Assume that I have vector shown in the figure below. By common sense, we can see that there are 2 values which suddenly depart from the trend of the vector.
How do I eliminate these sudden changes. I mean how do I automatically detect and replace these noise values by the average value of their neighbors.
Define a threshold, compute the average values, then compare the relative error between the values and the averages of their neighbors:
threshold = 5e-2;
averages = [v(1); (v(3:end) + v(1:end-2)) / 2; v(end)];
is_outlier = (v.^2 - averages.^2) > threshold^2 * averages.^2;
Then replace the outliers:
v(is_outlier) = averages(is_outlier);
Related
I'm working on fitting a model to data I have collected. Within the model, I identify the optimal parameter values that best fit participant data then adjust and reduce the viewing window (range) to get more precise parameter values. I'm using linspace to generate new values to test against the participant data. Initially using a prior adaptation I had no issues. Once I edited my script to include a conditional "If" section to prevent parameters from being less than zero, I encountered the "Error using linspace (line 22) Inputs must be scalar. If I'm understanding the problem correctly its becuase I'm passing a vector into the linspace function instead of a scalar, but I don't understand why that is occuring now and was not the case with my previous adaptation. I'm assuming it has to do with the " if min_ < 0....." addition. Could someone help me to better understand where I'm making my error and how to correct this?
"Error using linspace (line 22)
Inputs must be scalars."
"Error in Range_Adjustment (line 43)
alphas = linspace(min_aro, max_aro,10);"
%Finds coordinates of optimal parameters for Condition 0
[x0,y0] = find(all_SSD_cond0 == min(min(all_SSD_cond0)));
%%% Identifies alpha and beta values that correspond to the coordinates for
%%% optimal parameters%%%
alphas(y0)
betas(x0)
%%%% Adjusts range for Condition 0 %%%
%Adjusts Alpha Range%
alpha_range = alpha_range.*0.7 % Decreases alpha range by 30%
min_aro = alphas(y0)-(alpha_range./(0.5)) %Creates optimized minimum of alpha range half distance away from identified optimal alpha value from previous iteration.
if min_aro < 0 % Conditional statement if minimum alpha range is less than zero min_aro = 0
min_aro = 0
end
max_aro = alphas(y0)+(alpha_range./(0.5)) %Creates optimized maximum value of alpha range half distance away from identified optimal alpha value from previous iteration.
%Adjusts Beta Range%
beta_range = beta_range.*0.7;% Decreases beta range by 30%
min_bro = betas(x0)-(beta_range./(0.5)); %Creates optimized minimum of beta range half distance away from identified optimal beta value from previous iteration.
if min_bro < 0 % Conditional statement if minimum beta range is less than zero min_bro = 0
min_bro = 0
end
max_bro = betas(x0)+(beta_range./(0.5)); %Creates optimized minimum of beta range half distance away from identified optimal beta value from previous iteration.
%%% Adjusts viewing frame and creates optimized alpha and beta parameters
%%% for fitting process%%%
alphas = linspace(min_aro, max_aro,10);
betas = linspace(min_bro, max_bro,10);
%alphas = linspace(alphas(y0)-(alpha_range./2), alphas(y0)+(alpha_range./2),10);
%betas = linspace(betas(x0)-(beta_range./2), betas(x0)+(beta_range./2),10);
You need to change your code so that you guarantee that x0 and y0 only contain one index each. If there are multiple elements of the matrix that equal the minimum value, x0 and y0 will be arrays that propagate all the way to linspace().
How to do it depends on what behavior you want, if you for example want the first element of the matrix that equals the minimum value you can do this:
[x0, y0] = find(all_SSD_cond0 == min(min(all_SSD_cond0)), 1, 'first');
I have 2 nested loops which do the following:
Get two rows of a matrix
Check if indices meet a condition or not
If they do: calculate xcorr between the two rows and put it into new vector
Find the index of the maximum value of sub vector and replace element of LAG matrix with this value
I dont know how I can speed this code up by vectorizing or otherwise.
b=size(data,1);
F=size(data,2);
LAG= zeros(b,b);
for i=1:b
for j=1:b
if j>i
x=data(i,:);
y=data(j,:);
d=xcorr(x,y);
d=d(:,F:(2*F)-1);
[M,I] = max(d);
LAG(i,j)=I-1;
d=xcorr(y,x);
d=d(:,F:(2*F)-1);
[M,I] = max(d);
LAG(j,i)=I-1;
end
end
end
First, a note on floating point precision...
You mention in a comment that your data contains the integers 0, 1, and 2. You would therefore expect a cross-correlation to give integer results. However, since the calculation is being done in double-precision, there appears to be some floating-point error introduced. This error can cause the results to be ever so slightly larger or smaller than integer values.
Since your calculations involve looking for the location of the maxima, then you could get slightly different results if there are repeated maximal integer values with added precision errors. For example, let's say you expect the value 10 to be the maximum and appear in indices 2 and 4 of a vector d. You might calculate d one way and get d(2) = 10 and d(4) = 10.00000000000001, with some added precision error. The maximum would therefore be located in index 4. If you use a different method to calculate d, you might get d(2) = 10 and d(4) = 9.99999999999999, with the error going in the opposite direction, causing the maximum to be located in index 2.
The solution? Round your cross-correlation data first:
d = round(xcorr(x, y));
This will eliminate the floating-point errors and give you the integer results you expect.
Now, on to the actual solutions...
Solution 1: Non-loop option
You can pass a matrix to xcorr and it will perform the cross-correlation for every pairwise combination of columns. Using this, you can forego your loops altogether like so:
d = round(xcorr(data.'));
[~, I] = max(d(F:(2*F)-1,:), [], 1);
LAG = reshape(I-1, b, b).';
Solution 2: Improved loop option
There are limits to how large data can be for the above solution, since it will produce large intermediate and output variables that can exceed the maximum array size available. In such a case for loops may be unavoidable, but you can improve upon the for-loop solution above. Specifically, you can compute the cross-correlation once for a pair (x, y), then just flip the result for the pair (y, x):
% Loop over rows:
for row = 1:b
% Loop over upper matrix triangle:
for col = (row+1):b
% Cross-correlation for upper triangle:
d = round(xcorr(data(row, :), data(col, :)));
[~, I] = max(d(:, F:(2*F)-1));
LAG(row, col) = I-1;
% Cross-correlation for lower triangle:
d = fliplr(d);
[~, I] = max(d(:, F:(2*F)-1));
LAG(col, row) = I-1;
end
end
I'm trying to detect peak values in MATLAB. I'm trying to use the findpeaks function. The problem is that my data consists of 4200 rows and I just want to detect the minimum and maximum point in every 50 rows.After I'll use this code for real time accelerometer data.
This is my code:
[peaks,peaklocations] = findpeaks( filteredX, 'minpeakdistance', 50 );
plot( x, filteredX, x( peaklocations ), peaks, 'or' )
So you want to first reshape your vector into 50 sample rows and then compute the peaks for each row.
A = randn(4200,1);
B = reshape (A,[50,size(A,1)/50]); %//which gives B the structure of 50*84 Matrix
pks=zeros(50,size(A,1)/50); %//pre-define and set to zero/NaN for stability
pklocations = zeros(50,size(A,1)/50); %//pre-define and set to zero/NaN for stability
for i = 1: size(A,1)/50
[pks(1:size(findpeaks(B(:,i)),1),i),pklocations(1:size(findpeaks(B(:,i)),1),i)] = findpeaks(B(:,i)); %//this gives you your peak, you can alter the parameters of the findpeaks function.
end
This generates 2 matrices, pklocations and pks for each of your segments. The downside ofc is that since you do not know how many peaks you will get for each segment and your matrix must have the same length of each column, so I padded it with zero, you can pad it with NaN if you want.
EDIT, since the OP is looking for only 1 maximum and 1 minimum for each 50 samples, this can easily be satisfied by the min/max function in MATLAB.
A = randn(4200,1);
B = reshape (A,[50,size(A,1)/50]); %//which gives B the structure of 50*84 Matrix
[pks,pklocations] = max(B);
[trghs,trghlocations] = min(B);
I guess alternatively, you could do a max(pks), but it is simply making it complicated.
I have two signals and I calculated the local peaks of each signal and saved them in two different vectors for amplitude and another two for timing. I need to get the intersection between peaks. Each peak has a value and a time so I am trying to extract the peaks which has nearly the same amplitude at nearly the same time .. Any help??
My Code:
[svalue1,stime1] = findpeaks(O1);
[svalue2,stime2] = findpeaks(O2);
%note that the peaks count is different in each signal
% This is my try but it is not working
x = length(intersect(ceil(svalue1),ceil(svalue2)))/min(length(svalue1),length(svalue2));
It is my understanding what you want to determine those values in svalue1 and svalue2 that are similar to each other, and what's more important is that they are unequal in length.
What you can do is compare every value in svalue1 with every value in svalue2 and if the difference between a value in svalue1 and a value in svalue2 is less than a certain amount, then we would classify these two elements to be the same.
This can be achieved by bsxfun with the #minus function and eliminating any sign changes with abs. After, we can determine the locations where the values are below a certain amount.
Something like this:
tol = 0.5; %// Adjust if necessary
A = abs(bsxfun(#minus, svalue1(:), svalue2(:).')) <= tol;
[row,col] = find(A);
out = [row,col];
tol is the tolerance that we would use to define whether two values are close together. I chose this to be 0.5, but adjust this for your application. out is a 2D matrix that tells you which value in svalue1 was closest to svalue2. Rather than giving a verbose explanation, let's just show you an example of this working and we can explain along the way.
Let's try this on an example:
>> svalue1 = [0 0.1 1 2.2 3];
>> svalue2 = [0.1 0.2 2 3 4];
Running the above code, we get:
>> out
ans =
1 1
2 1
1 2
2 2
4 3
5 4
Now this makes sense. Each row tells you which value in svalue1 is close to svalue2. For example, the first row says that the first value in svalue1, or 0 is close to the second value in svalue2 or 0.1. The next row says that the second value of svalue1, or 0.2, is close to the first value of svalue2, or 0.
Obviously, this operation includes non-unique values. For example, the row with [1 2] and [2 1] are the same. I'm assuming this isn't a problem, so we'll leave that alone.
Now what I didn't cover is whether the peaks also happen within the same time value. This can be done by performing another bsxfun operation on the time vector values of stime1 and stime2 much like we did with svalue1 and svalue2, and performing a logical AND operation between the two matrices. Should the peaks be the same in both amplitude and time, then the result follows.... so something like this:
tol_amplitude = 5; %// Adjust if necessary
tol_time = 0.5;
A = abs(bsxfun(#minus, svalue1(:), svalue2(:).')) <= tol_amplitude;
Atime = abs(bsxfun(#minus, stime1(:), stime2(:).')) <= tol_time;
Afinal = A & Atime;
[row,col] = find(Afinal);
out = [row,col];
You'll notice that we have two thresholds for the time and the amplitudes. Adjust both if necessary. out will contain the results like we saw earlier, but these will give you those indices that are close in both time and amplitude. If you want to see what those are, you can do something like this:
peaks = [svalue1(out(:,1)) svalue2(out(:,2))];
times = [stime1(out(:,1)) stime2(out(:,2))];
peaks and times will give you what the corresponding peaks and times were that would be considered as "close" between the two signals. The first column denotes the peaks and times for the first signal and the second column is for the peaks and times for the second signal. The difference between columns should be less than their prescribed thresholds.
Below is some MATLAB code which is inside a loop that increments position every cycle. However after this code has run, the values in the vector F1 are all NaN. I print A to the screen and it outputs the expected values.
% 500 Hz Bands
total_energy = sum(F(1:(F_L*8)));
A = (sum(F(1:F_L))) /total_energy %% correct value printed to screen
F1(position) = (sum(F(1:F_L))) /total_energy;
Any help is appreciated
Assuming that F_L = position * interval, I suggest you use something like:
cumulative_energy_distribution = cumsum(abs(F).^2);
positions = 1:q;
F1 = cumulative_energy_distribution(positions * interval) ./ cumulative_energy_distribution(positions * 8 * interval);
sum-of-squares, which is the energy density (and also seen in Parseval's theorem) is monotonically increasing, so you don't need to worry about the energy going back down to zero.
In MATLAB, if you divide zero by zero, you get a NaN. Therefore, its always better to add an infinitesimal number to the denominator such that its addition doesn't change the final result, but it avoids division by zero. You can choose any small value as that infinitesimal number (such as 10^-10) but MATLAB already has a variable called eps.
eps(n) gives a distance to the next-largest number (whose precision is same as n) which could be represented in MATLAB. For example, if n=1, next double-precision number you could get to from 1 is 1+eps(1)=1+2.2204e-16. If n=10^10, then the next number is 10^10+1.9073e-06. eps is same as eps(1)=2.2204e-16. Adding eps doesn't change the output and avoids the situation of 0/0.