Finding the mean of every 24 elements of 3rd dimension of a 3d array

I have hourly weather data from 2000-2005 for an area giving a 241x121x52608 array. I am trying to condense the data down into the daily mean and so I'm looking to maintain the 241x121 matrix which denotes the location while condensing the 52608/24 into 2192 days in total (Including leap years)

I'm sure there is a simple way to do this but I'm new to Matlab so I'm struggling to come up with a solution.

Thanks

Answer 1

[Fangjun Jiang]

Something like this. Try simple example to make sure the dimension, row, column are right.

a=ones(2,3,10);
b=mean(reshape(a,2,3,2,[]),4)

b = 
b(:,:,1) =

     1     1     1
     1     1     1


b(:,:,2) =

     1     1     1
     1     1     1

[Liam]

I'm having trouble understanding the code, is it finding the mean of the first 5 and second 5 elements?

I'm testing it with the following code and the e value doesn't match the b(1,1,1) value

a=rand(2,3,10);

b=mean(reshape(a,2,3,2,[]),4);

c=a(1,1,:);

d=c(1,1,1)+c(1,1,2)+c(1,1,3)+c(1,1,4)+c(1,1,5);

e=d/5;

[Fangjun Jiang]

There are multiple ways. One example below. See doc for the arguments of calling reshape() and mean()

a=rand(2,3,10);
b=mean(reshape(a,2,3,5,[]),3);
b(1,1,1)

ans = 0.4785

c=a(1,1,:);
d=c(1,1,1)+c(1,1,2)+c(1,1,3)+c(1,1,4)+c(1,1,5);
e=d/5

e = 0.4785

[Liam]

That seemed to do the trick, thanks a million! New to this so I'm learning bit by bit!