[ncl-talk] delete 31 march for leap year

[Alan Brammer]

Depending on how long your dataset is, this may be more reliable than mod
4.
https://www.ncl.ucar.edu/Document/Functions/Built-in/isleapyear.shtml
e.g. 1900 was not a leap year.

On 17 April 2015 at 08:30, Marcus Morgan <mmorgan2009 at my.fit.edu> wrote:

> 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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