how to change the regularization parameter in keras layer without rebuild a new model in R
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I want to fine tuning my L2 parameter in my last keras layer using a for loop approach. My target is build a Extreme Machine Learning model. Now, I'm using the code below:
#possible values for L2...
k = 2^(seq(-20,-1,1))
#vectors with metrics
acc_vector = vector('numeric',length(k))
loss_vector = vector('numeric',length(k))
for(i in seq_along(k)){
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2') %>%
compile(loss = 'categorical_crossentropy',optimizer = optimizer_rmsprop(),
metrics = c('accuracy'))
model0 %>% fit(
x_train, y_train,
epochs = 5, batch_size = 512,
validation_split = 0.2,verbose=0)
eval = model0 %>% evaluate(x_test, y_test)
acc_vector[i] = eval$acc
loss_vector[i] = eval$loss
#I don't know why, but without the next 2 lines, my memory usage increase 2 times
rm(model0,eval)
gc()
}
So, here is my problem. With this aproach (run fast, at least), my weights start by random in each loop and the value of L2 doesn't make any sense. I tried other approachs like include weights = "weights" in the first layer and worked fine, except by the process time... it increased a lot! After this, I tried to pop the last layer and add a new layer with the new L2 as follow:
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2('any value'),name='dense2')
for(i in seq_along(k)){
model0 %>% pop_layer() %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2')
}
But doesn't work. The behavior of the last approach makes the models have just the first layer. I just want change the value of L2 to retrain the last layer of my model. How can I do that in a simple way?
r keras
add a comment |
up vote
0
down vote
favorite
I want to fine tuning my L2 parameter in my last keras layer using a for loop approach. My target is build a Extreme Machine Learning model. Now, I'm using the code below:
#possible values for L2...
k = 2^(seq(-20,-1,1))
#vectors with metrics
acc_vector = vector('numeric',length(k))
loss_vector = vector('numeric',length(k))
for(i in seq_along(k)){
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2') %>%
compile(loss = 'categorical_crossentropy',optimizer = optimizer_rmsprop(),
metrics = c('accuracy'))
model0 %>% fit(
x_train, y_train,
epochs = 5, batch_size = 512,
validation_split = 0.2,verbose=0)
eval = model0 %>% evaluate(x_test, y_test)
acc_vector[i] = eval$acc
loss_vector[i] = eval$loss
#I don't know why, but without the next 2 lines, my memory usage increase 2 times
rm(model0,eval)
gc()
}
So, here is my problem. With this aproach (run fast, at least), my weights start by random in each loop and the value of L2 doesn't make any sense. I tried other approachs like include weights = "weights" in the first layer and worked fine, except by the process time... it increased a lot! After this, I tried to pop the last layer and add a new layer with the new L2 as follow:
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2('any value'),name='dense2')
for(i in seq_along(k)){
model0 %>% pop_layer() %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2')
}
But doesn't work. The behavior of the last approach makes the models have just the first layer. I just want change the value of L2 to retrain the last layer of my model. How can I do that in a simple way?
r keras
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to fine tuning my L2 parameter in my last keras layer using a for loop approach. My target is build a Extreme Machine Learning model. Now, I'm using the code below:
#possible values for L2...
k = 2^(seq(-20,-1,1))
#vectors with metrics
acc_vector = vector('numeric',length(k))
loss_vector = vector('numeric',length(k))
for(i in seq_along(k)){
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2') %>%
compile(loss = 'categorical_crossentropy',optimizer = optimizer_rmsprop(),
metrics = c('accuracy'))
model0 %>% fit(
x_train, y_train,
epochs = 5, batch_size = 512,
validation_split = 0.2,verbose=0)
eval = model0 %>% evaluate(x_test, y_test)
acc_vector[i] = eval$acc
loss_vector[i] = eval$loss
#I don't know why, but without the next 2 lines, my memory usage increase 2 times
rm(model0,eval)
gc()
}
So, here is my problem. With this aproach (run fast, at least), my weights start by random in each loop and the value of L2 doesn't make any sense. I tried other approachs like include weights = "weights" in the first layer and worked fine, except by the process time... it increased a lot! After this, I tried to pop the last layer and add a new layer with the new L2 as follow:
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2('any value'),name='dense2')
for(i in seq_along(k)){
model0 %>% pop_layer() %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2')
}
But doesn't work. The behavior of the last approach makes the models have just the first layer. I just want change the value of L2 to retrain the last layer of my model. How can I do that in a simple way?
r keras
I want to fine tuning my L2 parameter in my last keras layer using a for loop approach. My target is build a Extreme Machine Learning model. Now, I'm using the code below:
#possible values for L2...
k = 2^(seq(-20,-1,1))
#vectors with metrics
acc_vector = vector('numeric',length(k))
loss_vector = vector('numeric',length(k))
for(i in seq_along(k)){
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2') %>%
compile(loss = 'categorical_crossentropy',optimizer = optimizer_rmsprop(),
metrics = c('accuracy'))
model0 %>% fit(
x_train, y_train,
epochs = 5, batch_size = 512,
validation_split = 0.2,verbose=0)
eval = model0 %>% evaluate(x_test, y_test)
acc_vector[i] = eval$acc
loss_vector[i] = eval$loss
#I don't know why, but without the next 2 lines, my memory usage increase 2 times
rm(model0,eval)
gc()
}
So, here is my problem. With this aproach (run fast, at least), my weights start by random in each loop and the value of L2 doesn't make any sense. I tried other approachs like include weights = "weights" in the first layer and worked fine, except by the process time... it increased a lot! After this, I tried to pop the last layer and add a new layer with the new L2 as follow:
model0 = keras_model_sequential() %>%
layer_dense(units = 500,activation = 'relu',input_shape = c(784),
trainable = F,name = 'dense1') %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2('any value'),name='dense2')
for(i in seq_along(k)){
model0 %>% pop_layer() %>%
layer_dense(units = 10, activation = 'softmax',
kernel_regularizer = regularizer_l2(k[i]),name='dense2')
}
But doesn't work. The behavior of the last approach makes the models have just the first layer. I just want change the value of L2 to retrain the last layer of my model. How can I do that in a simple way?
r keras
r keras
asked Nov 11 at 18:10
brunoroquette
464
464
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