How can i find cumulative mean inside a for loop?
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Meghana Balasubramanian
le 3 Sep 2019
Commenté : Meghana Balasubramanian
le 3 Sep 2019
Hey everyone,
I have a code which helps me to calculate the mean values. Each of the "cellList.meshData(j)" corresponds to a single column vector of values. I would like it to give me the total mean of all 25 cells. The following code gives me the mean for each cell.
for j=1:25
data=cellList.meshData(j);
if isempty(data{1,1})
continue
end
mean_int_val = mean(cellList.meshData{j}{1,1}.signal2);
end
1 commentaire
Johannes Fischer
le 3 Sep 2019
To imporve code-readability, I would not use 'continue' in this case:
for j = 1:25
data = cellList.meshData(j);
if ~isempty(data{1, 1})
mean_int_val = mean(cellList.meshData{j}{1, 1}.signal2);
end
end
Réponse acceptée
Nicolas B.
le 3 Sep 2019
Modifié(e) : Nicolas B.
le 3 Sep 2019
Hi,
For your situation, you should consider that
. So they are 2 situations:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/236811/image.png)
- All vectors have the same size
- Vectors can have different sizes
If you are in the first situation, I would simply keep all mean_int_val and simply recompute the the mean of all means.
nData = 25;
mean_int_val = NaN(1, nData);
for j=1:nData
data=cellList.meshData(j);
if isempty(data{1,1})
continue
end
mean_int_val(j) = mean(cellList.meshData{j}{1,1}.signal2);
end
mean_total = mean(mean_int_val, 'omitnan');
If you are in the second situation, I would also keem the number of samples and then compute the mean.
nData = 25;
mean_int_val = NaN(1, nData);
mean_size = NaN(1, nData);
for j=1:nData
data=cellList.meshData(j);
if isempty(data{1,1})
continue
end
mean_int_val(j) = mean(cellList.meshData{j}{1,1}.signal2);
mean_size(j) = numel(cellList.meshData{j}{1,1}.signal2);
end
mean_total = sum(mean_int_val .* mean_size) / sum(mean_size);
4 commentaires
Nicolas B.
le 3 Sep 2019
Thanks for the comment. I corrected it with j instead of i (bad habits).
In my suggestions, the mean of all means is saved in mean_total variable.
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