keras 分割网络自定义评估函数 - mean iou

爱被打了一巴掌 2023-02-12 10:28 1阅读 0赞

> Keras训练网络过程中需要实时观察性能，mean iou不是keras自带的评估函数，tf的又觉得不好用，自己写了一个，经过测试没有问题，本文记录自定义keras mean iou评估的实现方法。

## 计算 IoU ##

> 用numpy计算的，作为IoU的ground truth用作测试使用：

def iou_numpy(y_true,y_pred):
        
        intersection = np.sum(np.multiply(y_true.astype('bool'),y_pred == 1))
        union = np.sum((y_true.astype('bool')+y_pred.astype('bool'))>0)    
        
        return intersection/union

> keras metric IoU

def iou_keras(y_true, y_pred):
        """ Return the Intersection over Union (IoU). Args: y_true: the expected y values as a one-hot y_pred: the predicted y values as a one-hot or softmax output Returns: the IoU for the given label """
        label = 1
        # extract the label values using the argmax operator then
        # calculate equality of the predictions and truths to the label
        y_true = K.cast(K.equal(y_true, label), K.floatx())
        y_pred = K.cast(K.equal(y_pred, label), K.floatx())
        # calculate the |intersection| (AND) of the labels
        intersection = K.sum(y_true * y_pred)
        # calculate the |union| (OR) of the labels
        union = K.sum(y_true) + K.sum(y_pred) - intersection
        # avoid divide by zero - if the union is zero, return 1
        # otherwise, return the intersection over union
        return K.switch(K.equal(union, 0), 1.0, intersection / union)

## 计算 mean IoU ##

> mean IoU 简便起见，选取 (0,1,0.05) 作为不同的IoU阈值，计算平均IoU
> 
> numpy 真实值计算

def mean_iou_numpy(y_true,y_pred):
        
        iou_list = []
        for thre in list(np.arange(0.0000001,0.99,0.05)):
            y_pred_temp = y_pred >= thre
            iou = iou_numpy(y_true, y_pred_temp)
            iou_list.append(iou)
    
        return np.mean(iou_list)

> Keras mean IoU

def mean_iou_keras(y_true, y_pred):
        """ Return the mean Intersection over Union (IoU). Args: y_true: the expected y values as a one-hot y_pred: the predicted y values as a one-hot or softmax output Returns: the mean IoU """
        label = 1
        # extract the label values using the argmax operator then
        # calculate equality of the predictions and truths to the label
        y_true = K.cast(K.equal(y_true, label), K.floatx())
        
        mean_iou = K.variable(0)
        
        thre_list = list(np.arange(0.0000001,0.99,0.05))
        
        for thre in thre_list:
            
            y_pred_temp = K.cast(y_pred >= thre, K.floatx())
            y_pred_temp = K.cast(K.equal(y_pred_temp, label), K.floatx())
            # calculate the |intersection| (AND) of the labels
            intersection = K.sum(y_true * y_pred_temp)
            # calculate the |union| (OR) of the labels
            union = K.sum(y_true) + K.sum(y_pred_temp) - intersection
            iou = K.switch(K.equal(union, 0), 1.0, intersection / union)
            mean_iou = mean_iou + iou
        
        return mean_iou / len(thre_list)

## 测试 ##

## 随机生成预测值
    y_true_np = np.ones([10,10])
    y_pred_np = np.random.rand(10,10)
    
    ## 真实IoU值
    print(f' iou : {iou_numpy(y_true_np, y_pred_np)}')
    print(f' mean_iou_numpy : {mean_iou_numpy(y_true_np, y_pred_np)}')
    
    y_true = tf.Variable(y_true_np)
    y_pred = tf.Variable(y_pred_np)
    
    ## 计算节点
    iou_res = iou_keras (y_true, y_pred)
    m_iou_res = mean_iou_keras (y_true, y_pred)
    
    ## 变量初始化
    init_op = tf.global_variables_initializer()
    
    with tf.Session() as sess:
        sess.run(init_op)
        
        ## 由于存在误差，结果在0.0000001范围内即可认为相同
        
        result = sess.run(iou_res)
        print(f'result : {result} \nsame with ground truth: {abs(iou_numpy(y_true_np, y_pred_np) - result)< 0.0000001}')
        
        result = sess.run(m_iou_res)
        print(f'result : {result} \nsame with ground truth: {abs(mean_iou_numpy(y_true_np, y_pred_np) - result) < 0.0000001}')

> 输出：

iou : 0.0
    mean_iou_numpy : 0.5295
    result : 0.0 
    same with ground truth: True
    result : 0.5295000076293945 
    same with ground truth: True

### 源码下载 ###

https://github.com/zywvvd/Python\_Practise