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

Bruce Southey bsouthey@gmail....
Fri Jul 25 10:43:34 CDT 2008


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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>   
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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