Using shapely.geometry.polygon to calculate the IOU of any two quadrilaterals

The source is Mr. Bai Xiang of Huazhong University of science and technology.

import numpy as np 
import shapely
from shapely.geometry import Polygon,MultiPoint  #Polygon
line1=[2,0,2,2,0,0,0,0,2] #One-dimensional array representation of the coordinates of the four points of the quadrilateral, [x,y,x,y....]
a = np.array(line1).reshape(4, 2) # quadrilateral two-dimensional coordinate representation
poly1 = Polygon(a).convex_hull # python quadrilateral object, will automatically calculate four points, the last four points in the order of: top left bottom right bottom right top left top
print(Polygon(a).convex_hull) # you can print to see if this is the case

b=np.array(line2).reshape(4, 2)
poly2 = Polygon(b).convex_hull
union_poly = np.concatenate((a,b))   #Merge two box coordinates to become 8*2
print(MultiPoint(union_poly).convex_hull) # contains the smallest polygon point of the two quadrilaterals
if not poly1.intersects(poly2): #If the two quadrilaterals do not intersect
    iou = 0
        inter_area = poly1.intersection(poly2).area #intersection area
        #union_area = poly1.area + poly2.area - inter_area
        union_area = MultiPoint(union_poly).convex_hull.area
        if union_area == 0:
            iou= 0
        #iou = float(inter_area)/(union_area-inter_area)  #wrong
        # iou=float(inter_area) /(poly1.area+poly2.area-inter_area)
        # The source code gives two ways to calculate IOU, the first one is: intersection part / area of the smallest polygon containing two quadrilaterals  
        # The second one: intersection/merge (common way to calculate IOU of rectangular box) 
    except shapely.geos.TopologicalError:
        print('shapely.geos.TopologicalError occured, iou set to 0')
        iou = 0

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