Calculate average over each specific interval

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Saad Alqahtani 2021-7-7

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/874158-calculate-average-over-each-specific-interval

评论： Saad Alqahtani 2021-7-8

采纳的回答： dpb

Hi,

I'm trying to calculate the mean every 5 seconds of data till the end.

so x= (1,63369) % my data in Newton Meter which is about 30.9419 seconds long

t = 5*2048 % 5 seconds interval.

I'd like to have your help in (1- how to make it calculate the mean in increaments from start till the end 2- how to deal with the last posrtion of the data that shorter than 5 seconds)

any help would be appreciate it. Thanks in advance.

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Image Analyst 2021-7-7

在 MATLAB Online 中打开

Please attach your time and signal data (I'm guessing it's t and x) in a .mat file with the paperclip icon

save('answers.mat', 't', 'x');

so we can try some things.

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Saad Alqahtani 2021-7-7

answers.mat

Definitely. Please see attachment. Thanks!

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Answer 1

dpb 2021-7-7

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https://ww2.mathworks.cn/matlabcentral/answers/874158-calculate-average-over-each-specific-interval#answer_742138

编辑：dpb 2021-7-8

在 MATLAB Online 中打开

2) first -- how to deal with the last posrtion of the data that shorter than 5 seconds"

How do you WANT to deal with it? You can either ignore it and have fix(numel/5) samples or average the values that do have in the last segment -- it's your choice.

1) The classic MATLAB way is to reshape() and use mean() --

a) Ignore last partial segment --

N=5;                                                % number elements to average over
mnX=mean(reshape(x(1:N*fix(numel(x)/N)),N,[]));     % average over N elements ignoring end values

b) Deal with odd end

x=[x nan(1,N*ceil(numel(x)/N)-numel(x))];           % augment vector to make even
mnX=mean(reshape(x,N,[]),1,'omitnan');              % average over N elements ignoring NaN elements

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dpb 2021-7-8

在 MATLAB Online 中打开

Must've made a typo somewhere; works for me---

>> x=rand(1,63369);
>> numel(x)
ans =
       63369
>> N
N =
     5
>> ceil(numel(x)/5)
ans =
       12674
>> N*ceil(numel(x)/N)
ans =
       63370
>> ans/N
ans =
       12674
>> x=[x nan(1,N*ceil(numel(x)/N)-numel(x))];
>> numel(x)
ans =
       63370
>> mnX=mean(reshape(x(1:5*fix(numel(x)/5)),5,[]));
>> whos mnX
  Name      Size                Bytes  Class     Attributes
  mnX       1x12674            101392  double              
>> 