Date and times logic

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Adnan Jayyousi
Adnan Jayyousi on 24 Jun 2022
Answered: Steven Lord on 25 Jun 2022
Hello,
I am trying to make array that contains 8760 Hours, starting from 1/1/2021 ending in 31/12/2021,
in the second column in the same array I want to fill in the corresponding electricty tariff at that certain time.
the tariff values may vary according to the month/days/hours/holidays along the year.
My intention is to make a logic using while loop (8760 loops) using index "i".
I am having trouble to find easy logic that can be applied on the "DateNtime" array in order to fill the tariff.
I will give an example for one scenario in words :
If Season is 1 (one of months Jan/Feb/Dec) & the hour is between 8:00 and 13:00 & the day is normal week day (Sunday till thursday), the tariff wil be 0.7170 (put into column 2 -- > DateNtime(x,2) = 0.7170
any suggestions ?
%% Define date&time arrays :
t1 = datetime(2021,1,1,0,0,0);
t2 = datetime(2021,12,31,23,0,0);
DateNtime = (t1:hours(1):t2)'
%% Create Holidays arrays :
ht1 = datetime(2021,3,27,0,0,0);
ht2 = datetime(2021,3,28,23,0,0);
HT1 = (ht1:hours(1):ht2)'
%%
ht3 = datetime(2021,4,2,0,0,0);
ht4 = datetime(2021,4,3,23,0,0);
HT2 = (ht3:hours(1):ht4)'
%%
ht5 = datetime(2021,4,14,0,0,0);
ht6 = datetime(2021,4,15,23,0,0);
HT3 = (ht5:hours(1):ht6)'
%%
ht7 = datetime(2021,4,14,0,0,0);
ht8 = datetime(2021,4,15,23,0,0);
HT4 = (ht7:hours(1):ht8)'
%%
ht9 = datetime(2021,5,16,0,0,0);
ht10 = datetime(2021,5,17,23,0,0);
HT5 = (ht9:hours(1):ht10)'
%%
ht11 = datetime(2021,9,6,0,0,0);
ht12 = datetime(2021,9,7,23,0,0);
HT6 = (ht11:hours(1):ht12)'
%%
ht13 = datetime(2021,9,15,0,0,0);
ht14 = datetime(2021,9,16,23,0,0);
HT7 = (ht13:hours(1):ht14)'
%%
ht15 = datetime(2021,9,20,0,0,0);
ht16 = datetime(2021,9,20,23,0,0);
HT8 = (ht15:hours(1):ht16)'
%%
ht17 = datetime(2021,9,27,0,0,0);
ht18 = datetime(2021,9,27,23,0,0);
HT9 = (ht17:hours(1):ht18)'
%%
%% Holidays Array :
Hdays = [HT1 ; HT2 ; HT3 ; HT4 ; HT5 ; HT6 ; HT7 ; HT8 ; HT9];
%% Start Loop for filling tariffs :
i = 1:1:8760;
while ( i < 8760 )
????
end
%% Create tariff table according to seasons (season 1 (Dec,Feb,Jan), Season 2 (March,april,may,october,november),season 3(june,july,august,september) :
Season1Tariffs = [0.7170,0.2307];
array2table(Season1Tariffs, 'VariableNames', {'OnPeak','OffPeak'});
Season2Tariffs = [0.2578,0.2243];
array2table(Season2Tariffs, 'VariableNames', {'OnPeak','OffPeak'});
Season3Tariffs = [1.0789,0.2701];
array2table(Season3Tariffs, 'VariableNames', {'OnPeak','OffPeak'});
Thanks !
  2 Comments
Adnan Jayyousi
Adnan Jayyousi on 25 Jun 2022
Thanks Dyuman,
That helped me alot,
I have implemented this and have added more logical data, i've a feeling that i haven't done it in efficient way...but works.
%% Define date&time arrays :
t1 = datetime(2021,1,1,0,0,0);
t2 = datetime(2021,12,31,23,0,0);
dt = (t1:hours(1):t2)'
%holidyas date array
Holidays = [datetime(2021,3,27,0,0,0) datetime(2021,3,28,0,0,0) datetime(2021,4,2,0,0,0) datetime(2021,4,3,0,0,0) datetime(2021,4,14,0,0,0) datetime(2021,4,15,0,0,0) datetime(2021,5,16,0,0,0) datetime(2021,5,17,0,0,0) datetime(2021,9,6,0,0,0) datetime(2021,9,7,0,0,0) datetime(2021,9,15,0,0,0) datetime(2021,9,16,0,0,0) datetime(2021,9,20,0,0,0) datetime(2021,9,27,0,0,0)]
Holidays = Holidays'
DateNtime=zeros(numel(dt),2); %preallocating
%% Hours between 00:00 - 16:00 "Off Peak" - Days of week. (All Seasons)
for i=1:numel(dt)
if hour(dt(i))>=0 & hour(dt(i))<=16 & ismember(day(dt(i),'dayofweek'),[1 2 3 4 5])
switch month(dt(i))
case {12,1,2} %% Season #1
DateNtime(i,2)=0.2307;
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2243;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=0.2701;
end
end
end
%% Hours between 22:00 - 23:00 "Off Peak" - Days of week. (Season#1 only)
for i=1:numel(dt)
if hour(dt(i))>=22 & hour(dt(i))<=23 & ismember(day(dt(i),'dayofweek'),[1 2 3 4 5])
switch month(dt(i))
case {12,1,2} %% Season #1
DateNtime(i,2)=0.2307;
end
end
end
%% Hour 23:00 "Off Peak" - Days of week. (Season#2 & Season #3)
for i=1:numel(dt)
if hour(dt(i))== 23 & ismember(day(dt(i),'dayofweek'),[1 2 3 4 5])
switch month(dt(i))
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2243;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=0.2701;
end
end
end
%% Weekends. (All seasons)
for i=1:numel(dt)
if ismember(day(dt(i),'dayofweek'),[6 7])
switch month(dt(i))
case {12,1,2} %% Season #1
DateNtime(i,2)=0.2307;
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2243;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=0.2701;
end
end
end
%% On Peak - Days of week only (All Seasons)
for i=1:numel(dt)
if hour(dt(i))>=17 & hour(dt(i))<=21 & ismember(day(dt(i),'dayofweek'),[1 2 3 4 5])
switch month(dt(i))
case {12,1,2} %% Season #1
DateNtime(i,2)=0.7170;
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2578;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=1.0789;
end
end
end
%% Hour 22:00 "On Peak" - Days of week. (Season#2 & Season #3)
for i=1:numel(dt)
if hour(dt(i)) == 22 ismember(day(dt(i),'dayofweek'),[1 2 3 4 5])
switch month(dt(i))
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2578;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=1.0789;
end
end
end
%% Holidays
doy = day (Holidays(:,:),'dayofyear')
for i=1:numel(dt)
if ismember(day(dt(i),'dayofyear'),doy)
switch month(dt(i))
case {12,1,2} %% Season #1
DateNtime(i,2)=0.2307;
case {3,4,5,10,11} %% Season #2
DateNtime(i,2)=0.2243;
case {6,7,8,9} %% Season #3
DateNtime(i,2)=0.2701;
end
end
end
DateNtime

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Accepted Answer

Steven Lord
Steven Lord on 25 Jun 2022
Let's say you had some sample dates with time components.
rng default
d = datetime(2022, randi(12, 10, 1), randi(31, 10, 1), randi(24, 10, 1), randi([0 59], 10, 1), 0)
d = 10×1 datetime array
05-Oct-2022 16:42:00 01-Dec-2022 01:01:00 02-Mar-2022 21:16:00 16-Nov-2022 23:02:00 25-Aug-2022 17:05:00 05-Feb-2022 19:49:00 14-Apr-2022 18:41:00 29-Jul-2022 10:19:00 25-Dec-2022 16:57:00 30-Dec-2022 05:02:00
And you had a vector of holidays with the time component set to midnight of that day.
holidays = dateshift(d([2 7 4]), 'start', 'day')
holidays = 3×1 datetime array
01-Dec-2022 14-Apr-2022 16-Nov-2022
We can see which of the elements in d fall on the same days as those in holidays.
isaHoliday = ismember(day(d, 'dayofyear'), day(holidays, 'dayofyear'))
isaHoliday = 10×1 logical array
0 1 0 1 0 0 1 0 0 0
Or if you don't want to use day you could use dateshift to make a copy of d with the time components set to midnight.
isaHoliday2 = ismember(dateshift(d, 'start', 'day'), holidays)
isaHoliday2 = 10×1 logical array
0 1 0 1 0 0 1 0 0 0
If you need to consider only part of a day a holiday (say if you work for a company that closes at noon on the day before Thanksgiving in the US) you could use timeofday to extract the time components of d and use relational operators.
t = timeofday(d)
t = 10×1 duration array
16:42:00 01:01:00 21:16:00 23:02:00 17:05:00 19:49:00 18:41:00 10:19:00 16:57:00 05:02:00
isAfternoon = t > duration(12, 0, 0)
isAfternoon = 10×1 logical array
1 0 1 1 1 1 1 0 1 0

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