On Tue, Jan 29, 2013 at 5:55 PM, Maria Liukis <liu...@usc.edu> wrote:
> Hello,
>
> I tested the following code on my Mac laptop and our production Linux
> server both running matplotlib V1.0.1. Both machines observe the same
> output from the code, so I was wondering if somebody is aware of the
> problem or if it's some undocumented feature of "pnpoly()" function from
> matplotlib.nxutils?
>
> I use matplotlib.nxutils.pnpoly() function from matplotlib to determine
> if point belongs to the polygon.
> The following code:
>
> >>> import numpy as np
> >>> import matplotlib.nxutils as nx
> >>> coords = np.array([[4.0, 1.0], [4.0, 4.0], [5.0, 5.0], [6.0, 4.0],
> [5.0, 0.0]])
>
> >>> nx.pnpoly(4.0, 1.0, coords)
> 1
> >>> nx.pnpoly(4.0, 4.0, coords)
> 1
> >>> nx.pnpoly(5.0, 5.0, coords)
> 0
> >>> nx.pnpoly(6.0, 4.0, coords)
> 0
> >>> nx.pnpoly(5.0, 0.0, coords)
> 0
>
> The question is why first two vertexes are considered to be inside of
> defined polygon, and last 3 vertexes are not? My guess, it's treating the
> polygon as a semi-open set, and I wonder if it can be changed to make all
> vertexes inclusive?
>
> Any help would be greatly appreciated.
>
> Many thanks,
> Masha
>
The documentation for pnpoly() for that version states:
"""
A point on the boundary may be treated as inside or outside.
See `pnpoly <
http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html>`_
"""
Note that in version 1.2.0, the nxutils module was deprecated. pnpoly()
and points_inside_poly() are now merely wrappers around the polygon's
implementations of point-testing, which differs from the nxutils'
implementation, so you may get slightly different results.
Do note that the point-testing algorithm in matplotlib was more geared for
visualization purposes rather than for strict geometric needs. If you need
a more well-behaved point-tester (and faster if the polygon is "prepared"),
use the shapely package instead.
I hope that clears things up!
Cheers!
Ben Root
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