【发布时间】:2021-06-03 12:50:52
【问题描述】:
我使用的是tf.keras 2.5.0 版。
我将回归问题的层序列定义为一个函数,使用 KerasRegressor 类进行包装。这样做是为了让我能够执行 RandomizedSearchCV。
import os
import tensorflow as tf
import tensorflow_addons as tfa
from tensorflow.keras import Sequential
from tensorflow.keras.layers import Dense, Dropout
from tensorflow.keras.layers import Input, InputLayer
from tensorflow.keras.callbacks import EarlyStopping, ModelCheckpoint, TensorBoard
from keras.wrappers.scikit_learn import KerasRegressor
from sklearn.model_selection import RandomizedSearchCV
def build_model(n_hidden = 1, n_neurons = 32, input_shape = (X_train.shape[1],), dropout = 0):
model = Sequential()
model.add(InputLayer(input_shape = input_shape))
for i in range(n_hidden):
model.add(Dense(n_neurons, activation = 'relu'))
model.add(Dropout(dropout))
model.add(Dense(1))
model.compile(loss = 'mean_squared_error', optimizer = 'adam', metrics = [tfa.metrics.RSquare(y_shape=(1,))])
return model
keras_reg = KerasRegressor(build_fn = build_model)
history = keras_reg.fit(X_train, y_train, epochs = 200, batch_size = 64,
validation_data = (X_valid, y_valid),
callbacks = callbacks)
mse_test = keras_reg.score(X_test, y_test)
上面的代码有效。但是,当我尝试定义另一个名为 keras_reg_A 的 KerasRegressor 并覆盖 build_model 的默认值时,我收到错误消息“必须始终传递 Layer.call 的第一个参数”(我故意忽略了回调)
build_model_A = build_model(n_hidden = 2, n_neurons = 64, input_shape = (X_train.shape[1],), dropout = 0)
keras_reg_A = KerasRegressor(build_model_A)
history_A = keras_reg_A.fit(X_train, y_train, epochs = 200, batch_size = 64,
validation_data = (X_valid, y_valid))
mse_test = keras_reg_A.score(X_test, y_test)
谁能解释一下这是为什么?
【问题讨论】:
标签: tf.keras python tensorflow keras tf.keras