Algorythm for Average of excel data

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SATYA PAL on 15 Feb 2024
Edited: Jon on 23 Feb 2024
Dear Sir please suggest .. How can I get average of my attached data in excel at an interval of every 5 datas. like at time interval 0-0.5 sec then 0.6-1 sec, 1-1.5 sec and so on....
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Jon on 21 Feb 2024
Edited: Jon on 23 Feb 2024
You have a lot of what look like helpful answers to your question. Unless, there is something that has not been addressed in these answers, it would be good for you to now select one of them as the answer. This will allow the question to be marked as answered so that others will know that an answer is available.

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Accepted Answer

Dyuman Joshi on 15 Feb 2024
Edited: Dyuman Joshi on 15 Feb 2024
data = readtable('S1IA.csv')
data = 5758×3 table
Time A B ____ ________ _________ 0.1 -0.54042 0.0005 0.2 -0.54042 0.0013958 0.3 -0.54042 0.0022292 0.4 -0.50863 0.0030625 0.5 -0.50863 0.0038958 0.6 -0.50863 0.0046042 0.7 -0.50863 0.0054792 0.8 10.792 0.0062708 0.9 19.677 0.0071042 1 31.439 0.0080208 1.1 48.049 0.0088958 1.2 61.957 0.0097292 1.3 74.72 0.010563 1.4 87.818 0.011396 1.5 99.977 0.012229 1.6 112.66 0.013062
%define bins to distribute bins in
idx = 0:0.5:0.5*ceil(max(data.Time)/0.5);
%Get the mean of the rest of the columns for the specified bins
out = groupsummary(data, 1, idx, @mean, 'IncludedEdge', 'right')
out = 1152×4 table
disc_Time GroupCount fun1_A fun1_B _________ __________ _______ _________ [0, 0.5] 5 -0.5277 0.0022167 (0.5, 1] 5 12.178 0.0062958 (1, 1.5] 5 74.504 0.010562 (1.5, 2] 5 136.88 0.014729 (2, 2.5] 5 196.86 0.018896 (2.5, 3] 5 256.22 0.023062 (3, 3.5] 5 314.36 0.027229 (3.5, 4] 5 371.88 0.031396 (4, 4.5] 5 430.15 0.035562 (4.5, 5] 5 488.54 0.039733 (5, 5.5] 5 548.08 0.043904 (5.5, 6] 5 607.7 0.048062 (6, 6.5] 5 668.74 0.052229 (6.5, 7] 5 730.91 0.056396 (7, 7.5] 5 794.33 0.060562 (7.5, 8] 5 859.02 0.064729
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SATYA PAL on 21 Feb 2024
I used this approach and its working nicely
Dyuman Joshi on 22 Feb 2024
Glad to have helped!

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More Answers (4)

Jon on 15 Feb 2024
Edited: Jon on 15 Feb 2024
If you have the Statistics and Machine learning toolbox you could do it like this
% Parameters
grpIncr = 0.5 % time increment for group averages
grpIncr = 0.5000
% Read the data into a matrix
dat = readmatrix('S1IA.csv')
dat = 5758x3
0.1000 -0.5404 0.0005 0.2000 -0.5404 0.0014 0.3000 -0.5404 0.0022 0.4000 -0.5086 0.0031 0.5000 -0.5086 0.0039 0.6000 -0.5086 0.0046 0.7000 -0.5086 0.0055 0.8000 10.7924 0.0063 0.9000 19.6775 0.0071 1.0000 31.4395 0.0080
% Provide grouping variable that makes elements within a specified sampling
% interval have the same group value
grp = floor(dat(:,1)/0.5);
[dat grp]
ans = 5758x4
0.1000 -0.5404 0.0005 0 0.2000 -0.5404 0.0014 0 0.3000 -0.5404 0.0022 0 0.4000 -0.5086 0.0031 0 0.5000 -0.5086 0.0039 1.0000 0.6000 -0.5086 0.0046 1.0000 0.7000 -0.5086 0.0055 1.0000 0.8000 10.7924 0.0063 1.0000 0.9000 19.6775 0.0071 1.0000 1.0000 31.4395 0.0080 2.0000
% Calculate mean of each group
stats = grpstats(dat(:,2:3),grp)
stats = 1152x2
-0.5325 0.0018 5.7888 0.0055 60.7967 0.0097 124.6993 0.0139 184.9874 0.0181 244.3314 0.0222 302.8520 0.0264 360.3649 0.0306 418.3420 0.0347 476.8403 0.0389
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SATYA PAL on 21 Feb 2024
I used this approach also and its also working nicely
SATYA PAL on 21 Feb 2024
Thanks a lot

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Voss on 15 Feb 2024
Maybe something like this:
filename = 'S1IA.csv';
T = readtable(filename);
T.Time = seconds(T.Time);
T = table2timetable(T,'RowTimes','Time');
new_t = T.Time(1):seconds(0.5):T.Time(end);
T = retime(T,new_t,'mean')
T = 1152x2 timetable
Time A B _______ _______ _________ 0.1 sec -0.5277 0.0022167 0.6 sec 12.178 0.0062958 1.1 sec 74.504 0.010562 1.6 sec 136.88 0.014729 2.1 sec 196.86 0.018896 2.6 sec 256.22 0.023062 3.1 sec 314.36 0.027229 3.6 sec 371.88 0.031396 4.1 sec 430.15 0.035562 4.6 sec 488.54 0.039733 5.1 sec 548.08 0.043904 5.6 sec 607.7 0.048062 6.1 sec 668.74 0.052229 6.6 sec 730.91 0.056396 7.1 sec 794.33 0.060562 7.6 sec 859.02 0.064729
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Mathieu NOE on 15 Feb 2024
hello again
well, this is quite the same as my answer to your other post
adapted to your new data file , this becomes :
data = readmatrix('S1IA.csv'); % Time,A,B
t = data(:,1);
dt = mean(diff(t));
%% home made solution (you choose the amount of overlap)
buffer_size = round(0.5/dt); % how many samples for 0.5 seconds buffer ?
overlap = 0; % overlap expressed in samples
%%%% main loop %%%%
[new_time,data_out] = my_movmean(t,data(:,2:3),buffer_size,overlap);
figure(2),
plot(t,data(:,2),new_time,data_out(:,1),'*-r');
title('A');
legend('raw data','0.5s mean');
xlabel('Time(s)');
figure(3),
plot(t,data(:,3),new_time,data_out(:,2),'*-r');
title('B');
legend('raw data','0.5s mean');
xlabel('Time(s)');
%%%%%%%%%% my functions %%%%%%%%%%%%%%
function [new_time,data_out] = my_movmean(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,~] = 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);
data_out(k,:) = mean(data_in(start_index:stop_index,:),1,'omitnan'); %
end
new_time = t(x_index); % time values are computed at the center of the buffer
end
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SATYA PAL on 21 Feb 2024
Thanks a lot sir

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Alexander on 15 Feb 2024
A very easy approach (as allways):
%Algorythm for Average of excel data
%https://de.mathworks.com/matlabcentral/answers/2082483-algorythm-for-average-of-excel-data
clear; close all;
data = dlmread('S1IA.csv',',',1,0);
t = data(:,1);A = data(:,2);B = data(:,3);
dy = floor(length(A)/5)
t = t(1:dy*5); % maximum 4 samples lost!
tr = reshape(t,5,dy);
trMean = mean(tr);
A = A(1:dy*5); % maximum 4 samples lost!
Ar = reshape(A,5,dy);
ArMean = mean(Ar);
B = B(1:dy*5); % maximum 4 samples lost!
Br = reshape(B,5,dy);
BrMean = mean(Br);
subplot(211)
plot(trMean ,ArMean); grid minor; title('A')
subplot(212)
plot(trMean, BrMean); grid minor; title('B')
@SATYA PAL beautifying is up to you.
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SATYA PAL on 21 Feb 2024
Thanks a lot sir

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