These two checks are relatively simple however whether they are necessary is up to you and how you apply this test. There are some cases where a boundary point would not be considered for inclusion. Either way now you have an option. This function could even be modified to optionally check for boundary points.
# Improved point in polygon test which includes edge # and vertex points def point_in_poly(x,y,poly): # check if point is a vertex if (x,y) in poly: return "IN" # check if point is on a boundary for i in range(len(poly)): p1 = None p2 = None if i==0: p1 = poly p2 = poly else: p1 = poly[i-1] p2 = poly[i] if p1 == p2 and p1 == y and x > min(p1, p2) and x < max(p1, p2): return "IN" n = len(poly) inside = False p1x,p1y = poly for i in range(n+1): p2x,p2y = poly[i % n] if y > min(p1y,p2y): if y <= max(p1y,p2y): if x <= max(p1x,p2x): if p1y != p2y: xints = (y-p1y)*(p2x-p1x)/(p2y-p1y)+p1x if p1x == p2x or x <= xints: inside = not inside p1x,p1y = p2x,p2y if inside: return "IN" else: return "OUT" # Test a vertex for inclusion poligono = [(-33.416032,-70.593016), (-33.415370,-70.589604), (-33.417340,-70.589046), (-33.417949,-70.592351), (-33.416032,-70.593016)] lat= -33.416032 lon= -70.593016 print point_in_poly(lat, lon, poligono) # test a boundary point for inclusion poly2 = [(1,1), (5,1), (5,5), (1,5), (1,1)] x = 3 y = 1 print point_in_poly(x, y, poly2)You can download this script here.