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

 
line2=[1,1,4,1,4,4,1,4]
b=np.array(line2).reshape(4, 2)
poly2 = Polygon(b).convex_hull
print(Polygon(b).convex_hull)
 
union_poly = np.concatenate((a,b))   #Merge two box coordinates to become 8*2
#print(union_poly)
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
else:
    try:
        inter_area = poly1.intersection(poly2).area #intersection area
        print(inter_area)
        #union_area = poly1.area + poly2.area - inter_area
        union_area = MultiPoint(union_poly).convex_hull.area
        print(union_area)
        if union_area == 0:
            iou= 0
        #iou = float(inter_area)/(union_area-inter_area)  #wrong
        iou=float(inter_area)/union_area
        # 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
 
print(a)
 
print(iou)