About metric mistake, recall and precison

[luzhenyu]

Efficient-Segmentation-Networks/utils/metric/metric.py

Lines 43 to 56 in 0f0c32e

 # Pii为预测正确的数量，Pij和Pji分别被解释为假正和假负，尽管两者都是假正与假负之和 
 def recall(self):  # 预测为正确的像素中确认为正确像素的个数 
     recall = 0.0 
     for i in range(self.nclass): 
         recall += self.M[i, i] / np.sum(self.M[:, i]) 
  
     return recall / self.nclass 
  
 def accuracy(self):  # 分割正确的像素除以总像素 
     accuracy = 0.0 
     for i in range(self.nclass): 
         accuracy += self.M[i, i] / np.sum(self.M[i, :]) 
  
     return accuracy / self.nclass 

it's reverse.
` # 理解混淆矩阵生成的代码的关键，行为真实值，列为预测值
# Pii为预测正确的数量，Pij=FN, Pji=FP
# 每一列之和表示被预测为该类别的样本数量 = TP+FP, precision = TP/(TP+FP), 所有被预测为正类中真正正类的比例
# 每一行之和表示该类别的真实样本数量 = TP+FN, recall = TP/TP+FN , 所有正类中，被找出的正类的比例

def recall(self):
    recall = 0.0
    class_recall = []
    for i in range(self.nclass):
        recall_i = self.M[i, i] / np.sum(self.M[i, :])
        recall += recall_i
        class_recall.append(recall_i)
    return recall / self.nclass, class_recall

def accuracy(self):
    accuracy = 0.0
    class_accuracy = []
    for i in range(self.nclass):
        accuracy_i = self.M[i, i] / np.sum(self.M[:, i])
        accuracy += accuracy_i
        class_accuracy.append(accuracy_i)
    return accuracy / self.nclass, class_accuracy`

So, change

Efficient-Segmentation-Networks/utils/metric/metric.py

Line 47 in 0f0c32e

 recall += self.M[i, i] / np.sum(self.M[:, i]) 

and

Efficient-Segmentation-Networks/utils/metric/metric.py

Line 54 in 0f0c32e

 accuracy += self.M[i, i] / np.sum(self.M[i, :]) 

[pipi-hua]

@DCC-lzhy Yes, I think your analysis makes sense. But this project may still use mIoU, so this may be less noticed. And it should be called "precision" here.