Re: annual mean and contour

Excuse me. It was pointed out that I am wrong. The ispan call does indeed return 85 values. However, in spite of the fact that you have
nlat set to 85, there is nothing in the code that shows what is the actual dimensionality of the 'var' variable. You need a printVarSummary(var)
to verify that you have what you think you have for its shape.
 -dave

On Feb 7, 2012, at 2:04 PM, David Brown wrote:

> Well the error message is telling you what is wrong: ispan(-42,42,1) produces 84 values, whereas your nlat value is 85.
> However, you should put some 'printVarSummary' messages in your code to help show you what is the problem.
> -dave
>
> On Feb 7, 2012, at 1:45 PM, Xiaoyan Ma wrote:
>
>> Hi
>>
>> I want to calculate the annual mean from 12 months hdf files, and then make a contour, below is what I did.
>> There is a fatal error as below
>> ---------------------------------------------------------------------------------------------
>> fatal:Coordinate variables must be the same dimension as their dimension
>> fatal:Execute: Error occurred at or near line 43 in file
>> -------------------------------------------------------------------------------------------
>> line 43 is " var&lat = ispan(-42,42,1)*dlat "
>>
>> Also the values of var is not correct. can anyone help to fix the problem?
>>
>> Thanks,
>> Xiaoyan
>>
>> ================================
>> nlat = 85
>> nlon = 72
>> ntim = 12
>>
>> do nt=0,ntim-1
>> fname = m+sprinti("%0.2i", nt+1)+".hdf"
>> print(fname)
>> in = addfiles(fname, "r") ;; load all the files
>> var0 = in[:]->variable_name
>> end do
>> var = dim_avg_n(var0, 0)
>>
>> dlat = 170./nlat
>> dlon = 360./nlon
>>
>> var!0 = "lat"
>> var&lat = ispan(-42,42,1)*dlat
>> var&lat@units = "degrees_north"
>>
>> var!1 = "lon"
>> var&lon = ispan(-36,35,1)*dlon
>> var&lon@units = "degrees_east"
>>
>> ...
>>
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Received on Tue Feb 7 14:23:41 2012