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我正在训练一个模型来预测医学图像中的分割。在训练数据中，输入数据的类型为：numpy.float64，ground truth 标签的类型为：numpy.uint8。问题是由于某种原因，我的模型正在生成 numpy.float32 的输出类型。

图像显示： 数据类型示例

# Defining the model
segmenter = Model(input_img, segmenter(input_img))

# Training the model (type of train_ground is numpy.uint8)
segmenter_train = segmenter.fit(train_X, train_ground, batch_size=batch_size,epochs=epochs,verbose=1,validation_data=(valid_X, valid_ground))

型号定义：

def segmenter(input_img):
    #encoder
    #input = 28 x 28 x 1 (wide and thin)
    conv1 = Conv2D(32, (3, 3), activation='relu', padding='same')(input_img) #28 x 28 x 32
    conv1 = BatchNormalization()(conv1)
    conv1 = Conv2D(32, (3, 3), activation='relu', padding='same')(conv1)
    conv1 = BatchNormalization()(conv1)
    pool1 = MaxPooling2D(pool_size=(2, 2))(conv1) #14 x 14 x 32
    conv2 = Conv2D(64, (3, 3), activation='relu', padding='same')(pool1) #14 x 14 x 64
    conv2 = BatchNormalization()(conv2)
    conv2 = Conv2D(64, (3, 3), activation='relu', padding='same')(conv2)
    conv2 = BatchNormalization()(conv2)
    pool2 = MaxPooling2D(pool_size=(2, 2))(conv2) #7 x 7 x 64
    conv3 = Conv2D(128, (3, 3), activation='relu', padding='same')(pool2) #7 x 7 x 128 (small and thick)
    conv3 = BatchNormalization()(conv3)
    conv3 = Conv2D(128, (3, 3), activation='relu', padding='same')(conv3)
    conv3 = BatchNormalization()(conv3)


    #decoder
    conv4 = Conv2D(64, (3, 3), activation='relu', padding='same')(conv3) #7 x 7 x 128
    conv4 = BatchNormalization()(conv4)
    conv4 = Conv2D(64, (3, 3), activation='relu', padding='same')(conv4)
    conv4 = BatchNormalization()(conv4)
    up1 = UpSampling2D((2,2))(conv4) # 14 x 14 x 128
    conv5 = Conv2D(32, (3, 3), activation='relu', padding='same')(up1) # 14 x 14 x 64
    conv5 = BatchNormalization()(conv5)
    conv5 = Conv2D(32, (3, 3), activation='relu', padding='same')(conv5)
    conv5 = BatchNormalization()(conv5)
    up2 = UpSampling2D((2,2))(conv5) # 28 x 28 x 64

    conv6 = Conv2D(64, (3, 3), activation='relu', padding='same')(up2) #7 x 7 x 128
    conv6 = BatchNormalization()(conv6)
    conv6 = Conv2D(64, (3, 3), activation='relu', padding='same')(conv6)
    conv6 = BatchNormalization()(conv6)
    up3 = UpSampling2D((2,2))(conv6) # 14 x 14 x 128

    conv7 = Conv2D(64, (3, 3), activation='relu', padding='same')(up3) #7 x 7 x 128
    conv7 = BatchNormalization()(conv7)
    conv7 = Conv2D(64, (3, 3), activation='relu', padding='same')(conv7)
    conv7 = BatchNormalization()(conv7)
    up4 = UpSampling2D((2,2))(conv7) # 14 x 14 x 128

    decoded = Conv2D(1, (3, 3), activation='sigmoid', padding='same')(up4) # 28 x 28 x 1
    return decoded

在此先感谢您的帮助:)

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1 回答 1

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Sigmoid 返回一个实数

最后一层恰好是 sigmoid 激活函数。它返回一个从 0 到 1 的实数，而不是整数。

此外，重要的是错误度量，即正确答案和计算值之间的差异，是连续的而不是离散的，因为这是可微的，并且允许通过反向传播正确学习神经网络权重。

只需转换和四舍五入

为了训练网络，只需将真值标签转换为浮点值。

一旦你训练了网络并想要使用它的输出，只需将它们四舍五入以将它们转换为整数 - sigmoid 激活非常适合这种情况。

于 2019-03-26T09:27:52.723 回答