Finding Maximum Consecutive Dry Days in Daily Rainfall Data
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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
采纳的回答
Bjorn Gustavsson
2022-7-5
编辑：Bjorn Gustavsson
2022-7-5
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 个评论
Faisal Baig
2022-7-5
Bjorn Gustavsson
2022-7-5
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
2022-7-5
Bjorn Gustavsson
2022-7-5
You're welcome, happy that it helped.
更多回答（2 个）
Saksham Gupta
2022-7-5
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
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)
long_idx = durations == longest;
start_of_longest = starts(long_idx)
end_of_longest = stops(long_idx)
plot(drf);
yline(DryDayThresh)
hold on
%stairs(rdc, 'k')
xline(start_of_longest, 'r')
xline(end_of_longest, 'r')
1 个评论
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)
end_of_longest = start_of_longest + bounds_of_longest(:,3) - 1
plot(drf);
yline(DryDayThresh)
hold on
xline(start_of_longest, 'r')
xline(end_of_longest, 'r')
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