如何計算 pytorch 中關鍵點檢測 CNN model 的准確度？

Question

有人可以幫我解決這個問題嗎，

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

也許與歐幾里得距離：真正的關鍵點：（x，y）預測的關鍵點：（x_，y_）距離d：sqrt（（x_ - x）^2 +（y_ - y）^2）。 從中你必須得到一個百分比。 如果 d == 0，則該關鍵點的准確度為 100%。 但什么是 0%？ 我會說從真正的關鍵點到離該關鍵點最遠的圖像角落的距離。 讓我們稱這個距離為 R。 因此，您的觀點的准確性是 d / R。 對每個關鍵點都這樣做並取平均值。 我只是想出了這個，所以它可能有一些缺陷，但我認為你可以使用它並檢查它是否適合你。

Answer 2

這很有趣，我自己幾分鍾前剛剛在做這個，你可能已經意識到了。 簡單地計算兩組關鍵點之間的歐幾里得距離並不能很好地推廣到需要比較身體形狀和大小的情況，所以我建議使用 Object 關鍵點相似度分數。 它測量由人的尺度歸一化的身體關節距離。 如本博客所述，OKS 定義為：

這里（第 313 行 function computeOKS ）是 Facebook 研究的實現：

 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