How can I average a 4-D array every nth element without using too much loops

2 vues (au cours des 30 derniers jours)
Hi,
So I am currently working on a (476,238,1,2699) array using MatLab 2017a corresponding to (longitude, latitude, 1, time). I basically want to make a monthly average of the data (so every 5th point). The problem is I tried using a loop but it takes so much time. Here is the code I am using :
data_zeros = [];
for i = 1:2699
for j = 1:238
for k = 1:476
if (mod(i, 5) == 0)
data_zeros(k,j,1,i) = mean(data(k,j,1,i-4:i));
i
j
k
end
end
end
end
I need your insight, please help :) !
  3 commentaires
Isma Di Carlo
Isma Di Carlo le 22 Juin 2018
The data is in seconds starting from 15th January 1947. I have a time table. Like I know a point corresponds to that point in time. I just need to make a monthly average and I estimated that 5 points of data = 1 month.
(The 4D array is because the 3rd object is the depth. I just only took the 10 first meters of my dataset)
Walter Roberson
Walter Roberson le 22 Juin 2018
"I have a time table."
Use retime()

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Réponse acceptée

OCDER
OCDER le 21 Juin 2018
data = rand(476, 238, 1, 2699);
N = floor(2699/5);
data_zeros = zeros(476, 238, 1, N); %Preallocate this for speed
k = 1;
for i = 5:5:2699
data_zeros(:, :, 1, k) = mean(data(:, :, 1, i-4:i), 4); %Mean along 4th dimension
k = k+1;
end

Plus de réponses (1)

Walter Roberson
Walter Roberson le 21 Juin 2018
mean( reshape(data, size(data,1), size(data,2), 5, size(data,4)/5 ), 3 )
I take advantage here of the fact that the third dimension is length 1 to pull in groups of 5 into the third dimension and mean() along that dimension.
However, you have the difficulty that your data is not an exact multiple of 5 long in that dimension. How do you want to take the mean of the entries 2696:2699 ?
  1 commentaire
Isma Di Carlo
Isma Di Carlo le 22 Juin 2018
Works amazingly like the other answer ! I interpolated for the last data point !

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