我需要求两个矩形的交集和并集的面积，以及交集面积与并集面积的比率。做这个的最好方式是什么？ Picture
def intersection_over_union(a, b):
ax1 = a['left']
ay1 = a['top']
aw = a['width']
ah = a['height']
ax2 = ax1 + aw
ay2 = ay1 + ah
bx1 = b['left']
by1 = b['top']
bw = b['width']
bh = b['height']
bx2 = bx1 + bw
by2 = by1 + bh
delta_x = max(ax1, bx1) - min(ax2, bx2)
delta_y = max(ay1, by1) - min(ay2, by2)
overlap = delta_x * delta_y
union = aw * ah + bw * bh - overlap
print(overlap, union)
return overlap / union
最佳答案
假设您正在处理 AABB(轴对齐边界框)，这个小类定义了您需要的一切:
class Rect:
def __init__(self, x, y, w, h):
self.x = x
self.y = y
self.w = w
self.h = h
def bottom(self):
return self.y + self.h
def right(self):
return self.x + self.w
def area(self):
return self.w * self.h
def union(self, b):
posX = min(self.x, b.x)
posY = min(self.y, b.y)
return Rect(posX, posY, max(self.right(), b.right()) - posX, max(self.bottom(), b.bottom()) - posY)
def intersection(self, b):
posX = max(self.x, b.x)
posY = max(self.y, b.y)
candidate = Rect(posX, posY, min(self.right(), b.right()) - posX, min(self.bottom(), b.bottom()) - posY)
if candidate.w > 0 and candidate.h > 0:
return candidate
return Rect(0, 0, 0, 0)
def ratio(self, b):
return self.intersection(b).area() / self.union(b).area()
if __name__ == "__main__":
assert Rect(1, 1, 1, 1).ratio(Rect(1, 1, 1, 1)) == 1
assert round(Rect(1, 1, 2, 2).ratio(Rect(2, 2, 2, 2)), 4) == 0.1111
关于python - 求两个矩形的交集和并集，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/68180297/