[Numpy-discussion] Medians that ignore values

[Charles R Harris]

On Thu, Sep 18, 2008 at 11:31 AM, Peter Saffrey <pzs@dcs.gla.ac.uk> wrote:

> Pierre GM <pgmdevlist <at> gmail.com> writes:
>
> > Mmh, typo?
> >
>
> Yes, apologies. I was aiming for thorough, but ended up just careless. It's
> been
> a long day.
>
> > Ohoh. What version of numpy are you using ?
>
> The version in the Ubuntu package repository. It says 1:1.0.4-6ubuntu3.
>
> > if you don't give an axis
> > parameter, you should get the median of the flattened array, therefore a
> > scalar, not an array.
>
> Not for my version.
>
> >>> a = rand(10,3)
> >>> a
> array([[ 0.1269796 ,  0.43003978,  0.4700416 ],
>       [ 0.28867077,  0.85265318,  0.35908364],
>       [ 0.72967127,  0.41856028,  0.54724918],
>       [ 0.28821876,  0.69684144,  0.54647616],
>       [ 0.09592476,  0.83704808,  0.52425368],
>       [ 0.743552  ,  0.44433314,  0.7362179 ],
>       [ 0.4283931 ,  0.13305385,  0.68422292],
>       [ 0.68860674,  0.15057373,  0.99206493],
>       [ 0.31846329,  0.77237046,  0.986883  ],
>       [ 0.4578616 ,  0.4580833 ,  0.97754176]])
> >>> median(a.T)
> array([ 0.43003978,  0.35908364,  0.54724918,  0.54647616,  0.52425368,
>        0.7362179 ,  0.4283931 ,  0.68860674,  0.77237046,  0.4580833 ])
>
> > Anyway: you should use ma.median for masked arrays. Else, you're just
> keeping
> > the NaNs where they were.
> >
>
> That will be the problem. My version does not have median or mean methods
> for
> masked arrays, only the average() method.
>
> According to this page:
>
> http://www.scipy.org/Download
>
> 1.1.0 is the latest release. Do I need to use an SVN build to get the
> ma.median
> functionality?
>

1.1.1 is the latest release and 1.2 is coming out shortly.

Chuck
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