I have a class describing a Point (has 2 coordinates x and y) and a class describing a Polygon which has a list of Points which correspond to corners (self.corners) I need to check if a Point is in a Polygon
Here is the function that is supposed to check if the Point is in the Polygon. I am using the Ray Casting Method
def in_me(self, point):
result = False
n = len(self.corners)
p1x = int(self.corners[0].x)
p1y = int(self.corners[0].y)
for i in range(n+1):
p2x = int(self.corners[i % n].x)
p2y = int(self.corners[i % n].y)
if point.y > min(p1y,p2y):
if point.x <= max(p1x,p2x):
if p1y != p2y:
xinters = (point.y-p1y)*(p2x-p1x)/(p2y-p1y)+p1x
print xinters
if p1x == p2x or point.x <= xinters:
result = not result
p1x,p1y = p2x,p2y
return result
I run a test with following shape and point:
PG1 = (0,0), (0,2), (2,2), (2,0)
point = (1,1)
The script happily returns False even though the point it within the line. I am unable to find the mistake
How to check if a point is inside a polygon in Python. To perform a Point in Polygon (PIP) query in Python, we can resort to the Shapely library's functions . within(), to check if a point is within a polygon, or . contains(), to check if a polygon contains a point.
Use polygon. contains(point) to test if point is inside ( True ) or outside ( False ) the polygon.
To find out if a point (x, y) is on the graph of a line, we plug in the values and see if we get a true statement, such as 10 = 10. If we get something different, like 6 = 4, we know that the point is not on the line because it does not satisfy the equation.
I would suggest using the Path
class from matplotlib
import matplotlib.path as mplPath
import numpy as np
poly = [190, 50, 500, 310]
bbPath = mplPath.Path(np.array([[poly[0], poly[1]],
[poly[1], poly[2]],
[poly[2], poly[3]],
[poly[3], poly[0]]]))
bbPath.contains_point((200, 100))
(There is also a contains_points
function if you want to test for multiple points)
I'd like to suggest some other changes there:
def contains(self, point):
if not self.corners:
return False
def lines():
p0 = self.corners[-1]
for p1 in self.corners:
yield p0, p1
p0 = p1
for p1, p2 in lines():
... # perform actual checks here
Notes:
continue
or use and
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