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

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

[Shubham Gupta]

Above mentioned precip= :,:, 8760 is not a correct way to show the dimesions of a 3D array. If precip is a 3D array of ( r x c x p ) dimension, we need to know r, c & p where 'r' is rows 'c' is columns and 'p' is pages. Also, dimesion of the final output.

Now, my understanding is precip is of dimension (24 x 1 x 365 )? If so,

result = max(precip)

might be the necessary output of dimension (1 x 1 x 365) where each page contains the maximum of each input page. Let us know more details about precip. Cheers

[Shubham Gupta]

Though for you attached data which is of dimension (1 x 1 x 72), if this a data for first 3 days. You want to convert it to a 3d array of dimension (24 x 1 x 3) where each page represents a day. You can do by

y = reshape(mn2t_first_24hours,24,1,length(mn2t_first_24hours)/24);

Then to find max of each day

output = max(y)

will give you output of dimesion (1x1x3) where each page will represent maximum value of each day. If that's what you want?

[BN]

Thank you it's work really nice for my attached file which are for first 3 days. But when I try to use this code for original 3d array:

I got this error:

filename='ERA_max_min_daily_all_hours_1983.nc'
time=ncread(filename, 'time');
time_real=hours(time)+datetime(1900, 1, 1);
lon = ncread(filename,'longitude');
lat = ncread(filename,'latitude');     
mn2t=ncread(filename,'mn2t');

y = reshape(mn2t,24,1,length(mn2t)/24); %%%%% I got ERROR here
output = max(y)

thank you again...

[Shubham Gupta]

As I have mentioned in the comments, this code will only work for dimension of (1x1x8760) but seems like your "mn2t" array is of (49x41x8760) dimension which has 49 rows, 41 columns & 8760 pages. Can you please explain, what does these dimesions represent for your data? and along which dimension you want the maximum value?

By the way, this maybe the thing you want but I may be completely wrong.

yd = max(max(mn2t)); % converts it into 1x1x8760 array where each page of 'yd' is the maximum of each page in 'mn2t'
y = reshape(yd,24,1,length(yd)/24); % reshape it to 24x1x365;
output = max(y);

If above is not what you want, let us know what is the output that you are expecting using a simple example?

About the error (why it ocurred):

you have mn2t with (49*41*8760 = 17598840 elements) and you are reshaping it to (24*1*365 = 8760 elements) which is not the same number of elements as previous one, and this is not permissible in MATLAB

[BN]

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.

I uploaded a little sample of my data. here is the link if you can download 68.7 MB:

https://drive.google.com/open?id=1wAgjftEHVuzlZmMgrlpGKO-66gk63WOY

Thank you again

[Shubham Gupta]

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 = ??

[BN]

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

Answer 1

[Shubham Gupta]

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 !

[BN]

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?

Thank you - I will accept your answer.

[Shubham Gupta]

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.

Answer 2

[BN]

Thank you ?