r - 在 Keras 中实现自定义层（RStudio 界面）

Question

我在使用 R 的 Keras 接口在词嵌入模型中实现自定义层时遇到了一些麻烦。下面是我的代码的玩具版本，它将重现错误：

rm(list = ls())
library(keras)
library(tensorflow)

# ================================
# inputs
# ================================

input_target <- layer_input(shape = 1)
input_context <- layer_input(shape = 1)

# embedding matrix for mean vectors
embedding_mu <- layer_embedding(
  input_dim = 1000, 
  output_dim = 50, 
  embeddings_initializer = initializer_random_uniform(),
  input_length = 1, 
  name = "embedding_mu"
)

# embedding matrix for sigma vectors
embedding_sigma <- layer_embedding(
  input_dim = 1000, 
  output_dim = 50, 
  embeddings_initializer = initializer_random_uniform(),
  input_length = 1, 
  name = "embedding_sigma"
)

# select target mu from the mu embedding matrix
target_vector_mu <- input_target %>%
  embedding_mu() %>% 
  layer_flatten()

# select context mu from the mu embedding matrix
context_vector_mu <- input_context %>%
  embedding_mu() %>%
  layer_flatten()

# select target sigma from the sigma embedding matrix
target_vector_sigma <- input_target %>%
  embedding_sigma() %>% 
  layer_flatten()

# select context sigma from the sigma embedding matrix
context_vector_sigma <- input_context %>%
  embedding_sigma() %>%
  layer_flatten()

# ================================
# custom layer
# ================================
KLenergy <- function(args){ # args <- list(mu_w, mu_c, sigma_w, sigma_c)
  comp1 <- tf$reduce_sum(tf$div(args[[4]], args[[3]]), axis = integer(1))
  comp2 <- tf$reduce_sum(tf$div(tf$square(tf$subtract(args[[1]], args[[2]])), args[[3]]), axis = integer(1))
  comp3 <- tf$subtract(tf$log(tf$reduce_prod(args[[4]], axis = integer(1))), tf$log(tf$reduce_prod(args[[3]], axis = integer(1))))
  energy <- 0.5*(comp1 + comp2 - comp3)
  return(energy)
}

kl_energy <- layer_lambda(list(target_vector_mu, 
                                 context_vector_mu, 
                                 target_vector_sigma, 
                                 context_vector_sigma),
                            KLenergy)


output <- layer_dense(kl_energy, units = 1, activation = "relu")

# ================================
# model compile
# ================================
model <- keras_model(list(input_target, input_context), output)
model %>% compile(
  loss = "binary_crossentropy", 
  optimizer = "Adagrad")

summary(model)

执行“输出”层后出现以下错误：

Error in py_call_impl(callable, dots$args, dots$keywords) : 
ValueError: Input 0 is incompatible with layer dense_2: expected min_ndim=2, found ndim=1

Detailed traceback: 
  File "/anaconda3/envs/r-tensorflow/lib/python3.6/site-packages/keras/engine/base_layer.py", line 414, in __call__
self.assert_input_compatibility(inputs)
  File "/anaconda3/envs/r-tensorflow/lib/python3.6/site-packages/keras/engine/base_layer.py", line 327, in assert_input_compatibility
str(K.ndim(x)))

我希望 kl_energy 层具有形状（无，1），但我得到（无，）。

kl_energy
Tensor("lambda_5/Mul:0", shape=(?,), dtype=float32)

我在定义自定义层时是否遗漏了一个论点？我尝试设置“keepdims = TRUE”：

KLenergy <- function(args){ # args <- list(mu_w, mu_c, sigma_w, sigma_c)
  comp1 <- tf$reduce_sum(tf$div(args[[4]], args[[3]]), axis = as.integer(1), keepdims = TRUE)
  comp2 <- tf$reduce_sum(tf$div(tf$square(tf$subtract(args[[1]], args[[2]])), args[[3]]), axis = as.integer(1), keepdims = TRUE)
  comp3 <- tf$subtract(tf$log(tf$reduce_prod(args[[4]], axis = as.integer(1), keepdims = TRUE)), tf$log(tf$reduce_prod(args[[3]], axis = as.integer(1), keepdims = TRUE)))
  energy <- 0.5*(comp1 + comp2 - comp3)
  return(energy)
}

但这给了我一个形状为 (1, None) 的 kl_energy 层，这不是我想要的。最终，该层的输出应该与原始 word2vec 模型（使用 layer_dot -请参见此处）中的形状相同，但使用此自定义层：

kl_energy
Tensor("lambda_7/Mul:0", shape=(1, ?), dtype=float32)

任何指导将不胜感激。

工作代码（以下归功于 Daniel）：

# ================================
# inputs
# ================================

input_target <- layer_input(shape = 1)
input_context <- layer_input(shape = 1)

# embedding matrix for mean vectors
embedding_mu <- layer_embedding(
  input_dim = 1000, 
  output_dim = 50, 
  embeddings_initializer = initializer_random_uniform(),
  input_length = 1, 
  name = "embedding_mu"
)

# embedding matrix for sigma vectors
embedding_sigma <- layer_embedding(
  input_dim = 1000, 
  output_dim = 50, 
  embeddings_initializer = initializer_random_uniform(),
  input_length = 1, 
  name = "embedding_sigma"
)

# select target mu from the mu embedding matrix
target_vector_mu <- input_target %>%
  embedding_mu() %>% 
  layer_flatten()

# select context mu from the mu embedding matrix
context_vector_mu <- input_context %>%
  embedding_mu() %>%
  layer_flatten()

# select target sigma from the sigma embedding matrix
target_vector_sigma <- input_target %>%
  embedding_sigma() %>% 
  layer_flatten()

# select context sigma from the sigma embedding matrix
context_vector_sigma <- input_context %>%
  embedding_sigma() %>%
  layer_flatten()

# ================================
# custom layer
# ================================
KLenergy <- function(args){ # args <- list(mu_w, mu_c, sigma_w, sigma_c)
  comp1 <- tf$reduce_sum(tf$div(args[[4]], args[[3]]), axis = as.integer(1), keepdims = TRUE)
  comp2 <- tf$reduce_sum(tf$div(tf$square(tf$subtract(args[[1]], args[[2]])), args[[3]]), axis = as.integer(1), keepdims = TRUE)
  comp3 <- tf$subtract(tf$log(tf$reduce_prod(args[[4]], axis = as.integer(1), keepdims = TRUE)), tf$log(tf$reduce_prod(args[[3]], axis = as.integer(1), keepdims = TRUE)))
  energy <- 0.5*(comp1 + comp2 - comp3)
  return(energy)
}

kl_energy <- layer_lambda(list(target_vector_mu, 
                               context_vector_mu, 
                               target_vector_sigma, 
                               context_vector_sigma),
                          KLenergy)


output <- layer_dense(kl_energy, units = 1, activation = "relu")

# ================================
# model compile
# ================================
model <- keras_model(list(input_target, input_context), output)
model %>% compile(
   loss = "binary_crossentropy", 
  optimizer = "Adagrad")

summary(model)

标签： rtensorflowkeras

Answer 1

不要忘记使用keepdims=TRUEto return (None,1)。

如前所述，您应该使用as.integer(1), 或1LR 表示法。