Day of year to month

Question

Claire 2011-2-9

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I have a years worth of data in the form of 1-365 (in a column of my matrix), and I want to change it to the days in a month, so I can plot a graph that is quicker to read, rather than having to work out, what month day 100 etc.. belongs to.

I'm sure it's really easy to do but I'm completely out of my depth.

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Answer 1

Andreas Goser 2011-2-9

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I am not 100% sure, but maybe this example helps you

d2011=734504; % 734504 ist the first day of 2011 - example
v=d2011:d2011+100; % create a vector of 100 days
datestr(v) % see all the dates as strings - use for the figure
datevec(v) % can extract day of month

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Answer 2

Claire 2011-2-9

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That does help thanks, but my problem now is that my data is hourly and so I need both time and data calculated to plot on my graph. I keep getting told that my vectors are different lengths, and I don't know how to rectify it.

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Jan 2011-2-9

Please show us the code and the error messages.

Claire 2011-2-9

Sorry Doug, I'll change it,

It's currently as:

Day = Belmonte(:,2);

Year = Belmonte (:,1);

Hour = Belmonte (:,3);

Minute = Belmonte (:,4);

Second = Belmonte (:,5);

And what I want is Year, Month, Day, Hour, Minute, Second, and then to be able to do a date in the format of Y, M, D, H, M, S, so that I can plot it against my evapotranspiration data which has a result for every hour. Does that make any sense?

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Answer 3

Brett Shoelson 2011-2-9

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If you have 365 days, for instance, but no month information, you can simply increment the DAY position in datenum:

mydates = datenum(2011,1,1:365,h,m,s);

Or, if you want the y,m,d,h,m,s:

[y,m,d,h,m,s] = datevec(datenum(2011,1,1:365,h,m,s))

Cheers,

Brett

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Brett Shoelson 2011-2-9

The answer doesn't change; you can specify the hour in the variable h.

Brett

Claire 2011-2-10

I've changed my code to D= datevec(datenum(1992,1,1:365, 1:24 ,00,00) but I'm still only getting 365 results whereas I need the 365 x 24.

Thanks

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Answer 4

Brett Shoelson 2011-2-10

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How about this?

D = datevec(datenum(1992,1,1:1/24:365, 0,0,0));

Cheers,

Brett

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Claire 2011-2-10

Thanks, that's improved the situation somewhat, although it seems to be still not quite right.

I've used D = datevec(datenum(1992,1,1:1/24:366)); but my Evapotranspiration data has 8760 rows, and the code I've used results in 8761 rows, and only one hour of results for 31/12

i.e. 1992, 12, 31, 0, 0, 0. I know 1992 is a leap year, but the data appears to only have 365 days, which I suspect is confusing the matter, although of course 8760/365 is 24, so that's just confused me more.

Thanks again for all your help.

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