How to get daily minimum value of temperature from a three dimensional matrix?

Let us say we have a matrix (say 'a') with 129x135x721 dimension.i.e.lat x lon x time... time is 720 (24*30)for hourly data for one month values. Suppose this data is temperature data. Now i need to know at what time of the day temperature is minimum.So for that, first i need to find mean value of temperature at all 1 o'clock,2,3,4 all 24 hr. i.e.,mean of (1:25:720) Similarly mean of (2:26:720) and so on. Then finally for each lat,lon we get 24 values now, or we will get a final matrix of 129x135x24 and from this i need to find the minimum value time. Please help me how to to do it

[Matt J]

Is it 721 hours or 720? You have different numbers in different places.

Answer 1

[Matt J]

Make your array 4-dimensional,

 a=reshape(a, 129,135,24,30);

 hourlymeans=mean(a,4);

 [~,minhour]=min(hourlymeans,[],3);

[trailokya]

I also need the hourly means matrix that should be 129*135*24 But I am not getting this by running the above codes...

[Matt J]

Works fine for me,

>> a=rand(129,135,24,30);
>> hourlymeans=mean(a,4); whos hourlymeans

    Name               Size                  Bytes  Class     Attributes

    hourlymeans      129x135x24            3343680  double

Answer 2

[KSSV]

T = rand(129,135,721) ;
Min = zeros(721,1) ;
for i = 1:721
    Ti = T(:,:,i) ;
    Min(i) = min(Ti(:)) ;   % Minimum value at evry time step 
end
[val,idx] = min(Min) ;

[trailokya]

First i need the hourly average values (say for the whole month). The final matrix should be 129*135*24...how is it possible Sir?