Using Matlab, I have to transform the intensity of an image using this given graphic:
where s=T(r)
I absolutely have no idea about this transformation. It's for an homework, can someone help me at least to recognize the function?
This is just the equation of a line, y=ax+b...
in matlab notation:
r=linspace(0,30,1000); %this can be whatever r units are... r=0:0.1:100 etc.
T=2*(r-10).*(r>=10 & r<=20);
plot(r,T)
will yield the plot you just draw...
The solution was s = 2r - 20 on the interval of [10, 20] and 0 otherwise.
Then, using an existing image, it's possible to modify the intensity.
Related
I have an image I which pixel intensities fall within the range of 0-1. I can calculate the image histogram by normalizing it but I found the curves is not exactly the same as the histogram of raw data. This will cause some issue for the later peaks finding process(See attached two images).
My question is in Matlab, is there any way I can plot the image histogram without normalization the data so that I can keep the curve shape unchanged? This will benefit for those raw images when their pixel intensities are not within 0-1 ranges. Currently, I cannot calculate their histogram if I don't normalize the data.
The Matlab code for normalization and histogram calculation is attached. Any suggestion will be appreciated!
h = imhist(mat2gray(I));
Documentation of imhist tells us that the function checks the data type of the input and scale the values accordingly. Therefore, without testing with your attached data, this may work:
h = imhist(uint8(I));
An alternatively you may scale the integer-representation to floating-representation, by either using argument of mat2gray
h = imhist(mat2gray(I, [0,255]));
or just divide it.
h = imhist(I/255);
The imhist answer in this thread describing normalizing or casting is completely correctly. Alternatively, you could use the histogram function in MATLAB which will work with unnormalized floating point data:
A = 255*rand(500,500);
histogram(A);
I need to generate some random 2D points (for example 30 points) near the y=x line, insert them in a matrix, plot it and then calculate the SVD of the matrix. But since I'm new to MATLAB I don't know how can I generate my desired matrix.
Since this looks like homework I'll just post some general ideas here.
randi can be used to get semi-random integers. Using that you can create a 2D matrix by duplicating the array and putting them together. Thus: generate a 30x1 column and duplicate it to a 30x2 column. All rows will have the same two entries, i.e. x=y.
Noise can be added to this by creating a 30x2 matrix of random numbers, use rand for that and simply add that to the previously created matrix.
Check the documentation on svd to see how the singular-value decomposition works, it's fairly straight-forward if you know your linear algebra.
Finally for plotting you can use various tools such as image, imagesc, plot, surf and scatter, try them and see which works best for you.
Here is a quick example I made: https://saturnapi.com/fullstack/2d-points-randomly-near-line
%// Welcome to Saturn's MATLAB-Octave API.
%// Delete the sample code below these comments and write your own!'
x = 13 + 6.*rand(20,1);
y = x*0.7 + 0.5*rand(20,1);
[X,Y] = meshgrid(x,y)
figure(1);
plot(x,y,'.');
%// Print plot as PNG with resultion of 60 pixels per inch
print("MyPNG.png", "-dpng", "-r60");
I'm trying to transform my original gray image to mapped gray image using grey-scale mapping function. I have no idea how to get any two minima correspond to the grey-scale range [a,b] of the original histogram so that I can use these values for the equations below to get the mapped gray image.
f(x,y)=0 if [0,a),
f(x,y)=(255/(a-b))-a for [a,b],
f(x,y)=255 if (b,255]
Thank you!
So essentially you want to scale the histogram of your image to range from 0 - 255. All you need is the max and the min. The easiest way to find them is
a = min(I(:));
b = max(I(:));
Also I suspect you middle equation should actually be
f(x,y)=(255/(a-b))*(f(x,y)-a) for [a,b]
however that would eliminate the need for your first two equations. So it's possible that a and b are not the extrema in your case but that you are actually trying to accentuate some range of intensities that sit in the middle of your images histogram (and essentially discard all information outside of that range). In this case you have not given us enough information to suggest values for either a or b.
I know there are many plotting documents for Matlab online and I am pretty sure that it has been asked many times. I aplogize in advance for any inconvenience.
I am dealing with a new distribution and I need to draw 3D plot for different values of parameters (I can do it with Excel or any other programs, however, since my other graphs is drawn with MATLAB, and I need to put this 3D in Matlab, too, to publish it as an article). I calculated the result using MATLAB loops, however, plotting gives me the hardest time. I had no other choice but to ask for your assistance. I have these equations for different alphas and betas with a constant sigma and calculate Galton's Skewness and Moor's Kurtosis given with the last two equations.
median=sqrt(2*(sigma^2)*beta*gammaincinv(0.5,alpha));
q1=sqrt(2*(sigma^2)*beta*gammaincinv((6/8),alpha));
q3=sqrt(2*(sigma^2)*beta*gammaincinv((2/8),alpha));
q4=sqrt(2*(sigma^2)*beta*gammaincinv((7/8),alpha));
q5=sqrt(2*(sigma^2)*beta*gammaincinv((5/8),alpha));
q6=sqrt(2*(sigma^2)*beta*gammaincinv((3/8),alpha));
q7=sqrt(2*(sigma^2)*beta*gammaincinv((1/8),alpha));
galtonskewness=(q1-2*median+q3)/(q1-q3);
moorskurtosis=(q4-q5+q6-q7)/(q1-q3);
Let's assume that,
sigma=1
beta=[0.1 0.2 0.5 1 2 5];
alpha=[0.1 0.2 0.5 1 2 5];
I have used mesh(X,Y,Z) for the same range of alphas and betas with the same increment but I take the error "these values cannot be complex". I just want to draw something like the one below.
It must be something easy that I am missing out, but I do not understand where the mistake is. I appreciate any help. Thank you!
I ran the above code for a 2D mesh of points for alpha and beta between 0.1 and 5 for both dimensions and I got results for both.
I suspect it's due to your alpha and beta declaration. You are only providing a few points, and if you try to use mesh, it won't get good results. Therefore, define a meshgrid of points for both alpha and beta, then vectorize your MATLAB code to produce the kurotsis and skewness curves. Only under certain situations should you use for loops. In general, you should avoid using them whenever possible.
How meshgrid works is that given a range of X and Y values, it will produce two (or three if you want 3D co-ordinates) arrays where each location in each array gives you the spatial co-ordinate at that particular location. Therefore, if we did something like:
[X,Y] = meshgrid(1:3, 1:3);
This is what we get:
X =
1 2 3
1 2 3
1 2 3
Y =
1 1 1
2 2 2
3 3 3
Notice that in a 2D grid, for the top-left corner, (x,y) = (1,1), and so for the corresponding location in X, we get 1 and Y we get 1. If you do the same logic for any other position in the 2D grid, you simply look at the X and Y values in each array and it will tell you what the component is for each dimension.
As such, instead of looping through all possible points in your grid, generate them all using meshgrid, then vectorize the computation by calculating your values all at once rather than individually. Once you do this, you have the right structure to be able to put this into mesh.
Therefore, try doing this instead:
%// Define meshgrid of points
[alpha,beta] = meshgrid(0.1:0.1:5, 0.1:0.1:5);
%// From your code
sigma = 1;
%// Calculate quantities - Notice that this is all vectorized
med=sqrt(2*(sigma^2)*beta.*gammaincinv(0.5,alpha));
q1=sqrt(2*(sigma^2)*beta.*gammaincinv((6/8),alpha));
q3=sqrt(2*(sigma^2)*beta.*gammaincinv((2/8),alpha));
q4=sqrt(2*(sigma^2)*beta.*gammaincinv((7/8),alpha));
q5=sqrt(2*(sigma^2)*beta.*gammaincinv((5/8),alpha));
q6=sqrt(2*(sigma^2)*beta.*gammaincinv((3/8),alpha));
q7=sqrt(2*(sigma^2)*beta.*gammaincinv((1/8),alpha));
galtonskewness=(q1-2*med+q3)./(q1-q3);
moorskurtosis=(q4-q5+q6-q7)./(q1-q3);
%// Show our meshes
figure;
mesh(alpha, beta, galtonskewness);
figure;
mesh(alpha, beta, moorskurtosis);
Also take note that I renamed your median variable to med. MATLAB has a function called median and so you don't want to unintentionally shadow over this function with a variable of the same name.
This is what I get:
Take note that I'm not getting the plots that you have placed in your post. It may be because I'm choosing the wrong variables to define the mesh, or perhaps your equations may be incorrect. Double check what you know in theory to what you have here in code and try again.
This should hopefully give you enough to start with though!
I have the file data.txt with two columns and N rows, something like this:
0.009943796 0.4667975
0.009795735 0.46777886
0.009623984 0.46897832
0.009564759 0.46941447
0.009546991 0.4703958
0.009428543 0.47224948
0.009375241 0.47475737
0.009298249 0.4767201
[...]
Every couple of values in the file correspond to one point coordinates (x,y).
If plotted, this points generate a curve. I would like to calculate the area under curve (AUC) of this curve.
So I load the data:
data = load("data.txt");
X = data(:,1);
Y = data(:,2);
So, X contains all the x coordinates of the points, and Y all the y coordinates.
How could I calculate the area under curve (AUC) ?
Easiest way is the trapezoidal rule function trapz.
If your data is known to be smooth, you could try using Simpson's rule, but there's nothing built-in to MATLAB for integrating numerical data via Simpson's rule. (& I'm not sure how to use it for x/y data where x doesn't increase steadily)
just add
AUC = trapz(X,Y)
to your program
and you will get the area under the curve
You can do something like that:
AUC = sum((Y(1:end-1)+Y(2:end))/2.*...
(X(2:end)-X(1:end-1)));
Source: Link
An example in MATLAB to help you get your answer ...
x=[3 10 15 20 25 30];
y=[27 14.5 9.4 6.7 5.3 4.5];
trapz(x,y)
In case you have negative values in y, you can do like,
y=max(y,0)
[~,~,~,AUC] = perfcurve(labels,scores,posclass);
% posclass might be 1
http://www.mathworks.com/matlabcentral/newsreader/view_thread/252131
There are some options to trapz for the person ready to do some coding by themselves. This link shows the implementation of Simpson's rule, with python code included. There is also a File Exchange on simpsons rule.