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Calculate daily standard deviation from timetable
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Hello, I want to calculate the mean max and mean min standard deviation of temperature data from a daily timetable.
t=timetable(dt, Temperature);
temperaturePerDayMin = retime(t2, 'daily', 'min');
tmpPerDayMin_std= std(temperaturePerDayMin.Temperature);
temperaturePerDayMax = retime(t2, 'daily', 'max');
tmpPerDayMax_std= std(temperaturePerDayMax.Temperature);
Right now tmpPerDayMin_std and tmpPerDayMax_std return only one value. How can I calculate the standard deviation for each day?
Accepted Answer
Star Strider
on 21 Dec 2022
Try something like this —
DT = datetime('01-Dec-2022 00:00:00') + hours(0:720).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
figure
plot(TT1.DT, TT1.Temperature)
grid
DailyMax = retime(TT1, 'daily', 'max');
DailyMin = retime(TT1, 'daily', 'min');
DailySTD = retime(TT1, 'daily', @std);
TT2 = table(DailyMax.DT, DailyMax.Temperature, DailyMin.Temperature, DailySTD.Temperature, 'VariableNames',{'DT','Max','Min','StDv'})
The standard deviation of a single value is zero by definition, so it is only possible to calculate the standard deviation of all temperatures for a single day for that day.
It is possible to calculate the standard deviation for the maximum and minimum temperatures for several days (at least more than 1) and of course for the entire month.
.
28 Comments
Ancalagon8
on 21 Dec 2022
Works perfect! Thanks! In the initial table (TT1) for December, is it possible to calculate the number of days that the temperature was higher than a specific threshold?
Star Strider
on 21 Dec 2022
Thank you!
Yes!
Example —
DT = datetime('01-Dec-2022 00:00:00') + hours(0:720).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
TT1 = 721×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 00:00:00 -0.91782
01-Dec-2022 01:00:00 3.4756
01-Dec-2022 02:00:00 5.1349
01-Dec-2022 03:00:00 5.9696
01-Dec-2022 04:00:00 10.697
01-Dec-2022 05:00:00 13.404
01-Dec-2022 06:00:00 14.426
01-Dec-2022 07:00:00 14.551
01-Dec-2022 08:00:00 16.658
01-Dec-2022 09:00:00 16.201
01-Dec-2022 10:00:00 14.351
01-Dec-2022 11:00:00 11.749
01-Dec-2022 12:00:00 10.131
01-Dec-2022 13:00:00 8.4033
01-Dec-2022 14:00:00 0.49336
01-Dec-2022 15:00:00 0.1879
figure
plot(TT1.DT, TT1.Temperature)
grid
DailyMax = retime(TT1, 'daily', 'max');
DailyMin = retime(TT1, 'daily', 'min');
DailySTD = retime(TT1, 'daily', @std);
T2 = table(DailyMax.DT, DailyMax.Temperature, DailyMin.Temperature, DailySTD.Temperature, 'VariableNames',{'DT','Max','Min','StDv'})
T2 = 31×4 table
DT Max Min StDv
___________ ______ _______ ______
01-Dec-2022 16.658 -8.8893 8.2913
02-Dec-2022 16.515 -6.5274 8.2742
03-Dec-2022 18.799 -7.4772 8.7142
04-Dec-2022 17.145 -9.1443 8.4744
05-Dec-2022 18.821 -8.4418 8.5831
06-Dec-2022 18.391 -5.9905 8.9501
07-Dec-2022 15.976 -8.7264 8.6104
08-Dec-2022 15.344 -7.7227 8.062
09-Dec-2022 18.631 -9.265 8.6461
10-Dec-2022 19.929 -7.4301 9.0601
11-Dec-2022 17.767 -9.7136 8.8605
12-Dec-2022 17.554 -7.4757 8.7277
13-Dec-2022 18.164 -7.6267 9.016
14-Dec-2022 20.333 -8.9258 9.2568
15-Dec-2022 16.008 -9.7254 8.7095
16-Dec-2022 20.091 -7.3932 9.2243
DaysMeetingThreshold = nnz(T2.Max >= 18) % Days Equal To Or Higher Than 18°C
DaysMeetingThreshold = 17
Adjust the conditions to select either greater than (>) or greater than or equal to (>=) and the desired temperature threshold (here 18°C). I did not include that in the timetable because it is for the entire time period and so would have to have a column length equal to the entire timetable. This could also be done for calendar months or calendar weeks or for the entire table. That would require extra code and probably an additional timetable as well.
An example of doing that using retime wouild be:
Threshold_1 = retime(TT1, 'weekly', @(x)nnz(x>=18))
Threshold_1 = 5×1 timetable
DT Temperature
___________ ___________
27-Nov-2022 1
04-Dec-2022 6
11-Dec-2022 8
18-Dec-2022 4
25-Dec-2022 7
TT2 = table2timetable(T2);
Threshold2 = retime(TT2, 'weekly', @(x)nnz(x>=18))
Threshold2 = 5×3 timetable
DT Max Min StDv
___________ ___ ___ ____
27-Nov-2022 1 0 0
04-Dec-2022 4 0 0
11-Dec-2022 4 0 0
18-Dec-2022 3 0 0
25-Dec-2022 5 0 0
The ‘Threshold1_1’ timetable counts the number of times each week any ‘Temperature’ equalled or exceeded 18°C.
The ‘Threshold_2’ timetable counts the number of times the maximum ‘Temperature’ equalled or exceeded 18°C.
It could be added to the existing timetable, however that would require additional code and that code would not be trivial, since the lengths of the constant values would have to match the entries of the corresponding dates in position and length.
.
Ancalagon8
on 23 Dec 2022
Thank you for the detailed answer! Is it possible to put
DaysMeetingThreshold = nnz(T2.Max >= 18) % Days Equal To Or Higher Than 18°C
inside a loop and store the result in a variable or cell or table? And what about different steps in thresholds? (e.g. 20,25,30, 35,36,38,40,42,44).
Star Strider
on 23 Dec 2022
As always, my pleasure!
I am not certain that I understand what you want to do.
You would likely need a loop for the various thresholds. After that, putting them into a table is straightforward —
DT = datetime('01-Dec-2022 00:00:00') + hours(0:720).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
TT1 = 721×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 00:00:00 -2.4124
01-Dec-2022 01:00:00 1.8933
01-Dec-2022 02:00:00 2.5341
01-Dec-2022 03:00:00 6.1807
01-Dec-2022 04:00:00 9.8868
01-Dec-2022 05:00:00 14.528
01-Dec-2022 06:00:00 14.26
01-Dec-2022 07:00:00 15.974
01-Dec-2022 08:00:00 14.304
01-Dec-2022 09:00:00 17.543
01-Dec-2022 10:00:00 12.134
01-Dec-2022 11:00:00 14.879
01-Dec-2022 12:00:00 11.453
01-Dec-2022 13:00:00 6.3793
01-Dec-2022 14:00:00 5.644
01-Dec-2022 15:00:00 2.349
figure
plot(TT1.DT, TT1.Temperature)
grid
DailyMax = retime(TT1, 'daily', 'max');
DailyMin = retime(TT1, 'daily', 'min');
DailySTD = retime(TT1, 'daily', @std);
T2 = table(DailyMax.DT, DailyMax.Temperature, DailyMin.Temperature, DailySTD.Temperature, 'VariableNames',{'DT','Max','Min','StDv'})
T2 = 31×4 table
DT Max Min StDv
___________ ______ _______ ______
01-Dec-2022 17.543 -9.0929 8.8102
02-Dec-2022 19.193 -6.685 8.5592
03-Dec-2022 19.767 -11.037 9.8884
04-Dec-2022 18.123 -8.7118 8.8951
05-Dec-2022 17.171 -7.8953 8.8536
06-Dec-2022 18.443 -8.2063 8.3054
07-Dec-2022 18.364 -8.341 8.5502
08-Dec-2022 18.076 -6.3634 8.7735
09-Dec-2022 18.339 -8.4784 8.7009
10-Dec-2022 17.352 -8.1611 8.4984
11-Dec-2022 18.944 -7.9201 8.536
12-Dec-2022 17.78 -7.6084 8.6725
13-Dec-2022 18.489 -8.9344 9.1313
14-Dec-2022 19.052 -7.7588 8.8545
15-Dec-2022 17.135 -7.8149 8.897
16-Dec-2022 19.373 -8.0265 9.1146
Thresholdv = [16; 17; 18; 19; 20]; % Column Vector
for k = 1:numel(Thresholdv)
DaysMeetingThreshold(k,:) = nnz(T2.Max >= Thresholdv(k)); % Days Equal To Or Higher Than Thresholdv (°C)
end
ThresholdTable = table(Thresholdv, DaysMeetingThreshold)
ThresholdTable = 5×2 table
Thresholdv DaysMeetingThreshold
__________ ____________________
16 30
17 30
18 17
19 5
20 0
.
Ancalagon8
on 23 Dec 2022
Edited: Ancalagon8
on 30 Dec 2022
That was exactly what I needed. Once more thank you!
In:
DailyMax = retime(TT1, 'daily', 'max');
can I also determine the hour that the maximum value was observed?
Star Strider
on 23 Dec 2022
Edited: Star Strider
on 30 Dec 2022
As always, my pleasure!
EDIT — (30 Dec 2022 at 16:56)
Not easily.
DT = datetime('01-Dec-2022 00:00:00') + hours(0:720).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
TT1 = 721×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 00:00:00 -1.1057
01-Dec-2022 01:00:00 1.3984
01-Dec-2022 02:00:00 6.8774
01-Dec-2022 03:00:00 6.4973
01-Dec-2022 04:00:00 12.631
01-Dec-2022 05:00:00 14.64
01-Dec-2022 06:00:00 16.174
01-Dec-2022 07:00:00 17.047
01-Dec-2022 08:00:00 17.984
01-Dec-2022 09:00:00 13.908
01-Dec-2022 10:00:00 12.703
01-Dec-2022 11:00:00 12.548
01-Dec-2022 12:00:00 11.678
01-Dec-2022 13:00:00 8.3992
01-Dec-2022 14:00:00 4.2721
01-Dec-2022 15:00:00 1.7873
figure
plot(TT1.DT, TT1.Temperature)
grid
DailyMax = retime(TT1, 'daily', 'max');
HourlyMax = retime(TT1, 'hourly', 'max')
HourlyMax = 721×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 00:00:00 -1.1057
01-Dec-2022 01:00:00 1.3984
01-Dec-2022 02:00:00 6.8774
01-Dec-2022 03:00:00 6.4973
01-Dec-2022 04:00:00 12.631
01-Dec-2022 05:00:00 14.64
01-Dec-2022 06:00:00 16.174
01-Dec-2022 07:00:00 17.047
01-Dec-2022 08:00:00 17.984
01-Dec-2022 09:00:00 13.908
01-Dec-2022 10:00:00 12.703
01-Dec-2022 11:00:00 12.548
01-Dec-2022 12:00:00 11.678
01-Dec-2022 13:00:00 8.3992
01-Dec-2022 14:00:00 4.2721
01-Dec-2022 15:00:00 1.7873
DailyMin = retime(TT1, 'daily', 'min');
DailySTD = retime(TT1, 'daily', @std);
T2 = table(DailyMax.DT, DailyMax.Temperature, DailyMin.Temperature, DailySTD.Temperature, 'VariableNames',{'DT','Max','Min','StDv'})
T2 = 31×4 table
DT Max Min StDv
___________ ______ _______ ______
01-Dec-2022 17.984 -9.2499 9.0444
02-Dec-2022 18.206 -8.5237 9.1837
03-Dec-2022 15.431 -7.5063 8.2329
04-Dec-2022 19.696 -9.143 9.0854
05-Dec-2022 19.471 -7.37 8.7003
06-Dec-2022 19.677 -10.246 9.3604
07-Dec-2022 16.639 -6.4818 8.3036
08-Dec-2022 18.539 -6.9418 8.6188
09-Dec-2022 17.33 -9.3591 8.8737
10-Dec-2022 17.894 -7.7901 8.513
11-Dec-2022 18.26 -8.2763 8.9268
12-Dec-2022 18.52 -9.1745 9.0402
13-Dec-2022 18.13 -8.7439 8.5561
14-Dec-2022 19.723 -8.7907 8.8667
15-Dec-2022 19.534 -9.033 9.2443
16-Dec-2022 19.735 -7.8816 8.3044
Thresholdv = [16; 17; 18; 19; 20]; % Column Vector
for k = 1:numel(Thresholdv)
DaysMeetingThreshold(k,:) = nnz(T2.Max >= Thresholdv(k)); % Days Equal To Or Higher Than Thresholdv (°C)
end
ThresholdTable = table(Thresholdv, DaysMeetingThreshold)
ThresholdTable = 5×2 table
Thresholdv DaysMeetingThreshold
__________ ____________________
16 29
17 26
18 17
19 9
20 0
[yh,mh,dh] = ymd(HourlyMax.DT); % Return Hour At Which Maximum Temperature Occurred
for k = 1:numel(DailyMax)
[yd,md,dd] = ymd(DailyMax.DT(k));
Lv = ismember([yh,mh,dh], [yd,md,dd], 'rows');
idx = HourlyMax.Temperature(Lv) == DailyMax.Temperature(k);
HM = HourlyMax(Lv,:);
DayHourMax(k,:) = timetable2table(HM(idx,:));
end
DayHourMax % Table Of Hour In Each Day At Which Maximum Temperature Occurred
DayHourMax = 31×2 table
DT Temperature
____________________ ___________
01-Dec-2022 08:00:00 17.984
02-Dec-2022 07:00:00 18.206
03-Dec-2022 07:00:00 15.431
04-Dec-2022 08:00:00 19.696
05-Dec-2022 06:00:00 19.471
06-Dec-2022 08:00:00 19.677
07-Dec-2022 10:00:00 16.639
08-Dec-2022 09:00:00 18.539
09-Dec-2022 08:00:00 17.33
10-Dec-2022 08:00:00 17.894
11-Dec-2022 08:00:00 18.26
12-Dec-2022 09:00:00 18.52
13-Dec-2022 08:00:00 18.13
14-Dec-2022 08:00:00 19.723
15-Dec-2022 09:00:00 19.534
16-Dec-2022 07:00:00 19.735
This should be easier!
.
Ancalagon8
on 30 Dec 2022
Thank you for your answer, I receive the following error:
"To assign to or create a variable in a table, the number of rows must match the height of the table."
Star Strider
on 30 Dec 2022
My pleasure!
My code worked when I ran it, or I would not have posted it.
I have no idea what the problem could be.
Ancalagon8
on 28 Jan 2023
Having a further investigation in my data, I want to calculate the mean min and mean max Temperatures per month, as well as their mean standard deviations. So I guess that:
MonthlyMax = retime(TT1, 'monthly', 'max');
MonthlyMin = retime(TT1, 'monthly', 'min');
and then I have to calculate the MeanMonthlyMax and MeanMonthlyMin?
Star Strider
on 28 Jan 2023
Probably something like this —
MonthlyMax = retime(TT1, 'monthly', @(x)mean(max(x)));
MonthlyMin = retime(TT1, 'monthly', @(x)mean(min(x)));
although I am not certain what you want.
Testing that to see what it returns —
DT = datetime('01-Dec-2022 00:00:00') + caldays(1:364).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
TT1 = 364×1 timetable
DT Temperature
___________ ___________
02-Dec-2022 1.8812
03-Dec-2022 2.0567
04-Dec-2022 6.6711
05-Dec-2022 12.135
06-Dec-2022 10.489
07-Dec-2022 13.718
08-Dec-2022 13.921
09-Dec-2022 17.49
10-Dec-2022 16.473
11-Dec-2022 16.773
12-Dec-2022 16.449
13-Dec-2022 10.577
14-Dec-2022 9.2101
15-Dec-2022 9.5114
16-Dec-2022 5.1743
17-Dec-2022 0.10695
figure
plot(TT1.DT, TT1.Temperature)
grid
hold on
MonthlyMax = retime(TT1, 'monthly', @(x)mean(max(x)))
MonthlyMax = 12×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 17.49
01-Jan-2023 18.8
01-Feb-2023 20.47
01-Mar-2023 17.53
01-Apr-2023 18.049
01-May-2023 18.325
01-Jun-2023 19.46
01-Jul-2023 18.845
01-Aug-2023 19.449
01-Sep-2023 19.491
01-Oct-2023 20.38
01-Nov-2023 17.817
MonthlyMin = retime(TT1, 'monthly', @(x)mean(min(x)))
MonthlyMin = 12×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 -8.1543
01-Jan-2023 -8.6214
01-Feb-2023 -8.5488
01-Mar-2023 -8.1802
01-Apr-2023 -8.1751
01-May-2023 -9.2394
01-Jun-2023 -9.6803
01-Jul-2023 -7.3772
01-Aug-2023 -6.802
01-Sep-2023 -8.406
01-Oct-2023 -7.7957
01-Nov-2023 -7.0099
MthV = DT(1) + calmonths(0:11).';
hold on
plot(MthV, [MonthlyMax.Temperature, MonthlyMin.Temperature], '-r')
hold off
.
Ancalagon8
on 28 Jan 2023
Thank you! The mean min and mean max Temperatures per month are correct using:
MonthlyMax = retime(TT1, 'monthly', @(x)mean(max(x)));
MonthlyMin = retime(TT1, 'monthly', @(x)mean(min(x)));
How about their mean standard deviations?
Star Strider
on 28 Jan 2023
As always, my pleasure!
MonthlyMaxSD = retime(TT1, 'monthly', @(x)std(max(x)));
MonthlyMinSD = retime(TT1, 'monthly', @(x)std(min(x)));
That should work without further modification.
Star Strider
on 28 Jan 2023
I forgot about that. It’s taking the standard deviation of a scalar, and that’s zero by definition. The maximum and minimum of a standard deviation are going to be the same (since it’s also a scalar), and the standard deviation of the maxima or minima for the entire year is going to be scalars representing the entire year, plotting a straight line over all the months. so a different sort of metric may be necessary.
There may be other ways, for example the standard error of the mean for each month, presented as confidence intervals —
DT = datetime('01-Dec-2022 00:00:00') + caldays(1:364).';
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5;
TT1 = timetable(DT, Temperature)
TT1 = 364×1 timetable
DT Temperature
___________ ___________
02-Dec-2022 0.82051
03-Dec-2022 1.9947
04-Dec-2022 4.6657
05-Dec-2022 7.8428
06-Dec-2022 11.934
07-Dec-2022 14.881
08-Dec-2022 15.027
09-Dec-2022 15.659
10-Dec-2022 15.119
11-Dec-2022 16.932
12-Dec-2022 17.403
13-Dec-2022 13.246
14-Dec-2022 11.374
15-Dec-2022 7.0888
16-Dec-2022 4.5771
17-Dec-2022 -0.48706
figure
plot(TT1.DT, TT1.Temperature)
grid
hold on
MonthlyMax = retime(TT1, 'monthly', @(x)mean(x)+std(x)/sqrt(numel(x))*1.96)
MonthlyMax = 12×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 8.1303
01-Jan-2023 10.545
01-Feb-2023 8.4384
01-Mar-2023 5.0567
01-Apr-2023 8.5555
01-May-2023 10.352
01-Jun-2023 7.3369
01-Jul-2023 5.7303
01-Aug-2023 9.5159
01-Sep-2023 9.2639
01-Oct-2023 6.069
01-Nov-2023 7.6833
MonthlyMin = retime(TT1, 'monthly', @(x)mean(x)-std(x)/sqrt(numel(x))*1.96)
MonthlyMin = 12×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 2.3238
01-Jan-2023 4.3452
01-Feb-2023 2.2066
01-Mar-2023 -0.90149
01-Apr-2023 2.2013
01-May-2023 3.7726
01-Jun-2023 1.3024
01-Jul-2023 -0.49234
01-Aug-2023 3.4644
01-Sep-2023 3.3579
01-Oct-2023 0.20044
01-Nov-2023 1.7348
MthV = DT(1) + days(15) + calmonths(0:11).';
hold on
plot(MthV, [MonthlyMax.Temperature, MonthlyMin.Temperature], '-r')
hold off
Other options could be calculating the monthly standard deviation of the maxima and minima for each day during the month. That would initially require one aggregation to calculate the daily maxima and minima, and a second aggretation to calculate the standard deviation of the monthly values of the daily data. That may be the only way to get a standard deviation aggregated over a month.
The aggregation depends on what you want, and what you want to do with the aggregated data after calculating it.
.
Ancalagon8
on 28 Jan 2023
I understand. The standard error of the mean for each month, presented as confidence intervals is not what I need here. I am sure that the calculation of the monthly standard deviation of the maxima and minima for each day during the month would be enough. So first I calculate the daily maxima and minima like this:
DailyMin = retime(TT1, 'daily', 'min');
DailyMax = retime(TT1, 'daily', 'max');
How can i calculate the second aggretation (the standard deviation of the monthly values of the daily data)?
Thank you in advance.
Star Strider
on 28 Jan 2023
As always, my pleasure!
I would create ‘DailyMin’ and ‘DailyMax’ as additional temporary variables, probably in a separate timetable, and then aggregate over them as 'monthly' to create the monthly standard deviations. That may be the only way to do it.
Ancalagon8
on 28 Jan 2023
You mean:
MonthlyMin = retime(DailyMin, 'monthly', 'min');
MonthlyMax = retime(DailyMax, 'monthly', 'max');
Star Strider
on 28 Jan 2023
I was thinking of something like this —
DT = datetime('01-Dec-2022 00:00:00') + hours(1:364*24).'; % Hours
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5; % Temperatures
TT1 = timetable(DT, Temperature)
TT1 = 8736×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 01:00:00 -1.4502
01-Dec-2022 02:00:00 -0.14428
01-Dec-2022 03:00:00 5.977
01-Dec-2022 04:00:00 8.8143
01-Dec-2022 05:00:00 9.1807
01-Dec-2022 06:00:00 12.38
01-Dec-2022 07:00:00 15.92
01-Dec-2022 08:00:00 16.297
01-Dec-2022 09:00:00 16.229
01-Dec-2022 10:00:00 16.797
01-Dec-2022 11:00:00 15.728
01-Dec-2022 12:00:00 12.875
01-Dec-2022 13:00:00 9.361
01-Dec-2022 14:00:00 9.4571
01-Dec-2022 15:00:00 5.3876
01-Dec-2022 16:00:00 0.11136
DailyMax = retime(TT1, 'daily', 'max') % Daily Aggregation
DailyMax = 365×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 16.797
02-Dec-2022 17.7
03-Dec-2022 17.013
04-Dec-2022 16.641
05-Dec-2022 17.55
06-Dec-2022 18.994
07-Dec-2022 20.201
08-Dec-2022 17.164
09-Dec-2022 17.171
10-Dec-2022 18.355
11-Dec-2022 16.357
12-Dec-2022 15.898
13-Dec-2022 17.103
14-Dec-2022 17.685
15-Dec-2022 20.491
16-Dec-2022 20.725
DailyMin = retime(TT1, 'daily', 'min') % Daily Aggregation
DailyMin = 365×1 timetable
DT Temperature
___________ ___________
01-Dec-2022 -7.4328
02-Dec-2022 -5.7989
03-Dec-2022 -8.8891
04-Dec-2022 -8.6656
05-Dec-2022 -8.6631
06-Dec-2022 -8.6668
07-Dec-2022 -8.3697
08-Dec-2022 -9.3005
09-Dec-2022 -7.7954
10-Dec-2022 -8.7234
11-Dec-2022 -10.109
12-Dec-2022 -9.815
13-Dec-2022 -7.3838
14-Dec-2022 -8.13
15-Dec-2022 -7.6904
16-Dec-2022 -7.6138
DailyTT = DailyMax;
DailyTT = addvars(DailyTT, DailyMin.Temperature)
DailyTT = 365×2 timetable
DT Temperature Var2
___________ ___________ _______
01-Dec-2022 16.797 -7.4328
02-Dec-2022 17.7 -5.7989
03-Dec-2022 17.013 -8.8891
04-Dec-2022 16.641 -8.6656
05-Dec-2022 17.55 -8.6631
06-Dec-2022 18.994 -8.6668
07-Dec-2022 20.201 -8.3697
08-Dec-2022 17.164 -9.3005
09-Dec-2022 17.171 -7.7954
10-Dec-2022 18.355 -8.7234
11-Dec-2022 16.357 -10.109
12-Dec-2022 15.898 -9.815
13-Dec-2022 17.103 -7.3838
14-Dec-2022 17.685 -8.13
15-Dec-2022 20.491 -7.6904
16-Dec-2022 20.725 -7.6138
MonthlyTT = retime(DailyTT, 'monthly', @(x)std(x)) % Monthly Aggregation Of Daily Data
MonthlyTT = 12×2 timetable
DT Temperature Var2
___________ ___________ _______
01-Dec-2022 1.1873 1.0935
01-Jan-2023 1.0858 0.82949
01-Feb-2023 1.3895 0.80469
01-Mar-2023 1.2204 1.1097
01-Apr-2023 0.79565 1.0219
01-May-2023 0.94119 1.1451
01-Jun-2023 1.0995 1.0555
01-Jul-2023 0.9984 0.96406
01-Aug-2023 0.91378 1.1695
01-Sep-2023 1.1669 0.99959
01-Oct-2023 1.0167 0.82053
01-Nov-2023 3.8584 1.4976
MthV = DT(1) + calmonths(0:11).';
figure
hold on
plot(MthV, MonthlyTT{:,1}, '-r')
plot(MthV, MonthlyTT{:,2}, '-b')
hold off
legend('Max Monthly SD','Min Monthly SD', 'Location','best')
This plots the standard deviations, not the actual temperature values.
.
Ancalagon8
on 28 Jan 2023
Thank you for your answer. I am wondering if the part
DailyTT = DailyMax;
DailyTT = addvars(DailyTT, DailyMin.Temperature)
instead of DailyMin.Temperature need to have DailyMax.Temperature or I am wrong?
Star Strider
on 28 Jan 2023
The ‘DailyTT’ timetable is defined initially as having the daily max temperatures. The daily min temperatures were added to it as a new variable, so now ‘DailyTT’ has both as different variables. It’s just not possible to change their variable names easily to reflect that.
Ancalagon8
on 28 Jan 2023
Edited: Ancalagon8
on 28 Jan 2023
Thank you @Star Strider, it was my misunderstanding, everything is clear now. Having calculated the above, I also calculated the daily range like this:
DailyRange=DailyMax.Temperature-DailyMin.Temperature;
and added it to DailyTT like this:
DailyTT = addvars(DailyTT, DailyMin.Temperature,DailyRange);
So now I have a 365 timetable, where my final goal is to calculate the mean,min and max range for each month.Any ideas how to proceed?
Star Strider
on 28 Jan 2023
I’m not certain that I understand about ‘mean,min and max range for each month’. The mean for the month should be only one value, unless you intend the range of daily temperatures (max - min for each day) and the statistics on those values.
The coding depends on how you want to calculate the daily values and then aggregate them.
Ancalagon8
on 29 Jan 2023
Edited: Ancalagon8
on 29 Jan 2023
Indeed, the mean for each month is one value, so a 12X1. The min range for each month is the lower and upper limit of min values for each month, and the same for the max range. Any idea?
Star Strider
on 29 Jan 2023
Edited: Star Strider
on 29 Jan 2023
Thank you!
I just wanted to clarify my understanding of this.
EDIT — I’m kind of lost at this point. Isn’t that the result you’ree getting? That’s the result I got in the code in my earlier Comment. That dealt with the standard deviations, however calculating different parameters (min, max, and mean) should yield essentially the same sort of result. The only difference should be the last (function) argument in various aggregations:
MonthlyTT = retime(DailyTT, 'monthly', @(x)std(x)) % Monthly Aggregation Of Daily Data
to produce different results.
I’ll follow up on this tomorrow if there’s anything else to be coded.
.
Ancalagon8
on 30 Jan 2023
Edited: Ancalagon8
on 30 Jan 2023
What I mean with range is this:
DailyRange=DailyMax.Temperature-DailyMin.Temperature; %Daily range for the whole year
DailyRange_January = DailyRange(1:31); %Daily range for January
DailyRange_February = DailyRange(32:59); %Daily range for February
.
.
.
DaileRange_December = DailyRange(335:end); %Daily range for December
and then define the mean, min max for each month
MeanDailyRange_January=mean(DailyRange_January);
MinDailyRange_January=min(DailyRange_January);
MaxDailyRange_January=max(DailyRange_January);
And the same process for all the months. So I wonder if there is a more convenient way to improve the above code and how can I add the time information of the min and max.
Star Strider
on 30 Jan 2023
Try something like this —
DT = datetime('01-Dec-2022 00:00:00') + hours(1:364*24).'; % Hours
Temperature = sin(2*pi*(0:numel(DT)-1)/24-0.5).'*12+5 + randn(size(DT))*1.5; % Temperatures
TT1 = timetable(DT, Temperature)
TT1 = 8736×1 timetable
DT Temperature
____________________ ___________
01-Dec-2022 01:00:00 -0.37909
01-Dec-2022 02:00:00 0.28714
01-Dec-2022 03:00:00 6.4977
01-Dec-2022 04:00:00 7.7289
01-Dec-2022 05:00:00 12.292
01-Dec-2022 06:00:00 15.373
01-Dec-2022 07:00:00 14.676
01-Dec-2022 08:00:00 15.309
01-Dec-2022 09:00:00 15.836
01-Dec-2022 10:00:00 16.44
01-Dec-2022 11:00:00 16.672
01-Dec-2022 12:00:00 14.35
01-Dec-2022 13:00:00 8.6757
01-Dec-2022 14:00:00 9.1514
01-Dec-2022 15:00:00 6.0855
01-Dec-2022 16:00:00 2.4084
DailyMax = retime(TT1, 'daily', 'max'); % Daily Aggregation
DailyMin = retime(TT1, 'daily', 'min'); % Daily Aggregation
DailyMean = retime(TT1, 'daily', 'mean'); % Daily Aggregation
DailyRange = DailyMax.Temperature - DailyMin.Temperature; % Daily Aggregation
DailyTT = timetable(DailyMax.DT, DailyMax.Temperature, DailyMin.Temperature, DailyRange, DailyMean.Temperature, 'VariableNames',{'Max_Temperature','Min_Temperature','Range_Temperature','Mean_Temperature'})
DailyTT = 365×4 timetable
Time Max_Temperature Min_Temperature Range_Temperature Mean_Temperature
___________ _______________ _______________ _________________ ________________
01-Dec-2022 16.672 -9.4342 26.106 5.1988
02-Dec-2022 18.117 -8.7949 26.911 5.3819
03-Dec-2022 18.611 -7.9121 26.523 5.1666
04-Dec-2022 18.68 -8.1488 26.829 4.8373
05-Dec-2022 17.69 -7.1424 24.832 5.3975
06-Dec-2022 19.069 -8.1585 27.228 5.2481
07-Dec-2022 15.966 -9.9869 25.952 4.8057
08-Dec-2022 18.182 -9.3876 27.57 5.4564
09-Dec-2022 17.444 -8.3602 25.804 4.8884
10-Dec-2022 19.302 -7.5362 26.838 5.391
11-Dec-2022 16.696 -10.497 27.193 4.528
12-Dec-2022 18.3 -6.9914 25.291 5.1051
13-Dec-2022 18.151 -8.5995 26.75 5.4419
14-Dec-2022 17.836 -10.976 28.811 5.0321
15-Dec-2022 17.282 -9.1337 26.416 4.8376
16-Dec-2022 18.325 -8.0825 26.408 4.8119
MonthlyTTMax = retime(DailyTT, 'monthly', 'max')
MonthlyTTMax = 12×4 timetable
Time Max_Temperature Min_Temperature Range_Temperature Mean_Temperature
___________ _______________ _______________ _________________ ________________
01-Dec-2022 20.326 -6.2942 28.852 5.4564
01-Jan-2023 20.882 -6.5416 30.436 5.486
01-Feb-2023 21.771 -6.7069 29.895 5.42
01-Mar-2023 20.752 -6.3323 30.718 5.7682
01-Apr-2023 19.842 -6.1608 28.69 5.81
01-May-2023 19.771 -6.3754 29.726 5.3571
01-Jun-2023 20.162 -6.2147 29.415 5.839
01-Jul-2023 21.085 -6.1956 30.339 5.8512
01-Aug-2023 20.793 -5.192 29.184 5.6606
01-Sep-2023 20.346 -6.2424 29.367 5.5407
01-Oct-2023 20.827 -5.951 29.933 5.4544
01-Nov-2023 20.722 -5.9304 30.332 5.5216
MonthlyTTMin = retime(DailyTT, 'monthly', 'min')
MonthlyTTMin = 12×4 timetable
Time Max_Temperature Min_Temperature Range_Temperature Mean_Temperature
___________ _______________ _______________ _________________ ________________
01-Dec-2022 15.966 -10.976 22.838 4.4957
01-Jan-2023 16.409 -10.684 24.377 4.5691
01-Feb-2023 15.547 -10.91 24.217 4.3471
01-Mar-2023 15.748 -10.449 23.368 4.5982
01-Apr-2023 15.941 -10.591 22.533 4.2688
01-May-2023 16.111 -10.972 23.687 4.3221
01-Jun-2023 16.484 -11.179 23.418 4.3588
01-Jul-2023 16.605 -10.795 24.173 4.5544
01-Aug-2023 15.518 -10.379 22.288 4.3416
01-Sep-2023 15.359 -10.131 22.941 4.5646
01-Oct-2023 16.226 -10.036 22.756 4.4768
01-Nov-2023 -5.9304 -10.465 0 -5.9304
MonthlyTTMean = retime(DailyTT, 'monthly', 'mean')
MonthlyTTMean = 12×4 timetable
Time Max_Temperature Min_Temperature Range_Temperature Mean_Temperature
___________ _______________ _______________ _________________ ________________
01-Dec-2022 17.894 -8.3725 26.267 5.0415
01-Jan-2023 18.351 -8.3832 26.735 4.9523
01-Feb-2023 18.018 -8.3188 26.337 4.923
01-Mar-2023 18.023 -8.1983 26.222 5.0542
01-Apr-2023 17.919 -8.011 25.93 5.0405
01-May-2023 18.123 -8.3906 26.513 4.902
01-Jun-2023 17.953 -8.0177 25.971 5.0389
01-Jul-2023 18.193 -8.1407 26.334 5.0607
01-Aug-2023 18.041 -8.2389 26.28 4.9959
01-Sep-2023 18.051 -7.866 25.917 5.0096
01-Oct-2023 17.892 -7.9565 25.848 4.9462
01-Nov-2023 17.38 -8.1048 25.485 4.6034
MonthlyTTSD = retime(DailyTT, 'monthly', @(x)std(x)) % Monthly Aggregation Of Daily Data
MonthlyTTSD = 12×4 timetable
Time Max_Temperature Min_Temperature Range_Temperature Mean_Temperature
___________ _______________ _______________ _________________ ________________
01-Dec-2022 0.89012 1.0981 1.2396 0.28508
01-Jan-2023 1.1506 1.0405 1.3665 0.27683
01-Feb-2023 1.1575 1.0735 1.6071 0.29298
01-Mar-2023 1.2434 0.99746 1.8445 0.28534
01-Apr-2023 1 1.1311 1.6295 0.34489
01-May-2023 0.99102 1.1128 1.4308 0.28412
01-Jun-2023 0.8422 1.0872 1.5087 0.3929
01-Jul-2023 0.87463 1.033 1.4316 0.31694
01-Aug-2023 1.0816 1.1013 1.7648 0.32041
01-Sep-2023 1.1947 1.0865 1.6608 0.27119
01-Oct-2023 1.043 0.97543 1.5385 0.24107
01-Nov-2023 4.5398 1.0787 5.0632 2.007
MthV = DT(1) + calmonths(0:11).';
figure
hold on
plot(MthV, MonthlyTTSD{:,1}, '-r')
plot(MthV, MonthlyTTSD{:,2}, '-b')
hold off
legend('Max Monthly SD','Min Monthly SD', 'Location','best')
The ‘MonthlyTT’ set of timetable arrays should have everything you need. If necessary, it should be straightforward to abstract the various variables from them and import them to a new summary timetable, similar to what I did to create ‘DailyTT’.
Creating numbered variables such as:
MeanDailyRange_January=mean(DailyRange_January);
is generally discouraged. You can get the various monthly values from the timetable arrays that calculated them.
I am doing my best to provide what you want, however that is not always clear to me. The result is that we need to iterate until we converge on the correct arrays.
.
Ancalagon8
on 30 Jan 2023
Thank you @Star Strider. So far things are fine, but my concern is how can I add the time information of the min and max.
Star Strider
on 30 Jan 2023
As always, my pleasure!
I thought we already covered that problem at some point. By now I’ve lost track of where it is (although searching for it, I believe it was here).
It would be essentially the same approach.
.
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