Making a datetime vector with a leap year
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Hi all
I have some code making a 8760x1 datetime vector containing every hour of the year
dt = datetime(2021,1,1:365);
et = hours(0:23)';
t = repmat(dt,length(et),1) + repmat(et,1,length(dt));
t=t(:);
'01-Jan-2021 00:00:00'
'01-Jan-2021 01:00:00'
'01-Jan-2021 02:00:00'
'01-Jan-2021 03:00:00'
'01-Jan-2021 04:00:00'
'01-Jan-2021 05:00:00'
'01-Jan-2021 06:00:00'
'01-Jan-2021 07:00:00'
'01-Jan-2021 08:00:00'
However some years are leap years. So if it is a leap year I want the 29 of February to be inserted properly. Like:
if length(Data{:,1})==422 %regular year
dt = datetime(2021,1,1:365);
et = hours(0:23)';
t = repmat(dt,length(et),1) + repmat(et,1,length(dt));
t=t(:);
elseif length(Data{:,1})==423 %leap year
"The code I am asking for"
end
0 comentarios
Respuestas (2)
Fangjun Jiang
el 5 de Abr. de 2023
Editada: Fangjun Jiang
el 5 de Abr. de 2023
Do it directly by specifying the starting and ending datetime.
d1=datetime(2021,1,1,0,0,0);
d2=datetime(2021,12,31,23,0,0);
size(d1:hours(1):d2)
3 comentarios
Les Beckham
el 5 de Abr. de 2023
Yes, a leap year does have 8784 hours. But 2021 is not a leap year. 2024 is:
d1=datetime(2024,1,1,0,0,0);
d2=datetime(2024,12,31,23,0,0);
size(d1:hours(1):d2)
Steven Lord
el 5 de Abr. de 2023
2021 was not a leap year. But as an example, 2020 was a leap year.
y = 2020;
d1=datetime(y,1,1,0,0,0)
d2=datetime(y,12,31,23,0,0)
size(d1:hours(1):d2)
Another way to create d2 would be to shift d1 to the start of the next year then subtract an hour. This way the year information only appears in one place in the code, in the definition of d1.
d2Alternate = dateshift(d1, 'start', 'year', 'next') - hours(1)
d2 == d2Alternate
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