[Numpy-discussion] [Cdat-discussion] Arrays containing NaNs

Eric Firing efiring@hawaii....
Fri Jul 25 13:00:23 CDT 2008


Charles Doutriaux wrote:
> I mean not having to it myself.
> data is a numpy array with NaN in it
> masked_data = numpy.ma.array(data)
> returns a masked array with a mask where NaN were in data

Checking for nans is an expensive operation, so it makes sense to make 
it optional rather than impose the cost on all masked array creations. 
If you want the same effect, you can do this:

masked_data = numpy.ma.masked_invalid(data)

Eric


> 
> C.
> 
> Bruce Southey wrote:
>> Charles Doutriaux wrote:
>>   
>>> Hi Bruce,
>>>
>>> Thx for the reply, we're aware of this, basically the question was why 
>>> not mask NaN automatically when creating a nump.ma array?
>>>
>>> C.
>>>
>>> Bruce Southey wrote:
>>>   
>>>     
>>>> Charles Doutriaux wrote:
>>>>   
>>>>     
>>>>       
>>>>> Hi Stephane,
>>>>>
>>>>> This is a good suggestion, I'm ccing the numpy list on this. Because I'm 
>>>>> wondering if it wouldn't be a better fit to do it directly at the 
>>>>> numpy.ma level.
>>>>>
>>>>> I'm sure they already thought about this (and 'inf' values as well) and 
>>>>> if they don't do it , there's probably some good reason we didn't think 
>>>>> of yet.
>>>>> So before i go ahead and do it in MV2 I'd like to know the reason why 
>>>>> it's not in numpy.ma, they are probably valid for MVs too.
>>>>>
>>>>> C.
>>>>>
>>>>> Stephane Raynaud wrote:
>>>>>   
>>>>>     
>>>>>       
>>>>>         
>>>>>> Hi,
>>>>>>
>>>>>> how about automatically (or at least optionally) masking all NaN 
>>>>>> values when creating a MV array?
>>>>>>
>>>>>> On Thu, Jul 24, 2008 at 11:43 PM, Arthur M. Greene 
>>>>>> <amg@iri.columbia.edu <mailto:amg@iri.columbia.edu>> wrote:
>>>>>>
>>>>>>     Yup, this works. Thanks!
>>>>>>
>>>>>>     I guess it's time for me to dig deeper into numpy syntax and
>>>>>>     functions, now that CDAT is using the numpy core for array
>>>>>>     management...
>>>>>>
>>>>>>     Best,
>>>>>>
>>>>>>     Arthur
>>>>>>
>>>>>>
>>>>>>     Charles Doutriaux wrote:
>>>>>>
>>>>>>         Seems right to me,
>>>>>>
>>>>>>         Except that the syntax might scare a bit the new users :)
>>>>>>
>>>>>>         C.
>>>>>>
>>>>>>         Andrew.Dawson@uea.ac.uk <mailto:Andrew.Dawson@uea.ac.uk> wrote:
>>>>>>
>>>>>>             Hi,
>>>>>>
>>>>>>             I'm not sure if what I am about to suggest is a good idea
>>>>>>             or not, perhaps Charles will correct me if this is a bad
>>>>>>             idea for any reason.
>>>>>>
>>>>>>             Lets say you have a cdms variable called U with NaNs as
>>>>>>             the missing
>>>>>>              value. First we can replace the NaNs with 1e20:
>>>>>>
>>>>>>             U.data[numpy.where(numpy.isnan(U.data))] = 1e20
>>>>>>
>>>>>>             And remember to set the missing value of the variable
>>>>>>             appropriately:
>>>>>>
>>>>>>             U.setMissing(1e20)
>>>>>>
>>>>>>             I hope that helps, Andrew
>>>>>>
>>>>>>
>>>>>>
>>>>>>                 Hi Arthur,
>>>>>>
>>>>>>                 If i remember correctly the way i used to do it was:
>>>>>>                 a= MV2.greater(data,1.) b=MV2.less_equal(data,1)
>>>>>>                 c=MV2.logical_and(a,b) # Nan are the only one left
>>>>>>                 data=MV2.masked_where(c,data)
>>>>>>
>>>>>>                 BUT I believe numpy now has way to deal with nan I
>>>>>>                 believe it is numpy.nan_to_num But it replaces with 0
>>>>>>                 so it may not be what you
>>>>>>                  want
>>>>>>
>>>>>>                 C.
>>>>>>
>>>>>>
>>>>>>                 Arthur M. Greene wrote:
>>>>>>
>>>>>>                     A typical netcdf file is opened, and the single
>>>>>>                     variable extracted:
>>>>>>
>>>>>>
>>>>>>                                 fpr=cdms.open('prTS2p1_SEA_allmos.cdf')
>>>>>>                                 pr0=fpr('prcp') type(pr0)
>>>>>>
>>>>>>                     <class 'cdms2.tvariable.TransientVariable'>
>>>>>>
>>>>>>                     Masked values (indicating ocean in this case) show
>>>>>>                     up here as NaNs.
>>>>>>
>>>>>>
>>>>>>                                 pr0[0,-15:-5,0]
>>>>>>
>>>>>>                     prcp array([NaN NaN NaN NaN NaN NaN 0.37745094
>>>>>>                     0.3460784 0.21960783 0.19117641])
>>>>>>
>>>>>>                     So far this is all consistent. A map of the first
>>>>>>                     time step shows the proper land-ocean boundaries,
>>>>>>                     reasonable-looking values, and so on. But there
>>>>>>                     doesn't seem to be any way to mask
>>>>>>                      this array, so, e.g., an 'xy' average can be
>>>>>>                     computed (it
>>>>>>                     comes out all nans). NaN is not equal to anything
>>>>>>                     -- even
>>>>>>                     itself -- so there does not seem to be any
>>>>>>                     condition, among the
>>>>>>                      MV.masked_xxx options, that can be applied as a
>>>>>>                     test. Also, it
>>>>>>                      does not seem possible to compute seasonal averages,
>>>>>>                     anomalies, etc. -- they also produce just NaNs.
>>>>>>
>>>>>>                     The workaround I've come up with -- for now -- is
>>>>>>                     to first generate a new array of identical shape,
>>>>>>                     filled with 1.0E+20. One test I've found that can
>>>>>>                     detect NaNs is numpy.isnan:
>>>>>>
>>>>>>
>>>>>>                                 isnan(pr0[0,0,0])
>>>>>>
>>>>>>                     True
>>>>>>
>>>>>>                     So it is _possible_ to tediously loop through
>>>>>>                     every value in the old array, testing with isnan,
>>>>>>                     then copying to the new array if the test fails.
>>>>>>                     Then the axes have to be reset...
>>>>>>
>>>>>>                     isnan does not accept array arguments, so one
>>>>>>                     cannot do, e.g.,
>>>>>>
>>>>>>                     prmasked=MV.masked_where(isnan(pr0),pr0)
>>>>>>
>>>>>>                     The element-by-element conversion is quite slow.
>>>>>>                     (I'm still waiting for it to complete, in fact).
>>>>>>                     Any suggestions for dealing with NaN-infested data
>>>>>>                     objects?
>>>>>>
>>>>>>                     Thanks!
>>>>>>
>>>>>>                     AMG
>>>>>>
>>>>>>                     P.S. This is 5.0.0.beta, RHEL4.
>>>>>>
>>>>>>
>>>>>>     *^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*
>>>>>>     Arthur M. Greene, Ph.D.
>>>>>>     The International Research Institute for Climate and Society
>>>>>>     The Earth Institute, Columbia University, Lamont Campus
>>>>>>     Monell Building, 61 Route 9W, Palisades, NY  10964-8000 USA
>>>>>>     amg*at*iri-dot-columbia\dot\edu | http://  iri.columbia.edu
>>>>>>     *^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*^*~*
>>>>>>
>>>>>>
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>>>>>>
>>>>>>
>>>>>>
>>>>>> -- 
>>>>>> Stephane Raynaud
>>>>>> ------------------------------------------------------------------------
>>>>>>
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>>>>>>   
>>>>>>     
>>>>>>       
>>>>>>         
>>>>>>           
>>>>> _______________________________________________
>>>>> Numpy-discussion mailing list
>>>>> Numpy-discussion@scipy.org
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>>>>>
>>>>>   
>>>>>     
>>>>>       
>>>>>         
>>>> Please look the various NumPy functions to ignore NaN like nansum(). See 
>>>> the NumPy example list 
>>>> (http://  www.  scipy.org/Numpy_Example_List_With_Doc) for examples under 
>>>> nan or individual functions.
>>>>
>>>> To get the mean you can do something like:
>>>>
>>>> import numpy
>>>> x = numpy.array([2, numpy.nan, 1])
>>>> numpy.nansum(x)/(x.shape[0]-numpy.isnan(x).sum())
>>>> x_masked = numpy.ma.masked_where(numpy.isnan(x) , x)
>>>> x_masked.mean()
>>>>
>>>> The real advantage of masked arrays is that you have greater control 
>>>> over the filtering so you can also filter extreme values:
>>>>
>>>> y = numpy.array([2, numpy.nan, 1, 1000])
>>>> y_masked =numpy.ma.masked_where(numpy.isnan(y) , y)
>>>> y_masked =numpy.ma.masked_where(y_masked > 100 , y_masked)
>>>> y_masked.mean()
>>>>
>>>> Regards
>>>> Bruce
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>>>>
>>>>
>>>>   
>>>>     
>>>>       
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>>>
>>>   
>>>     
>> You mean like doing:
>>
>> import numpy
>> y=numpy.ma.MaskedArray([ 2., numpy.nan, 1., 1000.], numpy.isnan(y))
>>
>> ?
>>
>> Bruce
>>
>>
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>>
>>   
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