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Re: how can you convert CMIP5 model output time ? [message #89110 is a reply to message #89096]

Thu, 24 July 2014 17:00

siumtesfai
Messages: 62
Registered: April 2013

Member

On Thursday, July 24, 2014 4:44:45 AM UTC-4, Fabien wrote:
> On 24.07.2014 02:37, siumtesfai@gmail.com wrote:
>
>> Any solution about how to handle the time array in CMIP5 model daily output.
>
>
>
> You answered your own question by printing the variable's attributes:
>
>
>
> double time(time) ;
>
> time:bounds = "time_bnds" ;
>
> time:units = "days since 1850-01-01 00:00:00" ;
>
>
>
> And Michael gave you the "formula" to use.
>
>
>
> So, what did you try and why didn't it work as expected?

Yes I tried

time=time+julday(1850,1,1,0,0,0)
Caldat,month,year,day

It did not work for me.

I should be able to see time ranging from Jan 1850 to Dec 2005 ( 1872 months). I can create my own time ( time=findgen(1872)/12. + 1850)

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[Message index]

		how can you convert CMIP5 model output time ? By: siumtesfai on Wed, 23 July 2014 12:59
		Re: how can you convert CMIP5 model output time ? By: Michael Galloy on Wed, 23 July 2014 13:45
		Re: how can you convert CMIP5 model output time ? By: siumtesfai on Wed, 23 July 2014 14:00
		Re: how can you convert CMIP5 model output time ? By: siumtesfai on Wed, 23 July 2014 17:37
		Re: how can you convert CMIP5 model output time ? By: Fabzi on Thu, 24 July 2014 01:44
		Re: how can you convert CMIP5 model output time ? By: siumtesfai on Thu, 24 July 2014 17:00
		Re: how can you convert CMIP5 model output time ? By: Fabzi on Fri, 25 July 2014 01:51

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