I am working on pedestrian step detection algorithm (acceleration data), I need to calculate statistical features from my filtered signal not raw data. I have already calculated mean, var and std and now I want to calculate correlation coefficients and gradient from filtered data. My filter data is of 1x37205 double. I calculated these features using for loop with moving window size=2samples and 50% overlap of previous window. Below I am attaching the code I tried. But when I run this for corrcoef it gives 1's as output for the whole data and for gradient it gives error Assignment has more non-singleton rhs dimensions than non-singleton subscripts. I am unable to understand well. Could some one suggest me or provide any code help in matlab and/or how can I work on that?
windowsize=2;
%%C is used for corrcoef and G for gradient. %%Data_filtered is caclulated
%%from acceleration xyz.
function [C, G] = features(Data_filtered, window)
C=zeros(length(Data_filtered),1);
G=zeros(length(Data_filtered),1);
for i=window:(length(Data_filtered))
C(i,1)=corrcoef(Data_filtered(i+1-window:i));
end;
for i=window:(length(Data_filtered))
G(i,1)=gradient(Data_filtered(i+1-window:i));
end;
end
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I want to evaluate the grid quality where all coordinates differ in the real case.
Signal is of a ECG signal where average life-time is 75 years.
My task is to evaluate its age at the moment of measurement, which is an inverse problem.
I think 2D approximation of the 3D case is hard (done here by Abo-Zahhad) with with 3-leads (2 on chest and one at left leg - MIT-BIT arrhythmia database):
where f is a piecewise continuous function in R^2, \epsilon is the error matrix and A is a 2D matrix.
Now, I evaluate the average grid distance in x-axis (time) and average grid distance in y-axis (energy).
I think this can be done by Matlab's Image Analysis toolbox.
However, I am not sure how complete the toolbox's approaches are.
I think a transform approach must be used in the setting of uneven and noncontinuous grids. One approach is exact linear time euclidean distance transforms of grid line sampled shapes by Joakim Lindblad et all.
The method presents a distance transform (DT) which assigns to each image point its smallest distance to a selected subset of image points.
This kind of approach is often a basis of algorithms for many methods in image analysis.
I tested unsuccessfully the case with bwdist (Distance transform of binary image) with chessboard (returns empty square matrix), cityblock, euclidean and quasi-euclidean where the last three options return full matrix.
Another pseudocode
% https://stackoverflow.com/a/29956008/54964
%// retrieve picture
imgRGB = imread('dummy.png');
%// detect lines
imgHSV = rgb2hsv(imgRGB);
BW = (imgHSV(:,:,3) < 1);
BW = imclose(imclose(BW, strel('line',40,0)), strel('line',10,90));
%// clear those masked pixels by setting them to background white color
imgRGB2 = imgRGB;
imgRGB2(repmat(BW,[1 1 3])) = 255;
%// show extracted signal
imshow(imgRGB2)
where I think the approach will not work here because the grids are not necessarily continuous and not necessary ideal.
pdist based on the Lumbreras' answer
In the real examples, all coordinates differ such that pdist hamming and jaccard are always 1 with real data.
The options euclidean, cytoblock, minkowski, chebychev, mahalanobis, cosine, correlation, and spearman offer some descriptions of the data.
However, these options make me now little sense in such full matrices.
I want to estimate how long the signal can live.
Sources
J. Müller, and S. Siltanen. Linear and nonlinear inverse problems with practical applications.
EIT with the D-bar method: discontinuous heart-and-lungs phantom. http://wiki.helsinki.fi/display/mathstatHenkilokunta/EIT+with+the+D-bar+method%3A+discontinuous+heart-and-lungs+phantom Visited 29-Feb 2016.
There is a function in Matlab defined as pdist which computes the pairwisedistance between all row elements in a matrix and enables you to choose the type of distance you want to use (Euclidean, cityblock, correlation). Are you after something like this? Not sure I understood your question!
cheers!
Simply, do not do it in the post-processing. Those artifacts of the body can be about about raster images, about the viewer and/or ... Do quality assurance in the signal generation/processing step.
It is much easier to evaluate the original signal than its views.
I have an RGB image and I am trying to calculate its Gaussian derivative.
Image is a greyscale image and the Gaussian window is 5x5,st is the standard deviation
This is the code i am using in order to find a 2D Gaussian derivative,in Matlab:
N=2
[X,Y]=meshgrid(-N:N,-N:N)
G=exp(-(x.^2+y.^2)/(2*st^2))/(2*pi*st)
G_x = -x.*G/(st^2);
G_x_s = G_x/sum(G_x(:));
G_y = -y.*G/(st^2);
G_y_s = G_y/sum(G_y(:));
where st is the standard deviation i am using. Before I proceed to the convolution of the Image using G_x_s and G_y_s, i have the following problem. When I use a standard deviation that is an even number(2,4,6,8) the program works and gives results as expected. But when i use an odd number for standard deviation (3 or 5) then the G_y_s value becomes Inf because sum(G_y(:))=0. I do not understand that behavior and I was wondering if there is some problem with the code or if in the above formula the standard deviation can only be an even number. Any help will be greatly appreciated.
Thank you.
Your program doesn't work at all. The results you find when using an even number is just because of some numerical errors.
Your G will be a matrix symmetrical to the center. x and y are both point symmetrical to the center. So the multiplication (G times x or y) will result in a matrix with a sum of zero. So a division by that sum is a division by zero. Everything else you observe is because of some roundoff errors. Here, I see a sum og G_xof about 1.3e-17.
I think your error is in the multiplication x.*G and y.*G. I can not figure out why you would do that.
I assume you want to do edge detection rigth? You can use fspecialto create several edge filters. Laplace of gaussian for instance. You could also create two gaussian filters with different standard deviations and subtract them from another to get an edge filter.
Can someone explain to me the correlation function corr2 in MATLAB? I know that it is for 2D comparing similarities of objects, but in the equation I have doubts what it is A and B (probably matrices for comparison), and also Amn and Bmn.
I'm not sure how MATLAB executes this function, because I have found in several cases that the correlation is not executed for the entire image (matrix) but instead it divides the image into blocks and then compares blocks of one picture with blocks of another picture.
In MATLAB's documentation, the corr2 equation is not put as referral point to the way the equation itself is calculated, like in other functions in MATLAB's documentation, such as referring to what book it is taken from and where it is explained.
The correlation coefficient is a number representing the similarity between 2 images in relation with their respective pixel intensity.
As you pointed out this function is used to calculate this coefficient:
Here A and B are the images you are comparing, whereas the subscript indices m and n refer to the pixel location in the image. Basically what Matab does is to compute, for every pixel location in both images, the difference between the intensity value at that pixel and the mean intensity of the whole image, denoted as a letter with a straightline over it.
As Kostya pointed out, typing edit corr2 in the command window will show you the code used by Matlab to compute the correlation coefficient. The formula is basically this:
a = a - mean2(a);
b = b - mean2(b);
r = sum(sum(a.*b))/sqrt(sum(sum(a.*a))*sum(sum(b.*b)));
where:
a is the input image and b is the image you wish to compare to a.
If we break down the formula, we see that a - mean2(a) and b-mean2(b) are the elements in the numerator of the above equation. mean2(a) is equivalent to mean(mean(a)) or mean(a(:)), that is the mean intensity of the whole image. This is only calculated once.
The 3rd line of code calculates the coefficient. Here sum(sum(a.*b)) calculates the double-sum present in the formula element-wise, that is considering each pixel location separately. Be aware that using sum(a) calculates the sum in every column individually, hence in order to get a single value you need to apply sum twice.
That's pretty much the same happening in the denominator, however calculations are performed on a-mean2(a)^2 and b-mean2(b)^2. You can see this a some kind of normalization process in which you consider the pixel intensity difference among each individual image.
As for your last comment, you can break down an image into small blocks and calculate the correlation coefficient on them; that might save some time for very large images but since everything is vectorized the calculation is quite fast. It might be useful in distributed processing I guess. Of course the correlation coefficient between 2 blocks of images is not necessarily identical to that of the whole image.
For the sake of curiosity you can look at this paper which highlights some caveats in using the correlation coefficient for image comparison.
Hope that makes things a bit clearer!
In MATLAB I need to generate a second derivative of a gaussian window to apply to a vector representing the height of a curve. I need the second derivative in order to determine the locations of the inflection points and maxima along the curve. The vector representing the curve may be quite noise hence the use of the gaussian window.
What is the best way to generate this window?
Is it best to use the gausswin function to generate the gaussian window then take the second derivative of that?
Or to generate the window manually using the equation for the second derivative of the gaussian?
Or even is it best to apply the gaussian window to the data, then take the second derivative of it all? (I know these last two are mathematically the same, however with the discrete data points I do not know which will be more accurate)
The maximum length of the height vector is going to be around 100-200 elements.
Thanks
Chris
I would create a linear filter composed of the weights generated by the second derivative of a Gaussian function and convolve this with your vector.
The weights of a second derivative of a Gaussian are given by:
Where:
Tau is the time shift for the filter. If you are generating weights for a discrete filter of length T with an odd number of samples, set tau to zero and allow t to vary from [-T/2,T/2]
sigma - varies the scale of your operator. Set sigma to a value somewhere between T/6. If you are concerned about long filter length then this can be reduced to T/4
C is the normalising factor. This can be derived algebraically but in practice I always do this numerically after calculating the filter weights. For unity gain when smoothing periodic signals, I will set C = 1 / sum(G'').
In terms of your comment on the equivalence of smoothing first and taking a derivative later, I would say it is more involved than that. As which derivative operator would you use in the second step? A simple central difference would not yield the same results.
You can get an equivalent (but approximate) response to a second derivative of a Gaussian by filtering the data with two Gaussians of different scales and then taking the point-wise differences between the two resulting vectors. See Difference of Gaussians for that approach.
I'm trying to generate an AR(2) process with MATLAB's filter() function, as shown here:
A=[1 -2.7607 3.8106 -2.6535 0.9238];
% AR(4) coefficients
y=filter(1,A,0.2*randn(1024,1));
% Filter a white noise input to create AR(4) process
[ar_coeffs,nv] =arburg(y,4);
%compare the results in ar_coeffs to the vector A.
I have a time series data set and would like to approximately match the 'total' variance of the data in a simulated data set. When I use nv in place of 0.2 in the second line of code, I get a variance in the simulated that is much too small.
Can anyone help me rectify this situation to generate a look-alike simulated AR(N) data set?
Thanks,
Mark
If you look at nv in this example it is 0.0392, this is variance. To create a white noise with variance a^2 you need to multiply that sequence by a. If a^2 = 0.392 then a is 0.198 (very close to 0.2). So Colin T is right and you need to multiply your randn(1024,1) by sqrt(nv) not by nv.