peak average of consecutive values in a matrix array

4 次查看（过去 30 天）
Jorge Luis Paredes Estacio
评论： Mathieu NOE 2023-7-19
Hello, I would like to calculate the average of the peaks (positive and negative values) of consecutive values from a predifined number of elements from a Nx1 matrix, where N is the number of rows. For example, let's say I have a matrix with the following form (24x1):
A[1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21;22;23;24]
I want to obtain the peak and the average for every 5 elements,
B=[mean(peak(1,2,3,4,5)); mean(peak((2,3,4,5,6)); mean(peak((3,4,5,6,7));...........mean(peak((20,21,22,23,24))]
I am chosing every 5 elements to simplify the example. But, this number number of elements can be higher than 100 and the points vary dramatically from positive to negative as they come from acceleration earthquake signals from N higher than 245000.
I would appreciate the help.
  6 个评论
显示 4更早的评论
Dyuman Joshi
Dyuman Joshi 2023-7-19
Ran in:
Suppose this is the data in hand -
x = randi([-20 20],1,15)
x = 1×15
-5 -16 14 -18 19 -3 18 1 -10 -6 -12 16 5 8 0
Assume the moving window is 4, what should be the output in thise case? and what is the logic behind it?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Mathieu NOE
Mathieu NOE 2023-7-19
hello
maybe this ?
I am using peakseek , this fex submission is available here
PeakSeek - File Exchange - MATLAB Central (mathworks.com)
, but you can use the standard (slower) findpeaks function offered by TMW
the code provided below will buffer your data with length = 51 samples , then split the signal for each cahnnel into a positive and negative signal and find the positive and negative peaks , then take the mean of them (separately not mixing pos and neg peaks !)
example for X channel :
data = readmatrix('Signal.txt'); % acceleration earthquake signal in cm/sec2 ,X, Y and Z
[samples, channels] = size(data);
dt = 1; % sample rate (s)
t = dt*(0:samples-1); % time vector
%% home made solution (you choose the amount of overlap)
buffer_size = 51; % how many samples
overlap = 0; % overlap expressed in samples
%%%% main loop %%%%
[new_time,data_out_pos,data_out_neg] = my_peakmean(t,data,buffer_size,overlap);
figure(1),
plot(t,data(:,1),new_time,data_out_pos(:,1),'*-g',new_time,data_out_neg(:,1),'*-r');
title('X');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
figure(2),
plot(t,data(:,2),new_time,data_out_pos(:,2),'*-g',new_time,data_out_neg(:,2),'*-r');
title('Y');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
figure(3),
plot(t,data(:,3),new_time,data_out_pos(:,3),'*-g',new_time,data_out_neg(:,3),'*-r');
title('Z');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
%%%%%%%%%% my functions %%%%%%%%%%%%%%
function [new_time,data_out_pos,data_out_neg] = my_peakmean(t,data_in,buffer_size,overlap)
% NB : buffer size and overlap are integer numbers (samples)
% data (in , out) are 1D arrays (vectors)
shift = buffer_size-overlap; % nb of samples between 2 contiguous buffers
[samples,channels] = size(data_in);
nb_of_loops = fix((samples-buffer_size)/shift +1);
for k=1:nb_of_loops
start_index = 1+(k-1)*shift;
stop_index = min(start_index+ buffer_size-1,samples);
x_index(k) = round((start_index+stop_index)/2);
% find peaks
for n = 1:channels
tmp = data_in(start_index:stop_index,n);
tmp_pos = tmp;
tmp_pos(tmp_pos<0) = 0;
tmp_neg = tmp;
tmp_neg(tmp_neg>0) = 0;
[locsp, pksp]= peakseek(tmp_pos,1,max(tmp_pos)/10); % pos peaks above 10% of max pos amplitude
[locsn, pksn]= peakseek(-tmp_neg,1,max(-tmp_neg)/10); % neg peaks above 10% of max neg amplitude
data_out_pos(k,n) = mean(pksp); % mean of pos peaks
data_out_neg(k,n) = mean(-pksn); % mean of neg peaks
end
end
new_time = t(x_index); % time values are computed at the center of the buffer
end
  5 个评论
显示 3更早的评论
Jorge Luis Paredes Estacio
Jorge Luis Paredes Estacio 2023-7-19
I really appreaciate your help. it is sorted now. Thank you very much.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Descriptive Statistics 的更多信息

标签

产品


版本

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by