[Numpy-discussion] slicing / indexing question
Timothy W. Hilton
hilton@meteo.psu....
Thu Sep 23 11:53:54 CDT 2010
Hi Brett, Josef, Gökhan, and Anne,
Many thanks for the suggestions! This indexing problem was not as
straight-forward as I had anticipated. I have it working now, thanks
to your help.
Gökhan, I'm driving an ecosystem--atmosphere carbon dioxide flux model
with MODIS measurements, so this_par and pars are parameter values
that I need to combine with the measurements. The arrays represent
time series of model parameters and MODIS reflectances and land
surface products (enhanced vegetation index, land cover class,
vegetation dynamics) at 1 km resolution for 1200 km by 1200 km
"tiles".
There are lots of IDL and Matlab folks in my department too. I've been
using R and, more recently, Scipy... Being able to test out code
offline on my own machine without worrying about licenses (not only cost
but also talking to license servers, availability of site licenses,
etc.) is a big big help!
Along those lines, many thanks also to all who have put so much time
and energy into developing python and Scipy. It's a great platform!
-Tim
On Wed, Sep 2010, 22 at 02:22:50PM -0400, Anne Archibald wrote:
> On 21 September 2010 19:20, Timothy W. Hilton <hilton@meteo.psu.edu> wrote:
>
> > I have an 80x1200x1200 nd.array of floats this_par. I have a
> > 1200x1200 boolean array idx, and an 80-element float array pars. For
> > each element of idx that is True, I wish to replace the corresponding
> > 80x1x1 slice of this_par with the elements of pars.
> >
> > I've tried lots of variations on the theme of
> >>>>this_par[idx[np.newaxis, ...]] = pars[:, np.newaxis, np.newaxis]
> > but so far, no dice.
>
> How about this?
>
>
> In [1]: A = np.zeros((2,3,5))
>
> In [2]: B = np.array([1,2])
>
> In [3]: C = np.zeros((3,5), dtype=np.bool)
>
> In [4]: C[1,1] = True
>
> In [5]: C[2,3] = True
>
> In [6]: C
> Out[6]:
> array([[False, False, False, False, False],
> [False, True, False, False, False],
> [False, False, False, True, False]], dtype=bool)
>
> In [7]: A[:,C] = B[:,np.newaxis]
>
> In [8]: A
> Out[8]:
> array([[[ 0., 0., 0., 0., 0.],
> [ 0., 1., 0., 0., 0.],
> [ 0., 0., 0., 1., 0.]],
>
> [[ 0., 0., 0., 0., 0.],
> [ 0., 2., 0., 0., 0.],
> [ 0., 0., 0., 2., 0.]]])
>
> The key is that indexing with C replaces the two axes C is indexing
> with only one; boolean indexing necessarily flattens the relevant
> axes. You can check this with (e.g.) A[:,C].shape.
>
> Be careful with these "mixed" indexing modes (partly fancy indexing,
> partly slicing) as they can sometimes seem to reorder your axes for
> you:
>
> In [16]: np.zeros((2,3,7))[:,np.ones(5,dtype=int),np.ones(5,dtype=int)].shape
> Out[16]: (2, 5)
>
> In [17]: np.zeros((2,3,7))[np.ones(5,dtype=int),:,np.ones(5,dtype=int)].shape
> Out[17]: (5, 3)
>
> In [18]: np.zeros((2,3,7))[np.ones(5,dtype=int),np.ones(5,dtype=int),:].shape
> Out[18]: (5, 7)
>
> Anne
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