Question

我正在研究一种U-net架构，以10类进行分段。我想在每个时期之后为每个班级计算骰子系数。

我的网络的输出是形状为(b_size, rows, cols, num_classes)的每个类别的分段掩码的堆栈。在此输出中，我通过以下方式计算每个类别的骰子系数：

def dice_metric(ground_truth, prediction):
    # initialize list with dice scores for each category
    dice_score_list = list()
    # get list of tensors with shape (rows, cols)
    ground_truth_unstacked = reshape_ground_truth(ground_truth)
    prediction_unstacked = tf.unstack(prediction, axis=-1)
    for (ground_truth_map, prediction_map) in zip(ground_truth_unstacked, prediction_unstacked):
        # calculate dice score for every class
        dice_i = dice_score(ground_truth_map, prediction_map)
        dice_score_list.append(dice_i)
    return tf.reduce_mean(dice_score_list, axis=[0])

有什么办法可以打印骰子分数列表而不是平均值。因此，在每个时期中，输出为：

Epoch 107/200
- 13s - loss: 0.8896 - dice_metric: [dice_class_1, ... dice_class_10] - val_loss: 3.3417 - val_dice_metric: [val_dice_class_1, ... val_dice_class_10]

关于Custom Metrics的Keras文档仅考虑了单个张量值（即”“自定义指标可以在编译步骤中传递。该函数需要采用(y_true, y_pred) 作为参数并返回单个张量值。“

是否有任何方法/解决方法来输出具有多个值的指标？

Answer 1

要使keras输出所有通道，每个通道将需要一个度量。您可以创建一个包含索引并仅返回所需类的包装器：

#calculates dice considering an input with a single class
def dice_single(true,pred):
    true = K.batch_flatten(true)
    pred = K.batch_flatten(pred)
    pred = K.round(pred)

    intersection = K.sum(true * pred, axis=-1)
    true = K.sum(true, axis=-1)
    pred = K.sum(pred, axis=-1)

    return ((2*intersection) + K.epsilon()) / (true + pred + K.epsilon())

def dice_for_class(index):
    def dice_inner(true,pred):

        #get only the desired class
        true = true[:,:,:,index]
        pred = pred[:,:,:,index]

        #return dice per class
        return dice_single(true,pred)
    return dice_inner

那么您在模型中的指标将是`metrics = [dice_for_class（i）for i in range（10）]

提示：除非绝对必要，否则请不要迭代。

没有迭代的十个类的骰子示例

def dice_metric(ground_truth, prediction):

    #for metrics, it's good to round predictions:
    prediction = K.round(prediction)

    #intersection and totals per class per batch (considers channels last)
    intersection = ground_truth * prediction
    intersection = K.sum(intersection, axis=[1,2])
    ground_truth = K.sum(ground_truth, axis=[1,2])
    prediction = K.sum(prediciton, axis=[1,2])

    dice = ((2 * intersection) + K.epsilon()) / (ground_truth + prediction + K.epsilon())

有没有一种方法可以输出带有多个值的度量值（单位：keras）？

1 个答案: