[ncl-talk] Time Vs Lat/Lon

Kunal Bali kunal.bali9 at gmail.com
Thu Nov 1 12:05:27 MDT 2018


I tried as

    arr_min = *dim_minind* ( arr(time|:, lat|:, lon|:),0 )
    printVarSummary(arr_min)
    printMinMax(arr_min,0)

    arr_max = *dim_minind* ( arr(time|:, lat|:, lon|:),0 )
    printVarSummary(arr_max)
    printMinMax(arr_max,0)

    time_min = arr_min    ; arr_min contains the indices that were the
minimum at each lat/lon point
    time_max = arr_max
    printVarSummary(time_max)
    do gg = 0,dimsizes(arr_min&lat)-1
     do hh = 0,dimsizes(arr_min&lon)-1
          time_min(gg,hh)  = time(arr_min(gg,hh))
          time_max(gg,hh)  = time(arr_max(gg,hh))
      end do
    end do

error comes at red line

Assignment type mismatch, right hand side can't be coerced to type of left
hand side


---
Kunal Bali





On Thu, Nov 1, 2018 at 9:13 PM Adam Phillips <asphilli at ucar.edu> wrote:

> Hi Kunal,
> In NCL v6.5.0 you can specify -4 as the 2nd option in cd_calendar. It
> returns results that are the same as if you specify 0, so try specifying 0.
>
> Again, you need to use dim_minind
> <https://www.ncl.ucar.edu/Document/Functions/Contributed/dim_minind.shtml>
> and dim_maxind
> <https://www.ncl.ucar.edu/Document/Functions/Contributed/dim_maxind.shtml> to
> return the index of the minimum (/maximum) value, and not dim_max/dim_min.
> The latter two functions return the minimum and maximum values. You the
> indices to plug them into your time array within the double do loop I sent
> earlier.
> Adam
>
> On Thu, Nov 1, 2018 at 12:31 AM Kunal Bali <kunal.bali9 at gmail.com> wrote:
>
>> I tried this but not working
>>
>>     arr    = a->BCSMASS
>>     lat    = a->lat
>>     lon    = a->lon
>>
>>     printVarSummary(arr)
>>     timeT = cd_calendar(arr&time,-4)   [==> *getting a warning*:cd_calendar:
>> Unknown option, defaulting to 0.]
>>     time = timeT(:,3)
>>
>>     ;arr_min = dim_min( arr(lat|:, lon|:, time|:) )
>>     ;printVarSummary(arr_min)
>>     ;printMinMax(arr_min,0)
>>
>>     ;arr_max = dim_max( arr(lat|:, lon|:, time|:) )
>>     ;printVarSummary(arr_max)
>>     ;printMinMax(arr_max,0)
>>
>>     ;time_min = arr_min    ; arr_min contains the indices that were the
>> minimum at each lat/lon point
>>     ;time_max = arr_max
>>     ;printVarSummary(time_max)
>>     do gg = 0,dimsizes(arr_min&lat)-1
>>      do hh = 0,dimsizes(arr_min&lon)-1
>>           time_min(gg,hh) = time(arr_min(gg,hh))
>>           time_max(gg,hh) = time(arr_max(gg,hh))
>>       end do
>>     end do
>>
>> ---
>> Kunal Bali
>>
>>
>>
>>
>>
>>
>>
>> On Thu, Nov 1, 2018 at 3:05 AM Adam Phillips <asphilli at ucar.edu> wrote:
>>
>>> Hi Kunal,
>>> Note that I recommended using dim_minind/dim_maxind, and not
>>> minind/maxind. Looking at the file you attached, you would pass your
>>> BCSMASS array into those two functions, and operate on the time (0)
>>> dimension. As previously stated, that will return a 2D array dimensioned
>>> lat x lon containing the desired indices. Then you would apply the double
>>> do loop coding I sent to form arrays of the times themselves.  Again, the
>>> time array in my example coding needs to be set to something that is easily
>>> plotted, say the Hour. You can set up an hour array for your data like this:
>>> a = addfile("hourmean.nc","r")
>>> arr = a->BCSMASS
>>> timeT = cd_calendar(arr&time,-4)
>>> time = timeT(:,3)
>>>
>>> Adam
>>>
>>> On Tue, Oct 30, 2018 at 12:02 PM Kunal Bali <kunal.bali9 at gmail.com>
>>> wrote:
>>>
>>>> Thanks for this informations.
>>>>
>>>> I tried to get min mad max by using the link which you mentioned, but
>>>> the error comes with fatal:Argument type mismatch on argument (0) of
>>>> (maxind) can not coerce
>>>>
>>>> I also tried the code given below but facing the same error here.
>>>> (version-NCL 6.4.0)
>>>>
>>>> dims = *dimsizes* <https://www.ncl.ucar.edu/Document/Functions/Built-in/dimsizes.shtml>(X)
>>>>   x1d = *ndtooned* <https://www.ncl.ucar.edu/Document/Functions/Built-in/ndtooned.shtml>(X)      ; convert 2D array to 1D for use in maxind
>>>>   inds = *ind_resolve* <https://www.ncl.ucar.edu/Document/Functions/Built-in/ind_resolve.shtml>(*maxind* (x1d), dims)    ; convert 1D array back to 2D
>>>>   ilat = inds(0,0)        ; select the latitude index where the X array is at its' maximum
>>>>   ilon = inds(0,1)        ; select the longitude index where the X array is at its' maximum
>>>>   lat_max = X&lat(ilat)   ; insert the latitude index into the lat coordinate variable
>>>>   lon_max = X&lon(ilon)   ; insert the longitude index into the lon coordinate variable
>>>>   *print* <https://www.ncl.ucar.edu/Document/Functions/Built-in/print.shtml>("Maximum value located at "+lat_max+", "+lon_max)
>>>>
>>>> I also attached one sample file, in case if you need it.
>>>>
>>>>
>>>> ---
>>>> Kunal Bali
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On Tue, Oct 30, 2018 at 2:47 AM Adam Phillips <asphilli at ucar.edu>
>>>> wrote:
>>>>
>>>>> Hi Kunal,
>>>>> If I understand correctly what you are after, you'd like to plot the
>>>>> time (at each grid point) where the maximum (or minimum) value occurs.
>>>>> This will likely take some trial and error on your part. Here's what I
>>>>> think you need to do:
>>>>> 1) Use dim_minind and dim_maxind to isolate the index where the grid
>>>>> point value is at its min/max.
>>>>>
>>>>> https://www.ncl.ucar.edu/Document/Functions/Contributed/dim_maxind.shtml
>>>>>
>>>>> https://www.ncl.ucar.edu/Document/Functions/Contributed/dim_minind.shtml
>>>>>
>>>>> This will result in a 2-D array (dimensioned lat x lon) containing the
>>>>> minimum indices, and the same type of array containing the maximum indices.
>>>>>
>>>>> 2)  There might be a more elegant way to map the indexes onto the
>>>>> correct times, but the following double do loop over the lat/lon dimensions
>>>>> will work:
>>>>>
>>>>> ; time is an array of your times
>>>>> time_min = arr_min    ; arr_min contains the indices that were the
>>>>> minimum at each lat/lon point
>>>>> time_max = arr_max
>>>>> do gg = 0,dimsizes(arr_min&lat)-1
>>>>>      do hh = 0,dimsizes(arr_min&lon)-1
>>>>>           time_min(gg,hh) = time(arr_min(gg,hh))
>>>>>           time_max(gg,hh) = time(arr_max(gg,hh))
>>>>>      end do
>>>>> end do
>>>>>
>>>>> You will want to make sure that your time array is in a format that is
>>>>> conducive to being plotted. (18:30 doesn't work for instance, but 1830
>>>>> would.)
>>>>> I think that's it. Try starting with the above suggestions and see if
>>>>> you can get things to work. If you have further questions about this
>>>>> subject please reply to the ncl-talk email list.
>>>>> Adam
>>>>>
>>>>>
>>>>>
>>>>> On Mon, Oct 29, 2018 at 6:57 AM Kunal Bali <kunal.bali9 at gmail.com>
>>>>> wrote:
>>>>>
>>>>>> Dear NCL users,
>>>>>>
>>>>>> I have one file (2000-2017) with time, latitude, longitude and
>>>>>> variable. The time is given in hours(00:30 to 18:30). I need to plot a
>>>>>> spatial map of hours values at each grid cells, the contour map of hours
>>>>>> values should correspond to the variable values. So that I can see which
>>>>>> hour has high and low-value on the map. Some examples are given for time vs
>>>>>> latitude and time vs longitude, I don't need that.  I need to plot time vs
>>>>>> latitude longitude.
>>>>>>
>>>>>> I hope you got my point.
>>>>>> Any information on that?
>>>>>>
>>>>>> Thank You
>>>>>> ---
>>>>>> Kunal Bali
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> _______________________________________________
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>>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Adam Phillips
>>>>> Associate Scientist,  Climate and Global Dynamics Laboratory, NCAR
>>>>> www.cgd.ucar.edu/staff/asphilli/   303-497-1726
>>>>>
>>>>> <http://www.cgd.ucar.edu/staff/asphilli>
>>>>>
>>>> _______________________________________________
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>>>
>>>
>>> --
>>> Adam Phillips
>>> Associate Scientist,  Climate and Global Dynamics Laboratory, NCAR
>>> www.cgd.ucar.edu/staff/asphilli/   303-497-1726
>>>
>>> <http://www.cgd.ucar.edu/staff/asphilli>
>>>
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>
>
> --
> Adam Phillips
> Associate Scientist,  Climate and Global Dynamics Laboratory, NCAR
> www.cgd.ucar.edu/staff/asphilli/   303-497-1726
>
> <http://www.cgd.ucar.edu/staff/asphilli>
>
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