[SciPy-User] Re[SciPy-user] binning to polar coordinates
arsbbr
arsbbr@gmx....
Fri Aug 31 02:36:21 CDT 2012
Dear Jerome,
thank you for your tip! Unfortunately I'm getting big gaps for my real data
in the small rho region. The optimal algorithm should interpolate in the
small rho region and bin in the high rho region.
I guess your split pixel solution would be best. But this a whole API with a
lot of corrections and methods for real experimental data. Is it possible to
use your sophisticated algorithm only for the transformation from Cartesian
to a evenly stepped polar grid? Speed is not the issue with my data, so a
numpy solution would be fine.
Thanks a lot!
Jerome Kieffer wrote:
>
> On Tue, 28 Aug 2012 03:33:11 -0700 (PDT)
> arsbbr <arsbbr@gmx.net> wrote:
>
>>
>> Hi,
>>
>> I'm trying to rebin some uniform gridded data to polar coordinates. The
>> original data (Z) is noisy but on a fine grid.
>>
>> ### snip
>>
>> # define grid in cartesian coordinates
>> x = arange(-500, 500)
>> y = arange(-500, 500)
>> X, Y = meshgrid(x, y)
>> Z = X**2 + Y**2
>> Z += uniform(-5000, 5000, size=Z.shape))
>>
>> # transform to polar coordinates
>>
>> def cart2pol(x, y):
>> theta = arctan2(y, x)
>> rho = sqrt(x**2 + y**2)
>> return (theta, rho)
>>
>> THETA_xy, RHO_xy = cart2pol(X, Y)
>>
>> # define new polar grid
>> theta = arange(-pi, pi, 0.1)
>> rho = arange(0.1 ,500, 2.0)
>>
>> THETA, RHO = meshgrid(theta, rho)
>>
>> # Tried to use griddata, but it gives of course very jumpy results,
>> because
>> it
>> # does not take the average over the pixels in a patch (dTHETA, dRHO) but
>> an
>> interpolation at the
>> # exact new point.
>>
>> interpolate.griddata((THETA_xy.ravel(), RHO_xy.ravel()),
>> Z.ravel(), (THETA, RHO), method='linear')
>>
>> ### snip
>>
>> Is there a method that rebins the data to a new grid, possibly taking the
>> average
>> of the Z-values inside a patch (dTHETA, dRHO)?
>
> Try to use 2D-weighted histogram (divided by unweighted histograms).
> This unfortunately does dot split pixels over various bins so it works
> badly when the number of output bins is large.
>
> w=histogram2d(THETA_xy.ravel(),RHO_xy.ravel(),(int(2*pi/0.1),500//2),weights=Z.ravel())
> u=histogram2d(THETA_xy.ravel(),RHO_xy.ravel(),(int(2*pi/0.1),500//2))
> imshow (w[0]/u[0])
>
> If you want to split pixels, I am working on more sophisticated algorithm
> to do that in
> https://github.com/kif/pyFAI Tell me if you are interested.
>
> Cheers,
> --
> Jérôme Kieffer
> On-Line Data analysis / Software Group
> ISDD / ESRF
> tel +33 476 882 445
> _______________________________________________
> SciPy-User mailing list
> SciPy-User@scipy.org
> http://mail.scipy.org/mailman/listinfo/scipy-user
>
>
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