[ncl-talk] Coefficient_of Divergence

[Dennis Shea]

*Variable: CoD1*
*Dimensions and sizes: [3] x [660] x [988]*
*(0) min=0   max=0.961394*


step 2:

   CoD2 = sum(CoD1)/3           <===  No!
   printVarSummary(CoD2)
   printMinMax(CoD2, 0)

https://www.ncl.ucar.edu/Document/Functions/Contributed/dim_avg_n_Wrap.shtml

   CoD2 = dim_avg_n_Wrap(CoD1, 0)    ; (nlat,mlon)
   printVarSummary(CoD2)
   printMinMax(CoD2,0)

===
If you still have large numbers, the find one grid point where a large
number occurs, then print the values.

   print(CoD1(:,nn,mm))   ; print series where large numbers occur
(indices:  nn,mm)

See documentation for
   https://www.ncl.ucar.edu/Document/Functions/Built-in/ind.shtml
   https://www.ncl.ucar.edu/Document/Functions/Built-in/ind_resolve.shtml
See Examples 2 & 3






On Tue, Nov 7, 2017 at 1:32 PM, Kunal Bali <kunal.bali9 at gmail.com> wrote:

> Thanks for suggesting this.
>
> Actually, I am trying to calculate the COD at each grid cell.
>
> So as you suggested
>
>
> *DATA_SINGLE_3D = conform(data_subset, data_single, 0)    ; it worked*
>
>
>
> *Variable: DATA_SINGLE_3DType: floatTotal Size: 7824960 bytes
> 1956240 valuesNumber of Dimensions: 3Dimensions and sizes: [3] x [660] x
> [988]Coordinates: Number Of Attributes: 1  _FillValue : -28672*
> step 1:
> CoD1 = ((DATA_SINGLE_3D - data_subset)/(DATA_SINGLE_3D+ data_subset))^2
>    printVarSummary(CoD1)
>    printMinMax(CoD1, 0)
>
> *Variable: CoD1*
> *Type: float*
> *Total Size: 7824960 bytes*
> *            1956240 values*
> *Number of Dimensions: 3*
> *Dimensions and sizes: [3] x [660] x [988]*
> *Coordinates: *
> *Number Of Attributes: 1*
> *  _FillValue : -28672*
> *(0) min=0   max=0.961394*
>
>
> step 2:
>
>    CoD2 = sum(CoD1)/3
>    printVarSummary(CoD2)
>    printMinMax(CoD2, 0)
>
> *Variable: CoD2*
> *Type: float*
> *Total Size: 4 bytes*
> *            1 values*
> *Number of Dimensions: 1*
> *Dimensions and sizes: [1]*
> *Coordinates: *
> *(0) min=20691.8   max=20691.8   --------> values are very high? and also
> it's not giving at each grid cell, because of that I can't calculate the
> sqrt in the next step*
>
>
> step 3:
>
>   CoD3 = sqrt(CoD2)
>  printVarSummary(CoD3)
>  printMinMax(CoD2, 0)
>
>
> The final values are coming very high. It should be less than 0.3 at each
> grid cell.
>
> regards
> Kunal Bali
>
>
>
>
>
> On Wed, Nov 8, 2017 at 1:18 AM, Dennis Shea <shea at ucar.edu> wrote:
>
>> If you have a one-dimensional array (1D; eg: a 'time series') and a
>> three-dimensional array (3D; eg: time series at multiple grid points), you
>> *must* make the arrays 'conformant' for an array calculation
>>
>> You have:
>>    CoD1 = ((data_single(:) - data_subset(:,:,:))/(data_single(:) +
>> data_subset(:,:,:)))^2
>>
>> You did not include a 'printVarSummary' of these variables. Assuming
>> data_single(ntim) and data_subset(ntim,nlat,mlon) then you must replicate
>> (aka, propogate) the 1D array to the 3D array size. In NCL, this is called
>> 'array conformance' . Please see and carefully read:
>>
>> https://www.ncl.ucar.edu/Document/Functions/Built-in/conform.shtml
>> https://www.ncl.ucar.edu/Document/Functions/Built-in/conform_dims.shtml
>>
>> ---
>>   DATA_SINGLE_3D = conform(data_subset, data_single, 0)    ;
>> (ntim,nlat,mlon)
>>   printVarSummary(DATA_SINGLE_3D)   ; Look at data
>>
>> Do *not* use (:,:,:) in a computation... this is inefficient.
>>
>>   CoD1 = ((DATA_SINGLE_3D - data_subset)/(DATA_SINGLE_3D+ data_subset))^2
>>   printVarSummary(CoD1)
>>   printMinMax(CoD1, 0)
>>
>>
>>
>>
>>
>>
>> On Tue, Nov 7, 2017 at 11:15 AM, Kunal Bali <kunal.bali9 at gmail.com>
>> wrote:
>>
>>> Dear NCL user
>>>
>>> Suppose, I have 2 datasets
>>> e.g.
>>>
>>> data1 = of single point lat & lon (e.g 28N, 72E) with time steps 365
>>> (having filling values)
>>> data2 = of grid box (24-32N, 68-76E) with time steps 365  (having
>>> filling values)
>>>
>>> Now I want to calculate the CoD from one single point to 400km grid box
>>> area and then lot the spatial map of the calculated product.
>>>
>>> Formula for CoD is GIVEN BELOW
>>> ​
>>>
>>> So I calculated in NCL as
>>>
>>> CoD1 = ((data1 - data2)/(data1 + data2))^2
>>> CoD_final = (sqrt(sum(CoD1)))/3
>>>
>>> please let me know is it the right way to do?
>>>
>>> I tried to plot the final product CoD_final with the attached script,
>>> but error comes as
>>>
>>> fatal:Minus: Number of dimensions do not match, can't continue
>>> fatal:["Execute.c":8578]:Execute: Error occurred at or near line 116 in
>>> file plot.ncl
>>>
>>>
>>>
>>> regards
>>> Kunal Bali
>>>
>>>
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>>>
>>>
>>
>
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