I have a nonlinear curve (found using bicubic interpolation; see below) that describes the transformation of intensity values from the target image to reference image values. (Apologies if I'm not using the correct terminology here).
What is the best way of applying this curve to the target image?
I am basically looking for the fastest way to achieve this;
for i = 1:length(curve)
I(I==i) = curve(i);
end
which is fairly slow.
Your curve function as look-up table, the simplest way to execute look up table is:
lookuptable=[ 9 8 7 6 5 4 3 2 1 0 ];
I=[ 1 3 4;
5 3 8];
Itransformed=lookuptable(I)
Notice that the index of the lookuptable is accessed by the value of the pixel.
So if the pixel value range is 0-255, first you should use lookuptable with size of 256 use, and second remember to compensate the fact that matlab index is 1:256 so use:
Itransformed=lookuptable(I-1);
Related
I am wondering if anyone can explain the interpretation of the size (number of feature) in a time series? For example consider a simple script in Matlab
X= randn(2,5,2)
X(:,:,1) =
-0.5530 0.4291 0.3937 -1.2534 0.2811
-1.4926 -0.7019 -0.8305 -1.4034 1.9545
X(:,:,2) =
0.2004 0.1438 2.3655 -0.1589 0.7140
0.4905 0.2301 -0.7813 -0.6737 0.2552
Assume X is a time series with the following output
This generates 2 vectors of length 5 each has 2 rows. Can anyone tell me what is exactly the meaning of first 2 and 5?
In some websites it says a creating 5 vectors of length 5 and size 2. What does size mean here?
Is 2 like number of features and 5 is like number of time series. The reason for this confusion is because I do not understand how to interpret following sentence:
"Generate 2 vector-valued sequences of length 5; each vector has size
2."
What do size 2 and length 5 mean here?
This entirely depends on your data, and how you want to store this. If you have some 2D data over time, I find it convenient to have a data matrix with in the 1st and 2nd dimension the 2D data per time step, and in the 3rd dimension time.
Say I have a movie of 1920 by 1080 pixels with 100 frames, I'd store this as mov = rand(1080,1920,100) (1080 and 1920 swapped because of row, col order of indexing). So now mov(:,:,1) would give me the first frame etc.
BTW, your X is a normal array, not to be confused with the timeseries object.
I'm sure this is a trivial question for a signals person. I need to find the function in Matlab that outputs averaging of contiguous segments of windowsize= l of a vector, e.g.
origSignal: [1 2 3 4 5 6 7 8 9];
windowSize = 3;
output = [2 5 8]; % i.e. [(1+2+3)/3 (4+5+6)/3 (7+8+9)/3]
EDIT: Neither one of the options presented in How can I (efficiently) compute a moving average of a vector? seems to work because I need that the window of size 3 slides, and doesnt include any of the previous elements... Maybe I'm missing it. Take a look at my example...
Thanks!
If the size of the original data is always a multiple of widowsize:
mean(reshape(origSignal,windowSize,[]));
Else, in one line:
mean(reshape(origSignal(1:end-mod(length(origSignal),windowSize)),windowSize,[]))
This is the same as before, but the signal is only taken to the end minus the extra values less than windowsize.
I'm sure this is a very simple mistake by me somewhere! But when I use Matlab's graycomatrix function, I don't get the expected result. Instead of a matrix output I expect, I always get an 8 x 8 (nearly) zero matrix with one entry in the bottom right - usually equal to 16. I haven't changed the default settings or used 'offset', so I'm not too sure what the problem is.
That's because your image is not normalized!
Your image should be range 0-1, so:
I = [1 1 2; 2 2 3; 1 2 5]; %or any other I
glcm = graycomatrix(I/max(I(:))); % or I/255 , but it would not work for this example
should do the job.
In your case, Matlab interprets that everything avobe 1 is 1, therefore the co-occurrence matrix gives you a unique value in the max position.
I found the detail and implementation of Local Ternary Pattern (LTP) on Calculating the Local Ternary Pattern of an image?. I want to ask more details that what the best way to choose the threshold t and also I have confusion in understand the role of reorder_vector = [8 7 4 1 2 3 6 9];
Unfortunately there isn't a good way to figure out what the threshold is using LTPs. It's mostly trial and error or by experimentation. However, I could suggest to make the threshold adaptive. You can use Otsu's algorithm to dynamically determine the best threshold of your image. This is assuming that the distribution of your intensities in the image is bimodal. In other words, there is a clear separation between objects and background. MATLAB has an implementation of this by the graythresh function. However, this generates a threshold between 0 and 1, so you will need to multiply the result by 255, assuming that the type of your image is uint8.
Therefore, do:
t = 255*graythresh(im);
im is the image that you desire to compute the LTPs. Now, I can certainly provide insight on what the reorder_vector is doing. Look at the following figure on how to calculate LTPs:
(source: hindawi.com)
When we generate the ternary code matrix (matrix in the middle), we need to generate an 8 element sequence that doesn't include the middle of the neighbourhood. We start from the east most element (row 2, column 3), then traverse the elements in counter-clockwise order. The reorder_vector variable allows you to select those specific elements that respect that order. If you recall, MATLAB can access matrices using column-major linear indices. Specifically, given a 3 x 3 matrix, we can access an element using a number from 1 to 9 and the memory is laid out like so:
1 4 7
2 5 8
3 6 9
Therefore, the first element of reorder_vector is index 8, which is the east most element. Next is index 7, which is the top right element, then index 4 which is the north facing element, then 1, 2, 3, 6 and finally 9.
If you follow these numbers, you will determine how I got the reorder_vector:
reorder_vector = [8 7 4 1 2 3 6 9];
By using this variable for accessing each 3 x 3 local neighbourhood, we would thus generate the correct 8 element sequence that respects the ordering of the ternary code so that we can proceed with the next stage of the algorithm.
This is a question about Matlab/Octave.
I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.
EDIT:Sorry forgot to mention:I am using Octave for Windows 3.2.4. This is where i see this behavior.
Please see the questions below, and point if I am missing something.
1] I have a 1D data array b=[ 3 5 -8 6 0];
out=medfilt1(b,3);
I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]
How come? What is wrong here?
FYI-Help says it pads the data at boundaries by 0(zero).
2] How does medfilt2(2D median filter command in matlab) work.
Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".
For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?
Any pointers will help.
thank you. -AD
I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.
When finding the median in a vector of even length, one usually take the mean of the two median values,
median([1 3 4 5]) = (3+4)/2 = 3.5
This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be
median([0 3 5]) = (3+5)/2 = 4
and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.
For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!
A = [1 2 3
14 5 33
11 7 13];
median(median(A)) = 11
median(A(:)) = 7