matplotlib-devel Mailing List for matplotlib

On Sat, Jul 7, 2012 at 7:30 PM, Amit Aronovitch <aro...@gm...>wrote:
>
> The current implementation of Delaunay interpolator returns NaN for grids
> whose x or y has dimension 1 (i.e. when you try to interpolate along a
> horizontal/vertical line, or in a single point). See example below.
>
> By looking at the code, it seems that this can be fixed by simple
> rearrangement of calculations.
> Suggested patch provided here:
> https://github.com/AmitAronovitch/matplotlib/commit/f312d864da9c72681eb3db3b5920ae64793c713e(let me know if you want a pull request).
> The suggested implementation is almost identical. It might actually
> perform faster in some cases (there is one less multiplication op in the
> inner loop). There might be some differences in accuracy, but I believe
> they should only become observable in cases where the grid size is very
> large (which would probably cause memory problems anyway).
>
> Example (before suggested patch):
>
> >>> from matplotlib.delaunay import Triangulation
> >>> tri = Triangulation([0,10,10,0],[0,0,10,10])
> >>> lin = tri.linear_interpolator([1,10,5,2.0])
> >>> # 2x2 grid works fine
> >>> lin[3:6:2j,1:4:2j]
> array([[ 1.6, 3.1],
> [ 1.9, 2.8]])
> >>> # but not when 1x2, 2x1, 1x1:
> >>> lin[3:6:2j,1:1:1j]
> array([[ nan],
> [ nan]])
> >>> lin[3:3:1j,1:1:1j]
> array([[ nan]])
> >>>
>
> After suggested patch:
>
> >>> from matplotlib.delaunay import Triangulation
> >>> tri = Triangulation([0,10,10,0],[0,0,10,10])
> >>> lin = tri.linear_interpolator([1,10,5,2.0])
> >>> # 2x2 grid: same same
> >>> lin[3:6:2j,1:4:2j]
> array([[ 1.6, 3.1],
> [ 1.9, 2.8]])
> >>> # but these work now
> >>> lin[3:6:2j,1:1:1j]
> array([[ 1.6],
> [ 1.9]])
> >>> lin[3:3:1j,1:1:1j]
> array([[ 1.6]])
> >>>
>
>
I am always a fan of people who test and design their methods against edge
cases like these, so my hat is off to you. I would suggest putting
together a pull request so that we can properly test the potential impact
such a change could have.
Thanks!
Ben Root