Finding Maximum Consecutive Dry Days in Daily Rainfall Data

Hi All,

I am trying to find maximum consecutive dry days in a long time series of daily rainfall. I am using a script to do for one year of data. Can somebody help to modify the code such that it counts the Max CDD for each year separately. Whenever the data is less than 0.1, it is considered as dry day.

max_dry = 0; % initialize the maximum number of consecuitive dry days
counter = 0;
for k=1:366
    if a(k) == 0
        counter = counter + 1;
    else
        if counter > max_dry
            max_dry = counter;
            max_dry_position = k - counter;
        end
        counter = 0;
    end
end

[Abderrahim. B]

your code in incomplete! what is a ?

[Faisal Baig]

Sorry... a is my mxn matrix for daily rainfall. You can call it RR. in my case its 6210X1 matrix

Answer 1

[Bjorn Gustavsson]

Don't do it with loops, that is a lot of work. Use the vectorized functions to your advantage. Here's how to get at it with using diff and max:

drf = 3*randn(366,1); % mocking up some daily rainfall-data.
DryDayThresh = 0.1;
rdc = find(drf > DryDayThresh); % Find the rainy days
plot(diff(rdc)) % Plot the difference between rainy days, 1 means two consecutive rainy days.
[maxCDD] = max(diff(rdc))-1;

HTH

[Faisal Baig]

yes, thanks alot for this. But how can I get the CDD for each year separately considering i have 17 years of dailiy rain data?

[Bjorn Gustavsson]

You simply put this into a loop and repeat the process for the daily rainfall-data for each separate year. To solve your problem for you will not help you develop your programming and problem solving skills, but here goes anyway:

drf = 3*randn(366,91); % mocking up some daily rainfall-data, for 91 years because why not
DryDayThresh = 0.1;
for iYear = size(drf,2):-1:1 % loop down to avoid reallocation of arrays I couldn't be bothered to preallocate
  % In here we simply repeat the above process for each separate year.
  rdc = find(drf(:,iYear) > DryDayThresh); % Find the rainy days
  plot(diff(rdc)) % Plot the difference between rainy days, 1 means two consecutive rainy days.
  [maxCDD(iYear)] = max(diff(rdc))-1;
end

[Faisal Baig]

Thanks alot.. that helped.

[Bjorn Gustavsson]

You're welcome, happy that it helped.

Answer 2

[Saksham Gupta]

As per my understanding, you need assistance in writing code for your problem.

This code should be helpful as per the conditions shared by you.

max_dry = 0; % initialize the maximum number of consecuitive dry days
counter = 0;
% a is supposed to be having data of each day as vector, I am using length, as year can
% have either 365 or 366 days.
for k=1:length(a) 
    if a(k) < 0.1 % conditon as mentioned by you
        counter = counter + 1;
    else
        counter = 0;
    end
    max_dry=max(counter,max_dry);
end

[Faisal Baig]

yes that gives the total number of dry days out of my data. But I need the Max Consecutive Dry Days (CDD) for each individual year.

Answer 3

[Walter Roberson]

This code shows one way of finding all the runs of maximum length.

rng(12345)
drf = rand(366,1); % mocking up some daily rainfall-data.
DryDayThresh = 0.2;

drf = reshape(drf, 1, []);  %need it to be a row vector

rdc = drf < DryDayThresh; % locate the dry days
starts = strfind([0 rdc], [0 1]);
stops = strfind([rdc 0], [1 0]);
durations = stops - starts + 1;
longest = max(durations)

longest = 3

long_idx = durations == longest;
start_of_longest = starts(long_idx)

start_of_longest = 1×2

234 345

end_of_longest = stops(long_idx)

end_of_longest = 1×2

236 347

plot(drf);
yline(DryDayThresh)
hold on
%stairs(rdc, 'k')
xline(start_of_longest, 'r')
xline(end_of_longest, 'r')

[Walter Roberson]

rng(12345)
drf = rand(366,1); % mocking up some daily rainfall-data.
DryDayThresh = 0.2;

drf = reshape(drf, 1, []);  %need it to be a row vector
rdc = drf < DryDayThresh; % locate the dry days

props = regionprops(rdc, 'BoundingBox', 'Area');
durations = [props.Area];
longest = max(durations);
long_idx = durations == longest;
selected_props = props(long_idx);

bounds_of_longest = round(vertcat(selected_props.BoundingBox) + [.5 .5 0 0]);
start_of_longest = bounds_of_longest(:,1)

start_of_longest = 2×1

234 345

end_of_longest = start_of_longest + bounds_of_longest(:,3) - 1

end_of_longest = 2×1

236 347

plot(drf);
yline(DryDayThresh)
hold on

xline(start_of_longest, 'r')
xline(end_of_longest, 'r')