Mean of multiple large matrices with NaN's

1 Ansicht (letzte 30 Tage)
Matthew Cooper
Matthew Cooper am 8 Mai 2015
Bearbeitet: Matthew Cooper am 14 Mai 2015
I have 23 individual matrices each are 385x781x365. They are saved as .mat files, each are ~100mb. I would like to take the mean of all 23 along the third dimension. Unfortunately, there is too much data to load all 23, concatenate them, and use mean(concatenated_array,3).
As an alternative, I have resorted to a loop (dummy data for your use):
data = randn(385,781,365,23);
for n = 1:23
if n == 1
data_mean = data(:,:,:,n)./23;
else
data_mean = data_mean + data(:,:,:,n)./23;
end
end
data_mean2 = squeeze(mean(data,4));
If you look at the data, you will see that data_mean and data_mean2 are equivalent (isequal will return 0 because of rounding, though).
So, this works ok. However, a problem arises because the 23rd matrix is actually 385x781x274. Thus, on the final iteration the loop fails because the matrices are not the same size.
I have tried setting the 275:365 of the final matrix equal to the mean:
data = randn(385,781,365,23);
for n = 1:23
if n == 1
data_mean = data(:,:,:,n)./23;
elseif n == 23
data(:,:,275:365,n) = data_mean(:,:,275:365);
data_mean = data_mean + data(:,:,:,n)./23;
else
data_mean = data_mean + data(:,:,:,n)./23;
end
end
However you will notice that this is not right.
Ideally, I would like to set the 275:365 of the final matrix to nan, but then upon division the entire data_mean becomes NaN for 275:365.
Any ideas how to fix? Ideally a solution that does not require the loop. Thank you.

Antworten (0)

Kategorien

Mehr zu Resizing and Reshaping Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by