# How to find the maximum value for each 24 rows in an array?

4 views (last 30 days)
Behzad Navidi on 4 Nov 2019
Hello,
I have a 3d array, precip= :,:, 8760. in fact 8760 rows are available in one column. I want to find the maximum value in this column 24-by-24 in rows. And saving the bigger value and eliminate the smaller one. and do it for all 8760-row
so if the dimension before doing this is precip = :, :, 8760, after this work should be precip = :, :, 365.
I wanna practical this for a 3d array which the third dimension is what I talking about.
I'm attaching all my array. as the volume of original file is so big I cut first 72 rows and attach it
Thank you

Behzad Navidi on 13 Nov 2019
Dear Shubham,
Sorry for the delayed reply, I was really sick.Thank you for your comment.
Actually, this 3D dimension represents: longitude*latitude*time
all this data represents hourly maximum temperature for a certain area in a year. the time is hourly so it has 87600 values. but I want to have a daily maximum temperature so I need to pick the biggest value of 24 hours (it is a maximum temperature of all day)
I want to divide third dimension 24-by-24 (day by day) and then selecting the maximum of every 24 values. in this way, after doing that I have a 3d array in which the third dimension has 365 or 366(leap years) values.
Thank you again
Shubham Gupta on 13 Nov 2019
Let's take a example with an array of dimension 2x3x4. So time = 1hr, 2hr, 3hr & 4hr. At each hour let's assume random data for temp & assume that on this earth 1 day is of 2hr.
t = 1hr
T = [20 30 40
50 60 70];
t = 2hr
T = [30 50 70
80 90 75];
t = 3hr
T = [-20 40 10
15 900 80];
t = 4hr
T = [25 55 11
10 -45 9];
What is the output that you are expecting in this hypothetical situation? I understand your 3rd dimesion (pages) will be 2 but I still don't know what do you want to do with the 1st & 2nd dimesion. I am expecting your output to be:
Day = 1
T = ??
Day = 2
T = ??
Behzad Navidi on 13 Nov 2019
ok if assume that on this earth 1 day is of 2hr, then the output is:
Day = 1
T = 90
Day = 2
T = 900
I don't want to change the first and second dimensions.
input = :, :, 87600
output = :, :, 365 or 366

Shubham Gupta on 14 Nov 2019
Edited: Shubham Gupta on 14 Nov 2019
According to the comments, you have provided, this will produce the desired output :
yd = max(max(mn2t)); % first find maximum for each hour
y = reshape(yd,24,1,size(yd,3)/24); % reshape it by day
output = max(y); % find maximum of each day
Let me know if this works !

Behzad Navidi on 14 Nov 2019
Dear Shubham,
It's works perfect for the array. thank you so much. I gonna accept your answer but webpage says: An Error Occurred (Unable to complete the action because of changes made to the page. Reload the page to see its updated state)
I have faced this error previously, and I know a few hours later it will be fixed and I can click on accept answer.
but now just in a case do you know how to use this code when mn2t is cell (34*1 cell) (instead of an array) and using this code for all 34 rows of this cell (the cell consists of 34 of this 3d array?
Shubham Gupta on 14 Nov 2019
Simply use for loop mn2t:
output = cell(size(mn2t));
for i = 1:length(mn2t)
mn2t_mat = mn2t{i}; % extract matrix of ith element
yd = max(max(mn2t_mat)); % first find maximum for each hour
y = reshape(yd,24,1,size(yd,3)/24); % reshape it by day
output{i} = max(y); % find maximum of each day
end
Output will be cell vector with each cell containing maximum of each cell of mn2t.