using the standard equation (like above).
How do I finally compute Mutual Information?
To finally compute Mutual Information, you're going to need the entropy of the two images. You can use MATLAB's built-in entropy function, but this assumes that there are 256 unique levels. You probably want to apply this for the case of there being N distinct levels instead of 256, and so you can use what we did above with the joint histogram, then computing the histograms for each image in the aside code above, and then computing the entropy for each image. You would simply repeat the entropy calculation that was used jointly, but apply it to each image individually:
%// Find non-zero elements for first image's histogram
indNoZero = histogramImage1 ~= 0;
%// Extract them out and get the probabilities
prob1NoZero = histogramImage1(indNoZero);
prob1NoZero = prob1NoZero / sum(prob1NoZero);
%// Compute the entropy
entropy1 = -sum(prob1NoZero.*log2(prob1NoZero));
%// Repeat for the second image
indNoZero = histogramImage2 ~= 0;
prob2NoZero = histogramImage2(indNoZero);
prob2NoZero = prob2NoZero / sum(prob2NoZero);
entropy2 = -sum(prob2NoZero.*log2(prob2NoZero));
%// Now compute mutual information
mutualInformation = entropy1 + entropy2 - jointEntropy;
Hope this helps!
I have looked at many articles and answers to questions on how the Viola-Jones algorithm really works. I keep finding the answers saying the "sum of pixels" in a certain region subtracted by the "sum of pixels" in the adjacent region. I'm confused on what "sum of pixels" means. What is the value based on? Is it the number of pixels in the area? The intensity of the color?
Thanks in advance.
These are the definitions based on Viola-Jones paper on 'Robust Real-time Object Detection'
Integral Image: Integral Image(ii) at location x, y = ii(x,y)
ii(x,y) = > Sum of the pixels above and to the left of x, y inclusive
Here 'Sum of Pixels' implies the sum of pixels intensity values ( e.g., for a 8 bit gray scale image, a value between 0 and 255 ) at each pixel element to the above and to the left of pixel (x, y) and including the row/column x and y, considering a gray scale image in the representation.
Significance of the integral image is that it speeds up the computation of the sum of pixel intensities within any rectangular block of pixels. e.g. four array references.
And the integral image value by itself at each point given by ii(x,y) can be computed in one pass over the original image i(x,y)
using the below equations on each point during the pass as detailed in the reference paper:
s(x,y) = s(x,y-1) + i(x,y);
ii(x,y) = ii(x-1,y) + s(x,y);
where
s(x,y) = the cumulative row sum;
s(x,-1) = 0;
ii(-1,y) = 0;
These integral image values are then used to generate features to learn and later detect objects.
The original Viola-Jones algorithm uses "Haar-like" features, which are approximations of first and second Gaussian derivative filters.
Gaussian derivative filters look like this:
Haar-like filters look like this:
The reason Viola and Jones used Haar-like filters, is that they can be evaluated very efficiently. All you have to do is subtract the sum of pixels covered by the black region of the filter from the sum of pixels covered by the white region. And since the regions are rectangular, the sum of the pixels in each region can be efficiently calculated from the corresponding integral image.