# Finding Maximum Consecutive Dry Days in Daily Rainfall Data

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Faisal Baig el 5 de Jul. de 2022
Comentada: Bjorn Gustavsson el 5 de Jul. de 2022
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
##### 2 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Faisal Baig el 5 de Jul. de 2022
Sorry... a is my mxn matrix for daily rainfall. You can call it RR. in my case its 6210X1 matrix

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Bjorn Gustavsson el 5 de Jul. de 2022
Editada: Bjorn Gustavsson el 5 de Jul. de 2022
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
##### 4 comentariosMostrar 3 comentarios más antiguosOcultar 3 comentarios más antiguos
Bjorn Gustavsson el 5 de Jul. de 2022
You're welcome, happy that it helped.

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### Más respuestas (2)

Saksham Gupta el 5 de Jul. de 2022
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
##### 1 comentarioMostrar NingunoOcultar Ninguno
Faisal Baig el 5 de Jul. de 2022
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.

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Walter Roberson el 5 de Jul. de 2022
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') ##### 1 comentarioMostrar NingunoOcultar Ninguno
Walter Roberson el 5 de Jul. de 2022
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') Iniciar sesión para comentar.

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