Re: Question on the longitude coordinates transform?

From: Dennis Shea <shea_at_nyahnyahspammersnyahnyah>
Date: Thu Nov 24 2011 - 10:01:08 MST

PLEASE ... do *not* repeat post to ncl-talk.
            This is a 4-day national holiday.

You have your settings incorrect. They are reversed

     res@mpMinLonF = -150
     res@mpMaxLonF = 30

On 11/24/11 9:54 AM, xiang lin wrote:
> Hi,all
>
> I am fluent with the longitude coordinates of (0E, 360E), but now
> was a little puzzled when dealing with the coordinates of (-180E,175E) .
>
> Here below is parts of my scripts :
>
> ; read the hadslp data
> data = asciiread("/Work/smi/Hadslp2_185001-200412.dat",-1,"float")
> x = onedtond(data,(/72,37,12,155/))
> x!0 = "lon"
> x!1 = "lat"
> x!2 = "month"
> x!3 = "year"
> x&lon = fspan(-180.,175.,72) ; see more details in the
> hadslpr2's data read instruction
> x&lon@units = "degrees_east"
> x&lat = fspan(90.,-90.,37)
> x&lat@units = "degree_north"
> x&month = ispan(1,12,1)
> x&year = ispan(1850,2004,1)
> printVarSummary(x)
> ......
> res@mpMinLonF = 30
> res@mpMaxLonF = -150
>
> I want to pick out the sub-region eastward extending from 30E to 150W,
> but of course the above resource would cause errors for the reason that
> mpMinLonF must be less than
> mpMaxLonF.
>
> How could I set the region correctly for plot? Is there a way to
> transform the coordinates such as (-180E, 175E) to the ones such as
> (0E,360E)? I know the lonFlip and lonPivot,
> However, I don't find a corresponding function to the opposite job.
>
> Thanks!
>
> best wish!
>
>
> Lin
>
>
>
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Received on Thu Nov 24 10:01:15 2011

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