[Numpy-discussion] [Cdat-discussion] Arrays containing NaNs
Bruce Southey
bsouthey@gmail....
Fri Jul 25 11:04:21 CDT 2008
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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>>>
>>>
>> 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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