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Categorizing groups of data
조회 수: 4 (최근 30일)
이전 댓글 표시
I separated rainfall data into 250 events. Each rainfall event has different number of points. I want to classify all the events with 4 points together, 5 points together, 6 points together and so on.
댓글 수: 8
Queena Edwards
2022년 3월 5일
The rainfall events are separated by NaN. The first line is the number of the event. The following data is the rainfall amount per hour.
Image Analyst
2022년 3월 5일
편집: Image Analyst
2022년 3월 5일
Thanks for attaching it. So are we to assume that you tried what Star suggested below and it still doesn't work? A similar option is bwlabel(). Attach your findgroups() script where you read in the data and call findgroups() so we can see what you did wrong. Then did you want to do a histogram of the length or rain events? What do you want to do after you classify/label/number/identify each run of rain days?
Queena Edwards
2022년 3월 5일
편집: Queena Edwards
2022년 3월 5일
Error using matlab.internal.math.grp2idx (line 37)
Unable to use a character array as a grouping variable. First convert the character array to a cell array of character
vectors using cellstr. Then specify the cell array as a grouping variable.
Error in matlab.internal.math.mgrp2idx (line 66)
[ogroup,gnum] = matlab.internal.math.grp2idx(group{1,1},inclnan,inclempty);
Error in findgroups (line 82)
gnums = matlab.internal.math.mgrp2idx(groupVars,0,inclnan,inclempty);
Error in extract_events_onlyP (line 41)
T = findgroups('Pevents.txt')
I'm getting the error above.
After I'll like to do some calculations for each group of the rainfall events (That's all the events with the same duration/points). Split it into intervals. Find the total amount of rainfall in each interval. Ranking these intervals. I haven't gotten to this part as yet because I'm figuring it out as I go.
Image Analyst
2022년 3월 5일
편집: Image Analyst
2022년 3월 5일
findgroups() takes numerical data, not a character array (string). Read in your data, like with readmatrix() or importdata() or xlsread(), and then pass it to findgroups().
What does rank groups mean? You have groups of varying lengths, and varying number of groups. For example you might have 4 groups that are 5 days long, and 2 groups that are 10 days long, and 28 groups that are only a single day. So what does "rank" mean in that case?
Walter Roberson
2022년 3월 5일
T = findgroups('Pevents.txt')
That code does not ask MATLAB to read the content of the file Pevents.txt and find groups inside it.
That code asks MATLAB to use the literal character vector as the name of an option in calling findgroups() .
If you wanted to ask findgroups to find the (single) group of the character vector that happened to be ['P' 'e' 'v' 'e' 'n' 't' 's' '.' 't' 'x' 't'] then you would need to ask for
T = findgroups({'Pevents.txt'})
but of course the result would just be T = 1
Queena Edwards
2022년 3월 5일
Image Analyst the read matrix worked but the findgroups is not in the format I want.
Going back to the file I sent, the first event is 9h. The second event is 1h. The third event is 5h.
The seventh event is also 5h long. I would like to group the third, seventh and any other 5h events out of the 250.
Afterwards, using this 5h group. The rainfall would be summed for each hour and then divided into 9 intervals
This is repeated for the 1h group, 2h group …to the 24h group.
채택된 답변
Star Strider
2022년 3월 5일
편집: Star Strider
2022년 3월 5일
EDIT — (5 Mar 2022 at 17:36)
Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip')
Uz = 1×1 cell array
{'Rainfallevents/Pevents.txt'}
T1 = readtable(Uz{1})
T1 = 3807×1 table
Var1
____
NaN
1
0.51
0
0
0
0
1.27
0
0
1.02
NaN
2
3.3
NaN
3
[G,ID] = findgroups(T1{:,1})
G = 3807×1
NaN
5
3
1
1
1
1
7
1
1
ID = 318×1
0
0.2500
0.5100
0.7600
1.0000
1.0200
1.2700
1.5200
1.7800
2.0000
Gu = unique(G); % Unique Groups
This file at least has something in it.
What am I supposed to do with this information?
What is the desired result?
.
댓글 수: 17
Walter Roberson
2022년 3월 5일
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip';
Uz = unzip(filename)
Uz = 1×1 cell array
{'Rainfallevents/Pevents.txt'}
C = fileread(Uz{1})
C =
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3.0499999999999998e+00
1.7800000000000000e+00
1.7800000000000000e+00
1.7800000000000000e+00
1.0200000000000000e+00
5.1000000000000001e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.3500000000000000e+02
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.3600000000000000e+02
1.5200000000000000e+00
1.5200000000000000e+00
5.1000000000000001e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.3700000000000000e+02
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.3800000000000000e+02
5.1000000000000001e-01
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.3900000000000000e+02
3.8100000000000001e+00
1.2700000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4000000000000000e+02
5.1000000000000001e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4100000000000000e+02
3.8100000000000001e+00
7.6000000000000001e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4200000000000000e+02
5.1000000000000001e-01
7.6000000000000001e-01
5.1000000000000001e-01
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
NaN
1.4300000000000000e+02
2.5000000000000000e-01
5.1000000000000001e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4400000000000000e+02
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4500000000000000e+02
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4600000000000000e+02
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
1.4700000000000000e+02
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
0.0000000000000000e+00
2.5000000000000000e-01
I = imread(Uz{1},'jpeg');
Error using imread (line 434)
'/users/mss.system.oOci5w/Rainfallevents/Pevents.txt' is not a JPEG file.
'/users/mss.system.oOci5w/Rainfallevents/Pevents.txt' is not a JPEG file.
whos
Queena Edwards
2022년 3월 5일
편집: Queena Edwards
2022년 3월 5일
Star Strider Going back to the file I sent, the first event is 9h. The second event is 1h. The third event is 5h.
The seventh event is also 5h long. I would like to group the third, seventh and any other 5h events out of the 250 sets of events.
Afterwards, using this 5h group. The rainfall would be summed for each hour and then divided into 9 intervals
This is repeated for the 1h group, 2h group …to the 24h group.
That’s the first thing I’ll like to do.
Star Strider
2022년 3월 5일
‘... the first event is 9h. The second event is 1h. The third event is 5h.’
That information does not exist in the file. (I have no idea what this information is.) The first file had nothing in it except a short character vector, so it is essentially useless.
Queena Edwards
2022년 3월 5일
편집: Queena Edwards
2022년 3월 5일
The rainfall events are separated by NaN. The first line is the number of the event. The following data is the rainfall amount per hour.
So if there are 10 lines of data between the NaN, the first line is the number of the event and the other lines are the rainfall amount in an hour.
Example:
this is the first event. The first line 1.0 names the event. The rest of lines are the amount of rainfall per hour. Hence it is a 9 hour event
NaN
1.0000000000000000e+00
5.1000000000000001e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
1.2700000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
1.0200000000000000e+00
NaN
Star Strider
2022년 3월 5일
편집: Star Strider
2022년 3월 5일
Try this —
Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip');
M1 = readmatrix(Uz{1})
M1 = 3807×1
NaN
1.0000
0.5100
0
0
0
0
1.2700
0
0
NaNv = find(isnan(M1)) % Numeric Indices Of 'NaN' Values
NaNv = 250×1
1
12
15
22
32
38
43
50
54
60
Check1 = [numel(M1) NaNv(end)] % There Are Data AFter The Last 'NaN'
Check1 = 1×2
3807 3785
Events = 1:numel(NaNv)-1;
for k = Events
idxrng = NaNv(k)+1:NaNv(k+1)-1;
RF_Vector{k} = M1(idxrng); % Vector Of Rainfall Events (Between 'NaN' Elements)
RF_Hours = numel(idxrng); % Number Of Hours In Each Event
RF_Total(k) = sum(M1(idxrng)); % Sum Of Rainfrall Data
RF_Mean(k) = mean(M1(idxrng)); % Mean Rainfall Per Event
end
figure
yyaxis left
plot(Events, RF_Total)
ylabel('Total Rainfall/Event')
yyaxis right
plot(Events, RF_Mean)
ylabel('Mean Rainfall/Event')
grid
xlabel('Event Nr')
This plots the individual totals, not the cumulative total, so I cannot explain the trend.
EDIT — (5 Mar 2022 at 18:38)
Changed code to provide what I believe to be the desired result.
.
Queena Edwards
2022년 3월 5일
편집: Queena Edwards
2022년 3월 5일
Image Analyst Nope it’s not correct.
Simplest way to explain it is, excluding the first line of data after NaN which is the name of the data set. If there are 5 lines of numbers, every single group where there are 5 lines of numbers out of the 250 data set I’ll like to find the sum of each line.
This has to be done for if there is 1line all the way up to 24 lines.
The lines represent the rainfall per hour. Ideally I’ll have:
1h events = Total amount of rainfall for the events with a duration of 1h.
2h events = [x y] where x is the total amount of rainfall for the 1st hour and y is the total amount of rainfall for the 2nd hour for events with a duration of 2h
Image Analyst
2022년 3월 5일
Here are 3 groups of numbers
NaN
4.0000000000000000e+00
2.5000000000000000e-01
5.1000000000000001e-01
2.5000000000000000e-01
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
0.0000000000000000e+00
2.5000000000000000e-01
NaN
5.0000000000000000e+00
7.6000000000000001e-01
5.1000000000000001e-01
0.0000000000000000e+00
5.1000000000000001e-01
NaN
6.0000000000000000e+00
1.0200000000000000e+00
5.1000000000000001e-01
2.5000000000000000e-01
The first groups has 9 lines after the nan, the second group has 5 (as expected) and the third group has 4. So how can you say that there are exactly 5 lines (no more and no less) in each group?
Queena Edwards
2022년 3월 5일
Star Strider This seems to be the closest so far. The first adjustment
idxrng =
2 3 4 5 6 7 8 9 10 11
idxrng =
13 14
You see these two samples of data for idxrng, the first number i.e 2 and 13 should number be included because that's the events name. Next, I like what you did where you made it into a vector. so for these two above, it'll be a (9x1) and (1x1).
For all the (9x1) vectors i'll want to sum each row. Similarly for all (1x1), (2x1),...(24x1) vectors i'll want to sum the rows of the same vector.
Star Strider
2022년 3월 5일
I forgot to subscript the ‘RF_Hours’ vector. All the values are summed in the ‘RF_Total’ value for each event.
Subscripting it, and summing the rainfall for durations of equal length —
Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip');
M1 = readmatrix(Uz{1})
M1 = 3807×1
NaN
1.0000
0.5100
0
0
0
0
1.2700
0
0
NaNv = find(isnan(M1)); % Numeric Indices Of 'NaN' Values
% Check1 = [numel(M1) NaNv(end)] % There Are Data AFter The Last 'NaN'
Events = 1:numel(NaNv)-1;
for k = Events
idxrng = NaNv(k)+1:NaNv(k+1)-1;
RF_Vector{k,:} = M1(idxrng); % Vector Of Rainfall Events (Between 'NaN' Elements)
RF_Hours(k,:) = numel(idxrng); % Number Of Hours In Each Event
RF_Total(k,:) = sum(M1(idxrng)); % Sum Of Rainfrall Data
RF_Mean(k,:) = mean(M1(idxrng)); % Mean Rainfall Per Event
end
% RF_Vector
% [RF_Vector{:}]
[UHrs,ia,ib] = unique(RF_Hours);
EventSums = accumarray(ib, RF_Total, [], @sum);
T1 = table(UHrs,EventSums, 'VariableNames',{'Event Duration (Hours)','Total Rainfall'})
T1 = 39×2 table
Event Duration (Hours) Total Rainfall
______________________ ______________
2 2020.9
3 3496.1
4 2203.3
5 2077.1
6 2306.2
7 1399.5
8 1067.9
9 2187.2
10 499.8
11 1119.6
12 1870.2
13 1521.7
14 1206.6
15 628.93
16 600.8
17 668.74
figure
stairs(UHrs, EventSums)
grid
xlabel('Event Duration (Hours)')
ylabel('Total Rainfall as Function of Event Duration')
axis([min(UHrs) max(UHrs) 0 3600])
set(gca, 'XScale','log')
figure
yyaxis left
plot(Events, RF_Total)
ylabel('Total Rainfall/Event')
yyaxis right
plot(Events, RF_Mean)
ylabel('Mean Rainfall/Event')
grid
xlabel('Event Nr')
I am still not absolutely certain what the desired result is.
Is figure(1) close?
.
Queena Edwards
2022년 3월 5일
I'm going to check thru what you just sent but one thing to note is that my last event from the Pevent.txt is not included because it does not end with NaN. How do i fix that?
Also, there are still some steps I have to do to obtain the desired graph. This is just the first step, grouping the events. My desired graph since you asked is the SCS Type II curve.
Star Strider
2022년 3월 5일
‘... my last event from the Pevent.txt is not included because it does not end with NaN. How do i fix that?’
That is actually quite straightforward. The last element of ‘NaNv’ becomes the length of the vector.
Including that change, the code and results are now —
Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip');
M1 = readmatrix(Uz{1})
M1 = 3807×1
NaN
1.0000
0.5100
0
0
0
0
1.2700
0
0
NaNv = find(isnan(M1)); % Numeric Indices Of 'NaN' Values
NaNv = [NaNv; size(M1,1)]; % Last Index (Instead Of Last 'NaN') Is The End Of The Vector
Events = 1:numel(NaNv)-1;
for k = Events
idxrng = NaNv(k)+1:NaNv(k+1)-1;
RF_Vector{k,:} = M1(idxrng); % Vector Of Rainfall Events (Between 'NaN' Elements)
RF_Hours(k,:) = numel(idxrng); % Number Of Hours In Each Event
RF_Total(k,:) = sum(M1(idxrng)); % Sum Of Rainfrall Data
RF_Mean(k,:) = mean(M1(idxrng)); % Mean Rainfall Per Event
end
[UHrs,ia,ib] = unique(RF_Hours);
EventSums = accumarray(ib, RF_Total, [], @sum);
T1 = table(UHrs,EventSums, 'VariableNames',{'Event Duration (Hours)','Total Rainfall'})
T1 = 39×2 table
Event Duration (Hours) Total Rainfall
______________________ ______________
2 2020.9
3 3496.1
4 2203.3
5 2077.1
6 2306.2
7 1399.5
8 1067.9
9 2187.2
10 499.8
11 1119.6
12 1870.2
13 1521.7
14 1206.6
15 628.93
16 600.8
17 668.74
figure
stairs(UHrs, EventSums)
grid
xlabel('Event Duration (Hours)')
ylabel('Total Rainfall as Function of Event Duration')
axis([min(UHrs) max(UHrs) 0 3600])
set(gca, 'XScale','log')
figure
yyaxis left
plot(Events, RF_Total)
ylabel('Total Rainfall/Event')
yyaxis right
plot(Events, RF_Mean)
ylabel('Mean Rainfall/Event')
grid
xlabel('Event Nr')
‘My desired graph since you asked is the SCS Type II curve.’
Please provide a reference for that. I do not recognise it, at least by that name.
.
Queena Edwards
2022년 3월 5일
After the line RF_Vector {k:1}, this was my intended process: Example: Looking at a (3x1) vector, for all the (3x1) vectors out of the 250 data I’ll like to sum all the 3 rows so I’ll have a total resulting (3x1) vector. Similarly for the rest of vectors. Suppose A=[1;4;5] B= [7;2;2] C= [6;5;3] total = [14;11;10]
Star Strider
2022년 3월 5일
I believe that is what my code does, except that it sums the columns, not the rows. That would be a challenge to plot, so I will not, however the coee is relatively straightforward —
Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/915459/Rainfallevents.zip');
M1 = readmatrix(Uz{1})
M1 = 3807×1
NaN
1.0000
0.5100
0
0
0
0
1.2700
0
0
NaNv = find(isnan(M1)); % Numeric Indices Of 'NaN' Values
NaNv = [NaNv; size(M1,1)]; % Last Index (Instead Of Last 'NaN') Is The End Of The Vector
Events = 1:numel(NaNv)-1;
for k = Events
idxrng = NaNv(k)+1:NaNv(k+1)-1;
RF_Vector{k,:} = M1(idxrng); % Vector Of Rainfall Events (Between 'NaN' Elements)
RF_Hours(k,:) = numel(idxrng); % Number Of Hours In Each Event
RF_Total(k,:) = sum(M1(idxrng)); % Sum Of Rainfrall Data
RF_Mean(k,:) = mean(M1(idxrng)); % Mean Rainfall Per Event
end
[UHrs,ia,ib] = unique(RF_Hours); % Unique Values & Indices
for k = 1:numel(UHrs)
EventSums{k,:} = sum([RF_Vector{k==ib}],2);
end
T1 = table(UHrs,EventSums, 'VariableNames',{'Event Duration (Hours)','Total Rainfall'})
T1 = 39×2 table
Event Duration (Hours) Total Rainfall
______________________ ______________
2 { 2×1 double}
3 { 3×1 double}
4 { 4×1 double}
5 { 5×1 double}
6 { 6×1 double}
7 { 7×1 double}
8 { 8×1 double}
9 { 9×1 double}
10 {10×1 double}
11 {11×1 double}
12 {12×1 double}
13 {13×1 double}
14 {14×1 double}
15 {15×1 double}
16 {16×1 double}
17 {17×1 double}
figure
for k = 1:4
subplot(2,2,k)
stairs(cell2mat(T1{k*8,2}), 'LineWidth',2)
title(sprintf('Event Duration = %2d',k*8))
end
figure
yyaxis left
plot(Events, RF_Total)
ylabel('Total Rainfall/Event')
yyaxis right
plot(Events, RF_Mean)
ylabel('Mean Rainfall/Event')
grid
xlabel('Event Nr')
This is the best that I can do with these data.
.
Star Strider
2022년 3월 5일
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
추가 답변 (1개)
Image Analyst
2022년 3월 5일
Try this:
data = readmatrix('pevents.txt')
% Convert nans to zeros
data(isnan(data)) = 0;
% Set rain days to 1;
rainDays = logical(data)
% Measure lengths of all rain events.
props = regionprops(rainDays, 'Area')
% Get the lengths of all the rain events into a single vectore.
allLengths = [props.Area];
% Show a histogram of the lengths
histogram(allLengths);
title('Distribution of Rain Storms in Days');
xlabel('Length of Rain in Days');
ylabel('Number of Events of this Length')
grid on;
fprintf('Mean length of rainstorm = %.2f days.\n', mean(allLengths));
fprintf('Median length of rainstorm = %.2f days.\n', median(allLengths));
fprintf('Longest length of rainstorm = %.2f days.\n', max(allLengths));
Mean length of rainstorm = 2.80 days.
Median length of rainstorm = 2.00 days.
Longest length of rainstorm = 209.00 days.
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