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 Comments
Answers (2)
  Fangjun Jiang
      
      
 on 5 Apr 2023
        
      Edited: Fangjun Jiang
      
      
 on 5 Apr 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 Comments
  Les Beckham
      
 on 5 Apr 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
    
      
 on 5 Apr 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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