duration of rain. help

Good afternoon, I have a file. It is not possible to find the average monthly duration of rain and its intensity.

Maybe someone can help. Thank you in advance

[Dyuman Joshi]

It's not clear (atleast to me) what you have asked/want to do.

Please clearly specify what the objective is, mention the code that you have written yet and state the issue you are facing.

[Nikita]

I cannt't undenstand how calculation duration of rain in matlab code

Answer 1

[Mathieu NOE]

hello

this is one way to do it - just plain simple matlab functions .

Of course , someone fluent with timetables can do it also , probably in few lines

this is my "old fashioned" way....

I don't see how rain intensity appears in your data - can you explain what are columns 9 to 16 ?

for the time being I simply computed the total amount of minutes of rain for each month and each year

% store data (year / month / total rain duration (minutes)

code done so far

T = readmatrix('rain.xlsx',"NumHeaderLines",2);
% cols :    1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18		
%           id	Year	Mn	Da											Rain		hr.min_start	hr.min_finish		
years = T(:,2);
month = T(:,3);
hr_min_start = T(:,17);
hr_min_finish = T(:,18);

[year_u,ia,ic] = unique(years);

out = [];

% loop over the years
for k = 1:numel(year_u)
    % extract rows
    rows_y = find(abs(years - year_u(k))<1);  % more robust than == 0
    
    % loop over the months
    for mk = 1:12
        
        month_list = month(rows_y);
        % extract rows
        rows_m = find(abs(month_list - mk)<1);  % more robust than == 0
        
        if ~isempty(rows_m) % we have data
            hr_min_start_m = hr_min_start(rows_m); % extract start time 
            start_minutes = myconvert(hr_min_start_m); % convert to minutes 
            
            hr_min_finish_m = hr_min_finish(rows_m); % extract finish time 
             finish_minutes = myconvert(hr_min_finish_m); % convert to minutes
             
             duration_minutes = finish_minutes - start_minutes;
             
             % some duration appears negative simply because 0:00 is same
             % as 24:00 , so we have to complement with 24*60 = 1440
             % minutes
             ind = (duration_minutes<0);
             duration_minutes(ind) = duration_minutes(ind) + 1440;
             total_duration_minutes = sum(duration_minutes);
             flag_store = 1;
        else 
            total_duration_minutes = 0;
            flag_store = 0;
       
        end
        
        % store data (year / month  / duration(minustes)
        if flag_store~= 0
            out = [out; [year_u(k) mk total_duration_minutes]];
        end
            
    end
        
        
    
end


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function out_minutes = myconvert(time)

    % time format is like 22.40 (hours = 22 , minutes = 40)
    % convert this in minutes only
    hh = floor(time); % hours 
    mm = time - hh; % minutes

    out_minutes = mm+60*hh;



end
        

[Nikita]

Many thanks!!!

"I don't see how rain intensity appears in your data - can you explain what are columns 9 to 16 ?" - it's don't matter. 16 column - it's code (rain or snow).

rain intensity - sorry, it's not need. I'm wrong

How I understand - this m-code calculaction sum duration for each month?

[Mathieu NOE]

yes it's just a sum of minutes of rain each month

[Mathieu NOE]

for my own fun I added also a fourth column in out, which is the daily average duration of rain in minutes (again for each month & year)

clc
clearvars

T = readmatrix('rain.xlsx',"NumHeaderLines",2);
% cols :    1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18		
%           id	Year	Mn	Da											Rain		hr.min_start	hr.min_finish		
years = T(:,2);
month = T(:,3);
hr_min_start = T(:,17);
hr_min_finish = T(:,18);

[year_u,ia,ic] = unique(years);

out = [];
% loop over the years
for k = 1:numel(year_u)
    % extract rows
    rows_y = find(abs(years - year_u(k))<1);  % more robust than == 0
    
    % how many days per month this year ? 
    mm = (1:12); % monthes
    days_per_month = datenum(year_u(k), mm(:)+1, 1) - datenum(year_u(k), mm(:), 1); 
    
    % loop over the months
    for mk = 1:12
        
        month_list = month(rows_y);
        % extract rows
        rows_m = find(abs(month_list - mk)<1);  % more robust than == 0
        
        if ~isempty(rows_m) % we have data
            hr_min_start_m = hr_min_start(rows_m); % extract start time 
            start_minutes = myconvert(hr_min_start_m); % convert to minutes 
            
            hr_min_finish_m = hr_min_finish(rows_m); % extract finish time 
             finish_minutes = myconvert(hr_min_finish_m); % convert to minutes
             
             duration_minutes = finish_minutes - start_minutes;
             
             % some duration appears negative simply because 0:00 is same
             % as 24:00 , so we have to complement with 24*60 = 1440
             % minutes
             ind = (duration_minutes<0);
             duration_minutes(ind) = duration_minutes(ind) + 1440;
             total_duration_minutes = sum(duration_minutes);
             flag_store = 1;
        else 
            total_duration_minutes = 0;
            flag_store = 0;
        end
        
        % average daily duration in minutes
        average_daily_duration_minutes = total_duration_minutes/days_per_month(mk);
        
        % store data (year / month  / total_duration_minutes / average_daily_duration_minutes
        if flag_store~= 0
            out = [out; [year_u(k) mk total_duration_minutes average_daily_duration_minutes]];
        end
    end
    
end


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function out_minutes = myconvert(time)

    % time format is like 22.40 (hours = 22 , minutes = 40)
    % convert this in minutes only
    hh = floor(time); % hours 
    mm = time - hh; % minutes
    out_minutes = mm+60*hh;

end
        

[Nikita]

Many thanks

[Mathieu NOE]

do you mind accepting my answer ?

Tx !!

[Mathieu NOE]

do you mind accepting my answer ?

Tx !!

[Nikita]

Yes, of course. Many thanks

[Mathieu NOE]

Thank you !

Answer 2

[Image Analyst]

You might look into splitapply, grpstats, groupsummary, and findgroups. They are very useful functions to know about! 🙂

[Image Analyst]

If an Answer solves your original question, then could you please click the "Accept this answer" link/button to award the answerer with "reputation points" for their efforts in helping you? They'd appreciate it. Thanks in advance. 🙂 Note: you can only accept one answer (so pick the best one - Matthew's) but you can click the "Vote" icon for as many Answers as you want. Voting for an answer will also award reputation points.