Time and Date

I've written some time and date code, but I'm sure there's a better way to do it. I had my data set up as dd:mm:yyyy hh:mm:ss with each one in a separate column with the rest of my input variables, so I did as follows:

load belmontemini;
Day = data(:,2);  
Month = data(:,3);
Year = data (:,1);
Hour = data (:,4);
Minute = data (:,5);
Second = data (:,6);
.....
Time = [Day Month Year Hour Second Minute];
C= [Time Evap];

If anyone could point me in the right direction that would be great. I tried to work it out using datenum, but I just couldn't make it work. I'm very new to Matlab, and really confused sorry!!

Thank you.

[Claire]

Sorry it's actually Day of the Year, so I need to convert that to Day and Month as well if possible? Thanks

[Walter Roberson]

You need to clarify what output you expect.

Why do you have both month and day of the year?

Answer 1

[Oleg Komarov]

Quote: "I had my data set up as dd:mm:yyyy hh:mm:ss with each one in a separate column with the rest of my input variables, so I did as follows"

Not clear...did you have smt like:

• Dates as strings:

data = {'01:01:2004 20:25:33'; 
        '09:02:2005 20:25:42'}

• Dates as datevec output (individual components):

data = [01 01 2004 20 25 33;
        09 02 2005 20 25 42];

And what's the result you trying to achieve? Convert to serial date?

• In the first case:

datenum(data, 'dd:mm:yyyy HH:MM:SS')

• In the second case:

datenum(data(:,[3,2,1,4:6]))

Oleg

Answer 2

[Walter Roberson]

If you have the day of year but not the month, then

datenum([Year 0 DayOfYear Hour Minute Second])

will produce a serial date number with the Day Of Year information properly distributed into month and day, taking in to account leap years.

If you need a string output instead, use datestr() with the same input technique.

If you need the vector back again but want the DayOfYear adjusted to month + day, then you can datenum() and datevec() the result of that.