[ncl-talk] delete 31 march for leap year

[Marcus Morgan]

Here you would use the modulus function.  Every four years a leap year
occurs, so:


leap=ind(mod(Time(:,0), 4) .eq. 0 .and. TIME(:,1).eq.3 .and.
TIME(:,2).eq.31)        ; Chooses March 31st From those Leap Years

keep=ind(.not.Time(leap,0))
FOO = foo(keep,:,:)
delete(foo)

Haven't tested it, but it should work.
Good Luck


*Marcus N. Morgan*Graduate of Meteorology

*Department of Marine and Environmental Systems*   Florida Institute of
Technology
   150 W. University Blvd.
   Melbourne, FL 32901

On Fri, Apr 17, 2015 at 1:52 AM, Xi Chang <xi.chang01 at gmail.com> wrote:

> Thanks Dennis,
>
> Yes because i just need to analyze JFM (90 days) for 40 years. I know the
> above trick but i just want delete 31 march only for LEAP YEAR not for
> whole years. Do you know how to do that?
>
> thanks
>
>
> On Friday, April 17, 2015, Dennis Shea <shea at ucar.edu> wrote:
>
>> [1] Yes ... there is a way
>> [2] But ... why would ypu want to do that? It introduces a 'spectral gap'
>> in the time series.
>>
>> ===
>>   foo(time,lat,lon)
>>
>>   time has units understood by cd_calendar
>>   https://www.ncl.ucar.edu/Document/Functions/Built-in/cd_calendar.shtml
>>
>>   TIME = cd_calendar(time, 0)
>>
>>   ikeep = ind(.not.(TIME(:,1).eq.3 .and. TIME(:,2).eq.31) )
>>   FOO = foo(ikeep,:,:)
>>   delete(foo)
>>
>>
>>
>> On Thu, Apr 16, 2015 at 4:16 PM, Xi Chang <xi.chang01 at gmail.com> wrote:
>>
>>> Hi NCL,
>>> I have 40 years daily data, i need to delete 31 march only for the leap
>>> year, is there any quick way to do that?
>>>
>>> Regards
>>>
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>>
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