1. GIoU 计算

关于 GIoU 的实现，可直接参看原文给出的网站：
https://giou.stanford.edu/

$$GIoU=\frac{|A\cap B|}{|A\cup B|}-\frac{|C\setminus (A\cup B)|}{|C|}=IoU-\frac{|C\setminus (A\cup B)|}{|C|}$$

而 $Los{s}_{GIoU}$ 则：

$$Los{s}_{GIoU}=1-GIoU$$

而 $A$ 和 $B$ 分别是预测框和 GT 框. C 是包含 A 和 B 的最小凸包矩形.

2. IoU 计算

先来看下 IoU 的计算方式，摘自 PaddleDetection ：

def bbox_overlap(self, box1, box2, eps=1e-10):
    """calculate the iou of box1 and box2
    Args:
        box1 (Tensor): box1 with the shape (..., 4)
        box2 (Tensor): box1 with the shape (..., 4)
        eps (float): epsilon to avoid divide by zero
    Return:
        iou (Tensor): iou of box1 and box2
        overlap (Tensor): overlap of box1 and box2
        union (Tensor): union of box1 and box2
    """
    x1, y1, x2, y2 = box1
    x1g, y1g, x2g, y2g = box2

    xkis1 = paddle.maximum(x1, x1g)
    ykis1 = paddle.maximum(y1, y1g)
    xkis2 = paddle.minimum(x2, x2g)
    ykis2 = paddle.minimum(y2, y2g)
    w_inter = (xkis2 - xkis1).clip(0)
    h_inter = (ykis2 - ykis1).clip(0)
    overlap = w_inter * h_inter

    area1 = (x2 - x1) * (y2 - y1)
    area2 = (x2g - x1g) * (y2g - y1g)
    union = area1 + area2 - overlap + eps
    iou = overlap / union

    return iou, overlap, union

以上的 box1 和 box2 都是 xyxy 的列表，也就是 box* = [x1, y1, x2, y2], 而 x1 等列表元素为:

x1.shape 为 [..., 1]

可以简单考虑以下4种情况:

xkis1 = paddle.maximum(x1, x1g)
ykis1 = paddle.maximum(y1, y1g)
xkis2 = paddle.minimum(x2, x2g)
ykis2 = paddle.minimum(y2, y2g)

左上，左下，右下三种情况都会取得交集 box 正确的结果，[xkis1, ykis1, xkis2, ykis2] 是交集的左上角和右下角

而右上的图，得到的交集，左上角要比右下角大了，是错误的，所以下一步用来修正这个问题

w_inter = (xkis2 - xkis1).clip(0)
h_inter = (ykis2 - ykis1).clip(0)
overlap = w_inter * h_inter

如果，存在类似右上图的那种没有交集的情况，则 clip 为0，于是交集 overlap 为0

交集计算完毕，并集就好计算了，二者的面积之和减去交集，就是并集:

area1 = (x2- x1) * (y2 - y1)
area2 = (x2g - x1g) * (y2g - y1g)
union = area1 + area2 - overlap + eps

eps 防止潜在的除零错误，之后计算 IoU，并返回

iou = overlap / union
return iou, overlap, union

通过该函数，获取到 IoU, 交集 和 并集

3. GIoU loss

来看看 PaddleDetection GIoU Loss 部分

class GIoULoss(object):
    """
    Generalized Intersection over Union, see https://arxiv.org/abs/1902.09630
    Args:
        loss_weight (float): giou loss weight, default as 1
        eps (float): epsilon to avoid divide by zero, default as 1e-10
        reduction (string): Options are "none", "mean" and "sum". default as none
    """

    def __init__(self, loss_weight=1., eps=1e-10, reduction='none'):
        self.loss_weight = loss_weight
        self.eps = eps
        assert reduction in ('none', 'mean', 'sum')
        self.reduction = reduction

    def bbox_overlap(self, box1, box2, eps=1e-10):
        """calculate the iou of box1 and box2
        Args:
            box1 (Tensor): box1 with the shape (..., 4)
            box2 (Tensor): box1 with the shape (..., 4)
            eps (float): epsilon to avoid divide by zero
        Return:
            iou (Tensor): iou of box1 and box2
            overlap (Tensor): overlap of box1 and box2
            union (Tensor): union of box1 and box2
        """
        x1, y1, x2, y2 = box1
        x1g, y1g, x2g, y2g = box2

        xkis1 = paddle.maximum(x1, x1g)
        ykis1 = paddle.maximum(y1, y1g)
        xkis2 = paddle.minimum(x2, x2g)
        ykis2 = paddle.minimum(y2, y2g)
        w_inter = (xkis2 - xkis1).clip(0)
        h_inter = (ykis2 - ykis1).clip(0)
        overlap = w_inter * h_inter

        area1 = (x2 - x1) * (y2 - y1)
        area2 = (x2g - x1g) * (y2g - y1g)
        union = area1 + area2 - overlap + eps
        iou = overlap / union

        return iou, overlap, union

    def __call__(self, pbox, gbox, iou_weight=1., loc_reweight=None):
        x1, y1, x2, y2 = paddle.split(pbox, num_or_sections=4, axis=-1)
        x1g, y1g, x2g, y2g = paddle.split(gbox, num_or_sections=4, axis=-1)
        box1 = [x1, y1, x2, y2]
        box2 = [x1g, y1g, x2g, y2g]
        iou, overlap, union = self.bbox_overlap(box1, box2, self.eps)
        xc1 = paddle.minimum(x1, x1g)
        yc1 = paddle.minimum(y1, y1g)
        xc2 = paddle.maximum(x2, x2g)
        yc2 = paddle.maximum(y2, y2g)

        area_c = (xc2 - xc1) * (yc2 - yc1) + self.eps
        miou = iou - ((area_c - union) / area_c)
        if loc_reweight is not None:
            loc_reweight = paddle.reshape(loc_reweight, shape=(-1, 1))
            loc_thresh = 0.9
            giou = 1 - (1 - loc_thresh
                        ) * miou - loc_thresh * miou * loc_reweight
        else:
            giou = 1 - miou
        if self.reduction == 'none':
            loss = giou
        elif self.reduction == 'sum':
            loss = paddle.sum(giou * iou_weight)
        else:
            loss = paddle.mean(giou * iou_weight)
        return loss * self.loss_weight

bbox_overlap 部分，刚刚已经看过，接下来看看 __call__ 部分

接下来，将 bbox 的4维坐标向量分离，传入 self.bbox_overlap 中计算 IoU, 交集 和 并集

x1, y1, x2, y2 = paddle.split(pbox, num_or_sections=4, axis=-1)
x1g, y1g, x2g, y2g = paddle.split(gbox, num_or_sections=4, axis=-1)
box1 = [x1, y1, x2, y2]
box2 = [x1g, y1g, x2g, y2g]
iou, overlap, union = self.bbox_overlap(box1, box2, self.eps)

接下来计算包含预测框和GT框的最小矩形，也就是公式中的 $C$

xc1 = paddle.minimum(x1, x1g)
yc1 = paddle.minimum(y1, y1g)
xc2 = paddle.maximum(x2, x2g)
yc2 = paddle.maximum(y2, y2g)

接下来计算 $C$ 框的面积，以及计算 GIoU 即可

area_c = (xc2 - xc1) * (yc2 - yc1) + self.eps
miou = iou - ((area_c - union) / area_c)

最后计算 loss 以及 reduce 操作

giou = 1 - miou

if self.reduction == 'none':
    loss = giou
elif self.reduction == 'sum':
    loss = paddle.sum(giou * iou_weight)
else:
    loss = paddle.mean(giou * iou_weight)

最后乘以 loss 的权重

return loss * self.loss_weight

最后可能有个 squeeze 的操作，比如我的操作最后返回的 shape 是 [200, 21, 1]

也就是预测框有200个，GT框有21个，比如以下代码中最后的 squeeze 操作

# Compute the giou cost betwen boxes
cost_giou = self.giou_loss(
    bbox_cxcywh_to_xyxy(out_bbox.unsqueeze(1)),
    bbox_cxcywh_to_xyxy(tgt_bbox.unsqueeze(0))).squeeze(-1)

债说一下这里，此处用于给loss做 reweight 操作，我暂时没用到

if loc_reweight is not None:
    loc_reweight = paddle.reshape(loc_reweight, shape=(-1, 1))
    loc_thresh = 0.9
    giou = 1 - (1 - loc_thresh
                ) * miou - loc_thresh * miou * loc_reweight