Dynamic window for smoothdata
Show older comments
Say I have a single stream of data (vector) that has sections where I want different filter window lengths. For instance, I want specify a vector of window lengths equal to the length of my data, how? I don't want to break up my data stream.
The only option I see is to somehow dynamically scale the data using interpolation and then use a smoothdata with a fixed window.
3 Comments
Mathieu NOE
on 16 Mar 2021
hello Matt
there are plenty of smoothing methods with matlab . I have also this one in my pocket for special cases () . We could modify the code so that we can provide the length of the window (as a vector of same size as the data) - but what is your logic to generate the window length ?
Fs = 1000;
samples = 1000;
dt = 1/Fs;
t = (0:samples-1)*dt;
y = square(2*pi*3*t) + 0.1*randn(size(t));
% %%%%%%%%%%%%%%%%
figure(1)
N = 10;
ys = slidingavg(y, N);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with slidingavg' ]);
% %%%%%%%%%%%%%%%%
figure(2)
N = 10;
ys = medfilt1(y, N,'truncate');
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with medfilt1' ]);
grid on
%%%%%%%%%%%%%%%%
figure(3)
N = 10;
ys = sgolayfilt(y,3,51);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with sgolayfilt' ]);
grid on
%%%%%%%%%%%%%%%%
NN = 4;
Wn = 0.1;
[B,A] = butter(NN,Wn);
figure(4)
ys = filtfilt(B,A,y);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with butterworth LP' ]);
grid on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function out = slidingavg(in, N)
% OUTPUT_ARRAY = SLIDINGAVG(INPUT_ARRAY, N)
%
% The function 'slidingavg' implements a one-dimensional filtering, applying a sliding window to a sequence. Such filtering replaces the center value in
% the window with the average value of all the points within the window. When the sliding window is exceeding the lower or upper boundaries of the input
% vector INPUT_ARRAY, the average is computed among the available points. Indicating with nx the length of the the input sequence, we note that for values
% of N larger or equal to 2*(nx - 1), each value of the output data array are identical and equal to mean(in).
%
% * The input argument INPUT_ARRAY is the numerical data array to be processed.
% * The input argument N is the number of neighboring data points to average over for each point of IN.
%
% * The output argument OUTPUT_ARRAY is the output data array.
%
% © 2002 - Michele Giugliano, PhD and Maura Arsiero
% (Bern, Friday July 5th, 2002 - 21:10)
% (http://www.giugliano.info) (bug-reports to michele@giugliano.info)
%
% Two simple examples with second- and third-order filters are
% slidingavg([4 3 5 2 8 9 1],2)
% ans =
% 3.5000 4.0000 3.3333 5.0000 6.3333 6.0000 5.0000
%
% slidingavg([4 3 5 2 8 9 1],3)
% ans =
% 3.5000 4.0000 3.3333 5.0000 6.3333 6.0000 5.0000
%
if (isempty(in)) | (N<=0) % If the input array is empty or N is non-positive,
disp(sprintf('SlidingAvg: (Error) empty input data or N null.')); % an error is reported to the standard output and the
return; % execution of the routine is stopped.
end % if
if (N==1) % If the number of neighbouring points over which the sliding
out = in; % average will be performed is '1', then no average actually occur and
return; % OUTPUT_ARRAY will be the copy of INPUT_ARRAY and the execution of the routine
end % if % is stopped.
nx = length(in); % The length of the input data structure is acquired to later evaluate the 'mean' over the appropriate boundaries.
if (N>=(2*(nx-1))) % If the number of neighbouring points over which the sliding
out = mean(in)*ones(size(in)); % average will be performed is large enough, then the average actually covers all the points
return; % of INPUT_ARRAY, for each index of OUTPUT_ARRAY and some CPU time can be gained by such an approach.
end % if % The execution of the routine is stopped.
out = zeros(size(in)); % In all the other situations, the initialization of the output data structure is performed.
if rem(N,2)~=1 % When N is even, then we proceed in taking the half of it:
m = N/2; % m = N / 2.
else % Otherwise (N >= 3, N odd), N-1 is even ( N-1 >= 2) and we proceed taking the half of it:
m = (N-1)/2; % m = (N-1) / 2.
end % if
for i=1:nx, % For each element (i-th) contained in the input numerical array, a check must be performed:
if ((i-m) < 1) & ((i+m) <= nx) % If not enough points are available on the left of the i-th element..
out(i) = mean(in(1:i+m)); % then we proceed to evaluate the mean from the first element to the (i + m)-th.
elseif ((i-m) >= 1) & ((i+m) <= nx) % If enough points are available on the left and on the right of the i-th element..
out(i) = mean(in(i-m:i+m)); % then we proceed to evaluate the mean on 2*m elements centered on the i-th position.
elseif ((i-m) >= 1) & ((i+m) > nx) % If not enough points are available on the rigth of the i-th element..
out(i) = mean(in(i-m:nx)); % then we proceed to evaluate the mean from the element (i - m)-th to the last one.
elseif ((i-m) < 1) & ((i+m) > nx) % If not enough points are available on the left and on the rigth of the i-th element..
out(i) = mean(in(1:nx)); % then we proceed to evaluate the mean from the first element to the last.
end % if
end % for i
end
Matt Gaidica
on 16 Mar 2021
Matt Gaidica
on 16 Mar 2021
Accepted Answer
More Answers (0)
Categories
Find more on Data Preprocessing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!