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python - 如何处理 KerasTensor 和 Tensor?

转载 作者:行者123 更新时间:2023-12-04 17:21:18 28 4
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我正在尝试创建变分自动编码器,这意味着我需要自定义损失函数。问题是内部损失函数我有 2 个不同的损失 - mse 和发散。 mse 是 Tensor,而 divergence 是 KerasTensor(由于离散和 mu,我从编码器中退出)。我得到这样的错误:

TypeError: Cannot convert a symbolic Keras input/output to a numpyarray. This error may indicate that you're trying to pass a symbolicvalue to a NumPy call, which is not supported. Or, you may be tryingto pass Keras symbolic inputs/outputs to a TF API that does notregister dispatching, preventing Keras from automatically convertingthe API call to a lambda layer in the Functional Model.


所以这是我的架构:
import tensorflow.keras as keras
from keras.layers import Input, Dense, Flatten, Reshape
from keras.layers import Conv2D, MaxPooling2D, UpSampling2D, Conv2DTranspose
from keras.models import Model
import tensorflow as tf
import keras.backend as K


encoded_dim = 2

class Sampling(keras.layers.Layer):
"""Uses (z_mean, z_log_var) to sample z, the vector encoding a digit."""

def call(self, inputs):
z_mean, z_log_var = inputs
batch = tf.shape(z_mean)[0]
dim = tf.shape(z_mean)[1]
epsilon = K.random_normal(shape=(batch, dim))
return z_mean + tf.exp(0.5 * z_log_var) * epsilon


img = Input((28,28,1), name='img')

x = Conv2D(32, (3,3), padding='same', activation='relu')(img)
x = MaxPooling2D()(x)
x = Conv2D(64, (3,3), padding='same', activation='relu')(x)
x = MaxPooling2D()(x)
x = Flatten()(x)
x = Dense(16, activation='relu')(x)
mu = Dense(encoded_dim, name='mu')(x)
sigma = Dense(encoded_dim, name='sigma')(x)
z = Sampling()([mu,sigma])
# print(mu)
xx = Input((encoded_dim,))

x = Dense(7*7*64, activation='relu')(xx)
x = Reshape((7,7,64))(x)

x = Conv2DTranspose(64, 3, activation="relu", strides=2, padding="same")(x)
x = Conv2DTranspose(32, 3, activation="relu", strides=2, padding="same")(x)

out = Conv2DTranspose(1, 3, activation="sigmoid", padding="same")(x)

encoder = Model(img,z, name='encoder')
decoder = Model(xx,out,name='decoder')

autoencoder = Model(img,decoder(encoder(img)),name='autoencoder')
和损失函数:
def vae_loss(x, y):
loss = tf.reduce_mean(K.square(x-y))
kl_loss = -0.5 * tf.reduce_mean(1 + sigma - tf.square(mu) - tf.exp(sigma))
print(type(loss))
print(type(kl_loss))
return loss + kl_loss

autoencoder.compile(optimizer='adam',
loss = vae_loss)

autoencoder.fit(train,train,
epochs=1,
batch_size=60,
shuffle=True,
verbose = 2)
loss和lk_loss的类型:

class 'tensorflow.python.framework.ops.Tensor'

class 'tensorflow.python.keras.engine.keras_tensor.KerasTensor'

最佳答案

您需要通过musigma到你的损失函数。 vae_loss现在接受 4 个输入:

def vae_loss(x, y, mu, sigma):
loss = tf.reduce_mean(K.square(x-y))
kl_loss = -0.5 * tf.reduce_mean(1 + sigma - tf.square(mu) - tf.exp(sigma))
return loss + kl_loss
您只需使用 autoencoder.add_loss 即可在模型中使用它.
同样重要的是 encoder不仅返回 z还有 musigma .
z, mu, sigma = encoder(img)
out = decoder(z)
autoencoder = Model(img, out, name='autoencoder')
autoencoder.add_loss(vae_loss(img, out, mu, sigma)) # <======= add_loss
autoencoder.compile(loss=None, optimizer='adam')
这里是运行笔记本 https://colab.research.google.com/drive/1r5lMZ2Dc_Lq4KJDqiirXla1qfDVmdwxW?usp=sharing

关于python - 如何处理 KerasTensor 和 Tensor?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/66057733/

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