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Regarding Days to month

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Uday
Uday il 31 Lug 2011
I have some data from 2008-2011, which includes leap year 2008. my data sets has following values [year day hour ]
I have days in each year ( 1:365 or 1:366), i wanna to calculate the months.
Thank you

Risposte (3)

Jan
Jan il 31 Lug 2011
Standard:
datevec(datenum(2008, 1, 12))
>> [2008, 1, 12, 0, 0, 0]
Automatic months overflow:
datevec(datenum(2008, 1, 32))
>> [2008, 2, 1, 0, 0, 0]
Considers leap year:
datevec(datenum(2008, 1, 60))
>> [2008, 2, 29, 0, 0, 0]
datevec(datenum(2008, 1, 61))
>> [2008, 3, 1, 0, 0, 0]
  3 Commenti
Oleg Komarov
Oleg Komarov il 31 Lug 2011
And how the automatic month overflow that Jan shows doesn't solve your problem? Have you tried it?
Jan
Jan il 31 Lug 2011
@Pimple: I do not see the differnce. Let [Day] be the day number in the range 1:356 or 366. Then this gets the corresponding month: "D = datevec(datenum(Year, 1, Day, Hour, 0, 0)); Month=D(2)".
If this still does not help, please explain the wanted output exactly again.

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the cyclist
the cyclist il 31 Lug 2011
I would expect you to be able to accomplish what you want with some combination of the command datevec(), datenum(), and datestr(). If you can't figure it out, post the exact format of what you have, and what you want as output. Show us the code that you have tried yourself.

Walter Roberson
Walter Roberson il 31 Lug 2011
T = datevec(datenum(YourYear, 1, YourDays, YourHours, 0, 0));
TheMonth = T(:,2)
But if you want to calculate fraction of a year, you do not need to calculate the month number:
YearFraction = datenum(YourYear, 1, YourDays, YourHours, 0, 0) - datenum(YourYear, 1, 1, 0, 0, 0);
  2 Commenti
Uday
Uday il 1 Ago 2011
I used fractYear=year+ ((day-1)+(hour/24))./(365+(eomday(year,2)==29));
it works fine.
Jan
Jan il 1 Ago 2011
@Pimple: I can see just a weak connection to your statement "i wanna to calculate the months". Please save our time by asking more precisely.

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