Iris数据集上的可变自动编码器

Question

我正在尝试开发一个变分自动编码器，并在iris数据集上进行测试。在数据集上训练了我的模型后，我试图生成一个15x4的矩阵。

import keras
from keras import backend as K
from keras import layers
from keras import optimizers
from keras.datasets import mnist
from keras.models import Model
from sklearn.model_selection import train_test_split
import tensorflow as tf
import pandas as pd
original_dim = 4
latent_dim = 2
def encoder():
encoder_input=layers.Input(shape=(original_dim,))
x = layers.Dense(latent_dim)(encoder_input)
z_mean = layers.Dense(latent_dim)(x)
z_log_var = layers.Dense(latent_dim)(x)
return Model(encoder_input, [z_mean, z_log_var], name='encoder')
def decoder():
decoder_input = layers.Input(shape=(latent_dim,))
x = layers.Dense(original_dim,activation='relu')(decoder_input)
return Model(decoder_input, x, name='decoder')
def sampling(args):
z_mean,z_log_var = args
z_sigma = K.sqrt(K.exp(z_log_var))
epsilon = K.random_normal(shape=(K.shape(z_mean)),mean=0,stddev=1)
return z_mean + z_sigma*epsilon
def create_sampler():
return layers.Lambda(sampling, name='sampler')
encoder = encoder()
decoder = decoder()
sampler = create_sampler()
x = layers.Input(shape=(original_dim,))
z_mean, z_log_var = encoder(x)
z = sampling([z_mean, z_log_var])
z_decoded = decoder(z)
vae = Model(x, z_decoded, name='vae')
rc_loss = keras.losses.mean_squared_error(x,z_decoded)
kl_loss = -0.5 * K.sum(1+z_log_var-K.square(z_mean)-K.exp(z_log_var), axis=-1)
vae_loss = K.mean(rc_loss+kl_loss)
vae.add_loss(vae_loss)
vae.compile(optimizer='sgd')
iris = pd.read_csv("~/Documents/bioresource eng proj/iris.csv")
x_train, x_test, y_train, y_test = train_test_split(iris[['sepal.length','sepal.width','petal.length','petal.width']],iris['variety'],test_size=0.1, random_state=1)
vae.fit(x_train, x_train,epochs=100,batch_size=135,validation_data=(x_test,x_test))
x_test_encoded = encoder.predict(x_test)
x_test_decoded = decoder.predict(x_test_encoded)
print('Original Datapoints :')
print(x_test)
print('Reconstructed Datapoints :')

然而，由于某些原因，我的代码不能正常工作。有时输出将等于nan，或者值将非常小(即小于0.1(，或者一列(或多列(将等于零。其他时候，它也很好用。

我已经意识到，使用优化器和损失可能会产生影响，这是意料之中的。

Answer 1

它很晚了，但当您按照以下方式初始化权重时，它会更稳定：

x = layers.Dense(latent_dim, kernel_initializer=initializers.RandomNormal(stddev=0.01))(encoder_input)