[ncl-talk] Coefficient_of Divergence

Tue Nov 7 22:26:44 MST 2017

I followed your instruction, But now CoD1 and CoD2 values are showing the
same values, which is incorrect? and because of this values of the square
root of CoD2 is not correct.

;-----calculation of coefficient of divergence
   CoD1 = ((DATA_SINGLE_3D - data_subset)/(DATA_SINGLE_3D+ data_subset))^2
   printVarSummary(CoD1)
   printMinMax(CoD1, 0)

   CoD1 at _FillValue = -28672
   CoD2 = dim_avg_n_Wrap(CoD1, 0)    ; (nlat,mlon)
   printVarSummary(CoD2)
   printMinMax(CoD2,0)

   CoD3 = sqrt(CoD2)
   printVarSummary(CoD3)
   printMinMax(CoD3, 0)

*Variable: CoD1Type: floatTotal Size: 7824960 bytes            1956240
valuesNumber of Dimensions: 3Dimensions and sizes:    [3] x [660] x
[988]Coordinates: Number Of Attributes: 1  _FillValue :    -28672(0)
min=0   max=0.961394Variable: CoD2Type: floatTotal Size: 2608320
bytes            652080 valuesNumber of Dimensions: 2Dimensions and
sizes:    [660] x [988]Coordinates: Number Of Attributes: 2  _FillValue
:    -28672  average_op_ncl :    dim_avg_n function was applied(0)
min=0   max=0.961394Variable: CoD3Type: floatTotal Size: 2608320
bytes            652080 valuesNumber of Dimensions: 2Dimensions and
sizes:    [660] x [988]Coordinates: Number Of Attributes: 1  _FillValue
:    -28672(0)    min=0   max=0.980507*

regsrds
Kunal Bali

On Wed, Nov 8, 2017 at 3:55 AM, Dennis Shea <shea at ucar.edu> wrote:

>
> *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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