5

I'm trying to draw polygons like this:

In [1]: canvas = numpy.zeros((12, 12), dtype=int)

In [2]: mahotas.polygon.fill_polygon(
   ...: [(1, 1), (1, 10), (10, 10), (10, 1)],
   ...: canvas)

In [3]: canvas
Out[3]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])

I'd expect the following output though:

array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])

Why are [(10,2):(10:10)] still zeros? Is there another way to draw a filled polygon to an array?

nmichaels
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Attila O.
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3 Answers3

2

It is a strange result. I found that if you reverse the order of the points it draws the complete figure. In other words:

# this is broken
pts = [(1, 1), (1, 10), (10, 10), (10, 1)]
# this works
pts = [(1, 1), (10, 1), (10, 10), (1, 10)]

Here is a test program:

import numpy
import mahotas.polygon

def run(n, reverse=0):
    canvas = numpy.zeros((n, n), dtype=int)
    lim = n-2
    print '\n%d x %d, lim=%d reverse=%d' % (n, n, lim, reverse)
    pts = [(1, 1), (1, lim), (lim, lim), (lim, 1), (1, 1)]
    if reverse:
        pts.reverse()
    mahotas.polygon.fill_polygon(pts, canvas)
    return canvas

for rev in (0, 1):
    for n in range(3, 14):
        print run(n, rev)

Examples:

6 x 6, lim=4 reverse=0
[[0 0 0 0 0 0]
 [0 1 0 0 1 0]
 [0 1 1 1 1 0]
 [0 1 1 1 1 0]
 [0 1 0 0 0 0]
 [0 0 0 0 0 0]]

6 x 6, lim=4 reverse=1
[[0 0 0 0 0 0]
 [0 1 1 1 1 0]
 [0 1 1 1 1 0]
 [0 1 1 1 1 0]
 [0 1 1 1 1 0]
 [0 0 0 0 0 0]]
samplebias
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  • Impressive! The only thing is, I don't know the order in advance. What I could possibly do is draw the polygon both ways, and then take the version that has more `1`s in it as the correct result. Thanks for the answer! – Attila O. Apr 08 '11 at 14:26
0

I think the reversal is required because numpy references array coordinates as y,x (row, column) where most other programs issue coordinates as x,y.

GeospatialPython.com
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0

If the polygons are always rectangles, then we only need two points and:

import numpy
canvas = numpy.zeros((12, 12), dtype=int)
points = [(1, 1), (1, 10), (10, 10), (10, 1)]
start_pt, end_pt = min(points), max(points)
canvas[start_pt[1]:end_pt[1]+1, start_pt[0]:end_pt[0]+1] = 1
Benjamin
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