Recently read a lot of bloggers on the Internet to write the iou implementation method, but Giou's code seems to be relatively small, so they wrote a new hand, if there is any error, please correct me, words do not say, on the code:
def Iou(rec1,rec2):
x1,x2,y1,y2 = rec1 # are the coordinates of the first rectangle's left, right, top, and bottom.
x3,x4,y3,y4 = rec2 # are the coordinates of the second rectangle's left, right, top and bottom, respectively
area_1 = (x2-x1)*(y1-y2)
area_2 = (x4-x3)*(y3-y4)
sum_area = area_1 + area_2
w1 = x2 - x1# The width of the first rectangle
w2 = x4 - x3# The width of the second rectangle
h1 = y1 - y2
h2 = y3 - y4
W = min(x1,x2,x3,x4)+w1+w2-max(x1,x2,x3,x4)# Width of cross section
H = min(y1,y2,y3,y4)+h1+h2-max(y1,y2,y3,y4)# Cross section high
Area = W*H# Area of intersection
Iou = Area/(sum_area-Area)
return Iou
def Giou(rec1,rec2):
x1,x2,y1,y2 = rec1 # are the coordinates of the first rectangle's left, right, top, and bottom.
x3,x4,y3,y4 = rec2
iou = Iou(rec1,rec2)
area_C = (max(x1,x2,x3,x4)-min(x1,x2,x3,x4))*(max(y1,y2,y3,y4)-min(y1,y2,y3,y4))
area_1 = (x2-x1)*(y1-y2)
area_2 = (x4-x3)*(y3-y4)
sum_area = area_1 + area_2
w1 = x2 - x1# The width of the first rectangle
w2 = x4 - x3# The width of the second rectangle
h1 = y1 - y2
h2 = y3 - y4
W = min(x1,x2,x3,x4)+w1+w2-max(x1,x2,x3,x4)# Width of cross section
H = min(y1,y2,y3,y4)+h1+h2-max(y1,y2,y3,y4)# Cross section high
Area = W*H# Area of intersection
add_area = sum_area - Area # Area of the concatenation of two rectangles
end_area = (area_C - add_area)/area_C # Area of (c/(AUB))/c
giou = iou - end_area
return giou
rec1 = (27,47,130,90)
rec2 = (30,68,150,110)
iou = Iou(rec1,rec2)
giou = Giou(rec1,rec2)
print("Iou = {},Giou = {}".format(iou,giou))
This python implementation of Iou and Giou code above is all I have shared with you.