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目标检测损失函数IoU、GIoU、DIoU和CIoU

ML/DL 迷途小书童 0评论

简介

在目标检测任务中,分类和检测框的回归是核心问题,损失函数的选择对模型的表现效果具有较大影响,本文介绍常用的损失函数IoUGIoUDIouCIoU

IoU

IoU是使用最广泛的检测框损失函数,大部分的检测算法都是使用的这个方法。IoU也就是交并比(Intersection over Union),预测框和真实框相交区域面积和合并区域面积的比值,计算公式如下

IoU

IoU的代码实现

import numpy as np

def Iou(box1, box2, wh=False):
    if wh == False:
        xmin1, ymin1, xmax1, ymax1 = box1
        xmin2, ymin2, xmax2, ymax2 = box2
    else:
        xmin1, ymin1 = int(box1[0]-box1[2]/2.0), int(box1[1]-box1[3]/2.0)
        xmax1, ymax1 = int(box1[0]+box1[2]/2.0), int(box1[1]+box1[3]/2.0)
        xmin2, ymin2 = int(box2[0]-box2[2]/2.0), int(box2[1]-box2[3]/2.0)
        xmax2, ymax2 = int(box2[0]+box2[2]/2.0), int(box2[1]+box2[3]/2.0)

    # 获取矩形框交集对应的左上角和右下角的坐标
    xx1 = np.max([xmin1, xmin2])
    yy1 = np.max([ymin1, ymin2])
    xx2 = np.min([xmax1, xmax2])
    yy2 = np.min([ymax1, ymax2])    

    # 计算两个矩形框面积
    area1 = (xmax1-xmin1) * (ymax1-ymin1) 
    area2 = (xmax2-xmin2) * (ymax2-ymin2)

    # 计算交集面积
    inter_area = (np.max([0, xx2-xx1])) * (np.max([0, yy2-yy1])) 

    # 计算交并比
    iou = inter_area / (area1+area2-inter_area+1e-6)  

    return iou

IoU的缺点:

如果两个框没有重叠,那么IoU等于0,这时候就没有梯度的回传,就无法进行学习训练

GIoU

2019年CVPRGIoU在论文Generalized Intersection over Union: A Metric and A Loss for Bounding Box Regression中被提出

上述公式的含义:先计算两个框的最小闭包区域面积 (也同时包含了预测框和真实框的最小框的面积),再计算闭包区域中不属于两个框的区域占闭包区域的比重,最后用IoU减去这个比重就得到GIoU

GIoU的代码实现

def Giou(rec1,rec2):
    # 分别是第一个矩形左右上下的坐标
    x1,x2,y1,y2 = rec1 
    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

    # 第一个矩形的宽
    w1 = x2 - x1

    # 第二个矩形的宽
    w2 = x4 - x3   
    h1 = y1 - y2
    h2 = y3 - y4
    # 交叉部分的宽
    W = min(x1,x2,x3,x4)+w1+w2-max(x1,x2,x3,x4) 
    # 交叉部分的高  
    H = min(y1,y2,y3,y4)+h1+h2-max(y1,y2,y3,y4)  
    # 交叉的面积  
    Area = W*H    
    # 两矩形并集的面积
    add_area = sum_area - Area  

    # 闭包区域中不属于两个框的区域占闭包区域的比重
    end_area = (area_C - add_area)/area_C    
    giou = iou - end_area
    return giou

GIoU考虑到了当检测框和真实框没有出现重叠的情况,但是当检测框和真实框之间出现包含的现象的时候GIoU就和IoU是同样的效果了。

DIoU

DIoUDistance-IoU,计算公式如下

其中, b 分别代表了预测框和真实框的中心点,且 代表的是计算两个中心点间的欧式距离。 c代表的是能够同时包含预测框和真实框的最小闭包区域的对角线距离。

DIoU考虑到GIoU的缺点,也是增加了两个框的最小闭包,将真实框和预测框都包含了进来,但是DIoU计算的不是框之间的交并,而是计算每个检测框之间的欧氏距离,这样就可以解决GIoU包含时出现的问题,同时,DIoU最小化预测框间的中心点距离,从而能够达到快速收敛。

DIoU代码实现

def Diou(bboxes1, bboxes2):
    rows = bboxes1.shape[0]
    cols = bboxes2.shape[0]
    dious = torch.zeros((rows, cols))
    if rows * cols == 0:#
        return dious
    exchange = False
    if bboxes1.shape[0] > bboxes2.shape[0]:
        bboxes1, bboxes2 = bboxes2, bboxes1
        dious = torch.zeros((cols, rows))
        exchange = True

    # #xmin,ymin,xmax,ymax->[:,0],[:,1],[:,2],[:,3]
    w1 = bboxes1[:, 2] - bboxes1[:, 0]
    h1 = bboxes1[:, 3] - bboxes1[:, 1] 
    w2 = bboxes2[:, 2] - bboxes2[:, 0]
    h2 = bboxes2[:, 3] - bboxes2[:, 1]

    area1 = w1 * h1
    area2 = w2 * h2

    center_x1 = (bboxes1[:, 2] + bboxes1[:, 0]) / 2 
    center_y1 = (bboxes1[:, 3] + bboxes1[:, 1]) / 2 
    center_x2 = (bboxes2[:, 2] + bboxes2[:, 0]) / 2
    center_y2 = (bboxes2[:, 3] + bboxes2[:, 1]) / 2

    inter_max_xy = torch.min(bboxes1[:, 2:],bboxes2[:, 2:]) 
    inter_min_xy = torch.max(bboxes1[:, :2],bboxes2[:, :2]) 
    out_max_xy = torch.max(bboxes1[:, 2:],bboxes2[:, 2:]) 
    out_min_xy = torch.min(bboxes1[:, :2],bboxes2[:, :2])

    inter = torch.clamp((inter_max_xy - inter_min_xy), min=0)
    inter_area = inter[:, 0] * inter[:, 1]
    inter_diag = (center_x2 - center_x1)**2 + (center_y2 - center_y1)**2
    outer = torch.clamp((out_max_xy - out_min_xy), min=0)
    outer_diag = (outer[:, 0] ** 2) + (outer[:, 1] ** 2)
    union = area1+area2-inter_area
    dious = inter_area / union - (inter_diag) / outer_diag
    dious = torch.clamp(dious,min=-1.0,max = 1.0)
    if exchange:
        dious = dious.T
    return dious

CIoU

CIoUComplete-IoU,就是在DIoU的基础上增加了长和宽的损失,能够进一步地快速收敛和提升性能。

其中, 是权重函数,而 用来度量长宽比的相似性

而完整的CIoU损失函数定位为

CIoU的梯度类似于DIoU,但还要考虑 的梯度,在长宽在[0, 1]的情况下,会导致梯度爆炸。

CIoU代码实现

def bbox_overlaps_ciou(bboxes1, bboxes2):
    rows = bboxes1.shape[0]
    cols = bboxes2.shape[0]
    cious = torch.zeros((rows, cols))
    if rows * cols == 0:
        return cious
    exchange = False
    if bboxes1.shape[0] > bboxes2.shape[0]:
        bboxes1, bboxes2 = bboxes2, bboxes1
        cious = torch.zeros((cols, rows))
        exchange = True

    w1 = bboxes1[:, 2] - bboxes1[:, 0]
    h1 = bboxes1[:, 3] - bboxes1[:, 1]
    w2 = bboxes2[:, 2] - bboxes2[:, 0]
    h2 = bboxes2[:, 3] - bboxes2[:, 1]

    area1 = w1 * h1
    area2 = w2 * h2

    center_x1 = (bboxes1[:, 2] + bboxes1[:, 0]) / 2
    center_y1 = (bboxes1[:, 3] + bboxes1[:, 1]) / 2
    center_x2 = (bboxes2[:, 2] + bboxes2[:, 0]) / 2
    center_y2 = (bboxes2[:, 3] + bboxes2[:, 1]) / 2

    inter_max_xy = torch.min(bboxes1[:, 2:],bboxes2[:, 2:])
    inter_min_xy = torch.max(bboxes1[:, :2],bboxes2[:, :2])
    out_max_xy = torch.max(bboxes1[:, 2:],bboxes2[:, 2:])
    out_min_xy = torch.min(bboxes1[:, :2],bboxes2[:, :2])

    inter = torch.clamp((inter_max_xy - inter_min_xy), min=0)
    inter_area = inter[:, 0] * inter[:, 1]
    inter_diag = (center_x2 - center_x1)**2 + (center_y2 - center_y1)**2
    outer = torch.clamp((out_max_xy - out_min_xy), min=0)
    outer_diag = (outer[:, 0] ** 2) + (outer[:, 1] ** 2)
    union = area1+area2-inter_area
    u = (inter_diag) / outer_diag
    iou = inter_area / union
    with torch.no_grad():
        arctan = torch.atan(w2 / h2) - torch.atan(w1 / h1)
        v = (4 / (math.pi ** 2)) * torch.pow((torch.atan(w2 / h2) - torch.atan(w1 / h1)), 2)
        S = 1 - iou
        alpha = v / (S + v)
        w_temp = 2 * w1
    ar = (8 / (math.pi ** 2)) * arctan * ((w1 - w_temp) * h1)
    cious = iou - (u + alpha * ar)
    cious = torch.clamp(cious,min=-1.0,max = 1.0)
    if exchange:
        cious = cious.T
    return cious

参考资料

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