Checking if a point is inside a polygon

I would suggest using the Path class from matplotlib

import matplotlib.path as mplPathimport numpy as nppoly = [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:

• A polygon with 5 corners also has 5 bounding lines, not 6, your loop is one off.
• Using a separate generator expression makes clear that you are checking each line in turn.
• Checking for an empty number of lines was added. However, how to treat zero-length lines and polygons with a single corner is still open.
• I'd also consider making the lines() function a normal member instead of a nested utility.
• Instead of the many nested if structures, you could also check for the inverse and then continue or use and.

Steps:

• Iterate over all the segments in the polygon
• Check whether they intersect with a ray going in the increasing-x direction

Using the intersect function from This SO Question

def ccw(A,B,C):    return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)# Return true if line segments AB and CD intersectdef intersect(A,B,C,D):    return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)def point_in_polygon(pt, poly, inf):    result = False    for i in range(len(poly.corners)-1):        if intersect((poly.corners[i].x, poly.corners[i].y), ( poly.corners[i+1].x, poly.corners[i+1].y), (pt.x, pt.y), (inf, pt.y)):            result = not result    if intersect((poly.corners[-1].x, poly.corners[-1].y), (poly.corners[0].x, poly.corners[0].y), (pt.x, pt.y), (inf, pt.y)):        result = not result    return result

Please note that the inf parameter should be the maximum point in the x axis in your figure.