具有多个输出的自定义损耗函数，并使用add_loss

Question

我在keras有一个自动编码器，我需要一个损失函数，它是mse，一个二进制交叉熵和第三部分的组合，它试图使输出的像素数最小，而输出的值不是0或1。最终的损失应该是这样的：amse+bbinary_crossentry+c*L。我为此使用了以下代码，但它产生了以下错误：

Traceback (most recent call last): File "", line 134, in if (pred_w(i,j)>=0 & pred_w(i,j)<0.1)|(pred_w(i,j)<=1 & pred_w(i,j)>=0.9): TypeError: 'Tensor' object is not callable

你能告诉我怎么解决这个问题吗？谢谢你的帮助。在

wtm=Input((4,4,1))
image = Input((28, 28, 1))
conv1 = Conv2D(64, (5, 5), activation='relu', padding='same', name='convl1e')(image)
conv2 = Conv2D(64, (5, 5), activation='relu', padding='same', name='convl2e')(conv1)
conv3 = Conv2D(64, (5, 5), activation='relu', padding='same', name='convl3e')(conv2)
#conv3 = Conv2D(8, (3, 3), activation='relu', padding='same', name='convl3e', kernel_initializer='Orthogonal',bias_initializer='glorot_uniform')(conv2)
BN=BatchNormalization()(conv3)
encoded =  Conv2D(1, (5, 5), activation='relu', padding='same',name='encoded_I')(BN)

#-----------------------adding w---------------------------------------   

wpad=Kr.layers.Lambda(lambda xy: xy[0] + Kr.backend.spatial_2d_padding(xy[1], padding=((0, 24), (0, 24))))
encoded_merged=wpad([encoded,wtm])

#-----------------------decoder------------------------------------------------
#------------------------------------------------------------------------------
deconv1 = Conv2D(64, (5, 5), activation='elu', padding='same', name='convl1d')(encoded_merged)
deconv2 = Conv2D(64, (5, 5), activation='elu', padding='same', name='convl2d')(deconv1)
deconv3 = Conv2D(64, (5, 5), activation='elu',padding='same', name='convl3d')(deconv2)
deconv4 = Conv2D(64, (5, 5), activation='elu',padding='same', name='convl4d')(deconv3)
BNd=BatchNormalization()(deconv4)

decoded = Conv2D(1, (5, 5), activation='sigmoid', padding='same', name='decoder_output')(BNd)    
model=Model(inputs=[image,wtm],outputs=decoded)

decoded_noise = GaussianNoise(0.5)(decoded)

#----------------------w extraction------------------------------------
convw1 = Conv2D(64, (5,5), activation='relu', name='conl1w')(decoded_noise)#24
convw2 = Conv2D(64, (5,5), activation='relu', name='convl2w')(convw1)#20
convw3 = Conv2D(64, (5,5), activation='relu' ,name='conl3w')(convw2)#16
convw4 = Conv2D(64, (5,5), activation='relu' ,name='conl4w')(convw3)#12
convw5 = Conv2D(64, (5,5), activation='relu', name='conl5w')(convw4)#8
convw6 = Conv2D(64, (5,5), activation='relu', name='conl6w')(convw5)#4
convw7 = Conv2D(64, (5,5), activation='relu',padding='same', name='conl7w',dilation_rate=(2,2))(convw6)#4
convw8 = Conv2D(64, (5,5), activation='relu', padding='same',name='conl8w',dilation_rate=(2,2))(convw7)#4
convw9 = Conv2D(64, (5,5), activation='relu',padding='same', name='conl9w',dilation_rate=(2,2))(convw8)#4
convw10 = Conv2D(64, (5,5), activation='relu',padding='same', name='conl10w',dilation_rate=(2,2))(convw9)#4
BNed=BatchNormalization()(convw10)
pred_w = Conv2D(1, (1, 1), activation='sigmoid', padding='same', name='reconstructed_W',dilation_rate=(2,2))(BNed)  

w_extraction=Model(inputs=[image,wtm],outputs=[decoded,pred_w])

count=0
for i in range(28):
    for j in range(28):
        if (pred_w(i,j)>=0 & pred_w(i,j)<0.1)|(pred_w(i,j)<=1 & pred_w(i,j)>=0.9):
            count+=1

loss = K.sum(0.7*mse(decoded, image),binary_crossentropy(pred_w,wtm))+count
w_extraction.add_loss(loss)

Answer 1

从技术上讲，错误只是告诉你pred_w（i，j）应该是pred_w[i，j]，如果你想引用矩阵的一个条目。然而，要使这段代码按您的预期运行，需要进行更多的重写。在

要真正优化一个损失，最主要的是它需要在网络的权重方面是可微的。在这种情况下，可以使用以下示例：

• （x-0.5）^N表示相对较高的N

• 或对数屏障，即-log（x）-log（1-x）

实际计算数字（这对优化没有帮助）可以通过以下方法实现

count = tf.sum(pred_w<=0.1) + tf.sum(predictions_w>=0.9)

也许这有助于在训练或类似的情况下输出这个范围内的数字。在