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?
Peter
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