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Selecting a rain event from data series

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Me
Me 2023-8-29
编辑： Walter Roberson 2023-9-9
Hi everyone. I hope you can help me. I have a time series of data: rain data (h in mm) recorded every 10 min in a day (column vector of 144 data). From the data I must select an event preceded and followed by 6 hours (36 data) without rain (h=0). How do I select the data range?
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Dyuman Joshi
Dyuman Joshi 2023-8-29
编辑：Dyuman Joshi 2023-8-29
A scalar data value recorded every 10 min for a day would give 6*24 == 144 data points per day. (6 data point an hour * 24 hours in a day)
How many data points do you collect each hour? If 1, then how did you get 1440 elements? or did you collect data for 10 days?
"From the data I must select an event preceded and followed by 6 hours (36 data) without rain (h=0). How do I select the data range?"
Using conditional operations and logical indexing - Find Array Elements That Meet a Condition
Me
Me 2023-8-29
Sorry, the zero is a typo. In one hour I have 6 logged data, so in one day 144 data. How do I select the data?

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Alexander
Alexander 2023-8-31
编辑：Walter Roberson 2023-9-9
I think this should work now:
clear;
load rain2.txt; rain = rain2; % Only to reuse the code above
NoOfDry = 1;
rHmm = rain(:,2);rHmm = rHmm(:)'; % To force a row vector
tH = rain(:,1)/6; % Only to have a timeline in hours, doesn't needed here
figure(1);plot(rHmm);grid;
dryTime = zeros(1,36); % 6h dry, 36*10 minutes
yDry = strfind(rHmm, dryTime); % Now we are looking for the indices dry >= 6h
if(isempty(yDry))
fprintf('In your data set are NO period(s) with ge. 6h no rain\n');
NoOfDry = 0;
return;
end
yDryDiff = diff(yDry); % 1 means > 6h dry
iNoOffDry = find(yDryDiff > 1); % Here are the positions of dry >= 6h
NoOfDry = NoOfDry+length(iNoOffDry); % This is the number of occurences dry >= 6h
fprintf('In your data set are %i period(s) with ge. 6h no rain\n',NoOfDry);
commandwindow
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Alexander
Alexander 2023-8-31
I think it's a good idea to accept this answer if it fulfills your needs. ;-)

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更多回答（2 个）

Alexander
Alexander 2023-8-29
编辑：Alexander 2023-8-29
You can use strfind:
y = strfind(data, [0 0 0 0 0 0])
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Me
Me 2023-8-30
Thanks for your quick reply! "strfind" is not good for my problem.
I am attaching a txt file to better explain the problem. In the txt file you will find the indices in the first column, the rainfall height in the second. The data that I have to select are between the index 37 and 105, these data are preceded by 6 hours without rain (at least 36 data in which h=0) and are followed by 6 hours without rain (at least 36 data in which h = 0). How can I do?

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Alexander
Alexander 2023-8-30
编辑：Dyuman Joshi 2023-8-30
Ran in:
I have to contradict, strfind is perfect if you want to avoid loops. Have a look at my rapid programmed script:
load rain.txt
tH = rain(:,1)/6; tH = tH(:)';% I prefere hours
rHmm = rain(:,2);rHmm = rHmm(:)'; % only for convenience
tHobs = tH(36:108); % I modified the observation time: 06:00 - 18:00
rHmmObs = rHmm(36:108);
figure(1);plot(tH,rHmm);grid;
dryTime = zeros(1,36); % 6h dry
figure(2);plot(tHobs,rHmmObs);grid;
yDry = strfind(rHmmObs, dryTime)
yDry = []
% Result: yDry = [], which means that no 6h w/o rain between 6-18:00
% Let's reduce the dry time to one hour:
dryTime = zeros(1,6);
yDry = strfind(rHmmObs, dryTime);
disp(yDry)
15 16 17 18 19 20 21 22 23 24 25 26 51 60 61 62 63 64
% There are a lot of consecutive numbers in, which means that there are dry
% periods which are longer than one hour. Let's test these:
yDryDiff = diff(yDry);
disp(yDryDiff)
1 1 1 1 1 1 1 1 1 1 1 25 9 1 1 1 1
% This means, you have one period of (6+11)*10 minutes w/o rain
% you have 2 periods with exact one hour w/o rain
% you have one period of (6+4)*10 minutes w/o rain
% To check item 1:
dryTime = zeros(1,17);
yDry = strfind(rHmmObs, dryTime);
disp(yDry)
15
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Alexander
Alexander 2023-8-30
Ran in:
I think I don't understand your problem. Are you looking for exact 6h dry (=0)? Or >= 6h dry? Just some code:
clear;
load rain2.txt; rain = rain2; % Only to reuse the code above
Error using load
Unable to find file or directory 'rain2.txt'.
rHmm = rain(:,2);rHmm = rHmm(:)'; % To force a row vector
tH = rain(:,1)/6; % Only to have a timeline in hours, doesn't needed here
figure(1);plot(rHmm);grid;
dryTime = zeros(1,36); % 6h dry, 36*10 minutes
yDry = strfind(rHmm, dryTime); % Now we are looking for the indices dry >= 6h
yDryDiff = diff(yDry); % 1 means > 6h dry
iNoOffDry = find(yDryDiff > 1); % Here are the positions of dry >= 6h
NoOffDry = length(iNoOffDry); % This is the number of occurences dry >= 6h
fprintf('In your data set are %i period(s) with ge. 6h no rain\n',NoOffDry)
commandwindow
If you want to get the exact time or index you have just work off the indices backward.
Just a very small piece of advice: If you really want to use the large advantages of Matlab, you should get a little bit more familiar with it. ;-)
Alexander
Alexander 2023-8-30
There is an exception I've forgotten: the first matching pattern. But I'm currently not at my PC. Hence, I will update my code tomorrow. Sorry.

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