如何计算 pytorch 中关键点检测 CNN model 的准确度？

Question

有人可以帮我解决这个问题吗，

def train_net(n_epochs):
    valid_loss_min = np.Inf    
    history = {'train_loss': [], 'valid_loss': [], 'epoch': []}

    for epoch in range(n_epochs):  
        train_loss = 0.0
        valid_loss = 0.0  
        net.train()
        running_loss = 0.0
        for batch_i, data in enumerate(train_loader):
            images = data['image']
            key_pts = data['keypoints']
            key_pts = key_pts.view(key_pts.size(0), -1)
            key_pts = key_pts.type(torch.FloatTensor).to(device)
            images = images.type(torch.FloatTensor).to(device)
            output_pts = net(images)
            loss = criterion(output_pts, key_pts)
            optimizer.zero_grad()
            loss.backward()
            optimizer.step()
            train_loss += loss.item()*images.data.size(0)      
        net.eval() 

        with torch.no_grad():
            for batch_i, data in enumerate(test_loader):
                images = data['image']
                key_pts = data['keypoints']
                key_pts = key_pts.view(key_pts.size(0), -1)
                key_pts = key_pts.type(torch.FloatTensor).to(device)
                images = images.type(torch.FloatTensor).to(device)
                output_pts = net(images)
                loss = criterion(output_pts, key_pts)          
                valid_loss += loss.item()*images.data.size(0) 
        train_loss = train_loss/len(train_loader.dataset)
        valid_loss = valid_loss/len(test_loader.dataset) 
        print('Epoch: {} \tTraining Loss: {:.6f}'.format(epoch+1,train_loss,valid_loss))

        if valid_loss <= valid_loss_min:
            print('Validation loss decreased ({:.6f} --> {:.6f}).  Saving model ...'.format(valid_loss_min,valid_loss))    
            torch.save(net,f'X:\\xxxx\\xxx\\xxx\\epoch{epoch + 1}_loss{valid_loss}.pth')
            valid_loss_min = valid_loss
        history['epoch'].append(epoch + 1)
        history['train_loss'].append(train_loss)
        history['valid_loss'].append(valid_loss)
    print('Finished Training')
    return history
'''

Above is the training code for reference!

Answer 1

也许与欧几里得距离：真正的关键点：（x，y）预测的关键点：（x_，y_）距离d：sqrt（（x_ - x）^2 +（y_ - y）^2）。 从中你必须得到一个百分比。 如果 d == 0，则该关键点的准确度为 100%。 但什么是 0%？ 我会说从真正的关键点到离该关键点最远的图像角落的距离。 让我们称这个距离为 R。 因此，您的观点的准确性是 d / R。 对每个关键点都这样做并取平均值。 我只是想出了这个，所以它可能有一些缺陷，但我认为你可以使用它并检查它是否适合你。

Answer 2

这很有趣，我自己几分钟前刚刚在做这个，你可能已经意识到了。 简单地计算两组关键点之间的欧几里得距离并不能很好地推广到需要比较身体形状和大小的情况，所以我建议使用 Object 关键点相似度分数。 它测量由人的尺度归一化的身体关节距离。 如本博客所述，OKS 定义为：

这里（第 313 行 function computeOKS ）是 Facebook 研究的实现：

 def computeOks(self, imgId, catId): p = self.params # dimention here should be Nxm gts = self._gts[imgId, catId] dts = self._dts[imgId, catId] inds = np.argsort([-d['score'] for d in dts], kind='mergesort') dts = [dts[i] for i in inds] if len(dts) > p.maxDets[-1]: dts = dts[0:p.maxDets[-1]] # if len(gts) == 0 and len(dts) == 0: if len(gts) == 0 or len(dts) == 0: return [] ious = np.zeros((len(dts), len(gts))) sigmas = np.array([.26, .25, .25, .35, .35, .79, .79, .72, .72, .62,.62, 1.07, 1.07, .87, .87, .89, .89])/10.0 vars = (sigmas * 2)**2 k = len(sigmas) # compute oks between each detection and ground truth object for j, gt in enumerate(gts): # create bounds for ignore regions(double the gt bbox) g = np.array(gt['keypoints']) xg = g[0::3]; yg = g[1::3]; vg = g[2::3] k1 = np.count_nonzero(vg > 0) bb = gt['bbox'] x0 = bb[0] - bb[2]; x1 = bb[0] + bb[2] * 2 y0 = bb[1] - bb[3]; y1 = bb[1] + bb[3] * 2 for i, dt in enumerate(dts): d = np.array(dt['keypoints']) xd = d[0::3]; yd = d[1::3] if k1>0: # measure the per-keypoint distance if keypoints visible dx = xd - xg dy = yd - yg else: # measure minimum distance to keypoints in (x0,y0) & (x1,y1) z = np.zeros((k)) dx = np.max((z, x0-xd), axis=0) + np.max((z, xd-x1), axis=0) dy = np.max((z, y0-yd), axis=0) + np.max((z, yd-y1), axis=0) e = (dx**2 + dy**2) / vars / (gt['area'] + np.spacing(1)) / 2 if k1 > 0: e=e[vg > 0] ious[i, j] = np.sum(np.exp(-e)) / e.shape[0] return ious