边界框回归(Bounding Box Regression,BBR)在目标检测和实例分割中被广泛应用,是定位目标的重要步骤。然而,大多数现有的边界框回归损失函数在预测框与实际标注框具有相同的宽高比但宽度和高度值完全不同的情况下无法进行优化。为了解决上述问题,作者充分探索了水平矩形的几何特征,提出了一种基于最小点距离的边界框相似度比较度量——MPDIoU,其中包含了现有损失函数中考虑的所有相关因素,例如重叠或非重叠面积、中心点距离以及宽度和高度的偏差,同时简化了计算过程。在此基础上,作者提出了一种基于MPDIoU的边界框回归损失函数。
关于MPDIoU:
MPDIoU: A Loss for Efficient and Accurate Bounding BoxRegression--论文学习笔记_athrunsunny的博客-CSDN博客
指标如下
根据下图的公式,以及原理图进行复现,如有问题还请大佬不吝赐教,相互交流
在metrics.py中修改bbox_iou函数
def bbox_iou(box1, box2, xywh=True, GIoU=False, DIoU=False, CIoU=False, MDPIoU=False, feat_h=640, feat_w=640,
eps=1e-7):
# Returns Intersection over Union (IoU) of box1(1,4) to box2(n,4)
# Get the coordinates of bounding boxes
if xywh: # transform from xywh to xyxy
(x1, y1, w1, h1), (x2, y2, w2, h2) = box1.chunk(4, 1), box2.chunk(4, 1)
w1_, h1_, w2_, h2_ = w1 / 2, h1 / 2, w2 / 2, h2 / 2
b1_x1, b1_x2, b1_y1, b1_y2 = x1 - w1_, x1 + w1_, y1 - h1_, y1 + h1_
b2_x1, b2_x2, b2_y1, b2_y2 = x2 - w2_, x2 + w2_, y2 - h2_, y2 + h2_
else: # x1, y1, x2, y2 = box1
b1_x1, b1_y1, b1_x2, b1_y2 = box1.chunk(4, 1)
b2_x1, b2_y1, b2_x2, b2_y2 = box2.chunk(4, 1)
w1, h1 = b1_x2 - b1_x1, b1_y2 - b1_y1
w2, h2 = b2_x2 - b2_x1, b2_y2 - b2_y1
# Intersection area
inter = (torch.min(b1_x2, b2_x2) - torch.max(b1_x1, b2_x1)).clamp(0) * \
(torch.min(b1_y2, b2_y2) - torch.max(b1_y1, b2_y1)).clamp(0)
# Union Area
union = w1 * h1 + w2 * h2 - inter + eps
# IoU
iou = inter / union
if CIoU or DIoU or GIoU:
cw = torch.max(b1_x2, b2_x2) - torch.min(b1_x1, b2_x1) # convex (smallest enclosing box) width
ch = torch.max(b1_y2, b2_y2) - torch.min(b1_y1, b2_y1) # convex height
if CIoU or DIoU: # Distance or Complete IoU https://arxiv.org/abs/1911.08287v1
c2 = cw ** 2 + ch ** 2 + eps # convex diagonal squared
rho2 = ((b2_x1 + b2_x2 - b1_x1 - b1_x2) ** 2 + (b2_y1 + b2_y2 - b1_y1 - b1_y2) ** 2) / 4 # center dist ** 2
if CIoU: # https://github.com/Zzh-tju/DIoU-SSD-pytorch/blob/master/utils/box/box_utils.py#L47
v = (4 / math.pi ** 2) * torch.pow(torch.atan(w2 / (h2 + eps)) - torch.atan(w1 / (h1 + eps)), 2)
with torch.no_grad():
alpha = v / (v - iou + (1 + eps))
return iou - (rho2 / c2 + v * alpha) # CIoU
return iou - rho2 / c2 # DIoU
c_area = cw * ch + eps # convex area
return iou - (c_area - union) / c_area # GIoU https://arxiv.org/pdf/1902.09630.pdf
elif MDPIoU:
d1 = (b2_x1 - b1_x1) ** 2 + (b2_y1 - b1_y1) ** 2
d2 = (b2_x2 - b1_x2) ** 2 + (b2_y2 - b1_y2) ** 2
mpdiou_hw_pow = feat_h ** 2 + feat_w ** 2
return iou - d1 / mpdiou_hw_pow - d2 / mpdiou_hw_pow # MPDIoU
return iou # IoU
在loss.py中修改iou的计算方式
iou = bbox_iou(pbox, tbox[i], MDPIoU=True, feat_h=tobj.size()[2], feat_w=tobj.size()[3]).squeeze()