激活参数在 GridSearch 中不起作用答案

【问题标题】：Activation parameter not working in GridSearch激活参数在 GridSearch 中不起作用
【发布时间】：2022-01-11 23:35:41
【问题描述】：

我正在尝试为最佳参数创建一个 GridSearch，如下所示：

def MultiPerceptron(optimizer = 'adam', loss = 'binary_cross_entropy', kernel_initializer = 'random_uniform', activation = 'relu', units = 16):
  model = Sequential()
  model.add(InputLayer(30))
  model.add(Dense(units = units, activation = activation, kernel_initializer = kernel_initializer))
  model.add(Dense(units = units, activation = activation, kernel_initializer = kernel_initializer))
  model.add(Dense(units = 1, activation = 'sigmoid'))
  model.compile(optimizer = optimizer, loss = loss, metrics =['binary_accuracy'])
  return model

classifier = KerasClassifier(build_fn = MultiPerceptron, validation_split = 0.1, validation_batch_size = 50)
param = {'batch_size': [10, 30],
         'epochs': [50, 100],
         'optimizer': ['adam', 'sgd'],
         'loss': ['binary_crossentropy', 'hinge'],
         'kernel_initializer': ['random_uniform', 'normal'],
         'activation': ['relu', 'tanh'],
         'units': [16, 8]}

search = GridSearchCV(estimator = classifier, param_grid = param, scoring = 'accuracy', cv = 5)
search = search.fit(x,y)

我收到以下错误：

ValueError: Invalid parameter activation for estimator KerasClassifier.
This issue can likely be resolved by setting this parameter in the KerasClassifier constructor:
`KerasClassifier(activation=relu)`
Check the list of available parameters with `estimator.get_params().keys()`

【问题讨论】：

标签： keras grid-search

【解决方案1】：

我认为他们改变了一些东西，因为我只能将activation=relu 参数传递给KerasClassifier。

那里不需要其他参数。

【讨论】：

【解决方案2】：

我遇到了同样的问题。以下代码在使用 keras.wrappers 时运行完美

def build_model(lambda_parameter):
    model = Sequential()
    model.add(Dense(10, input_dim=X.shape[1], activation='relu', 
    kernel_regularizer=l2(lambda_parameter)))
    model.add(Dense(6, activation='relu', 
    kernel_regularizer=l2(lambda_parameter)))
    model.add(Dense(4, activation='relu', 
    kernel_regularizer=l2(lambda_parameter)))
    model.add(Dense(1, activation='sigmoid'))
    model.compile(loss='binary_crossentropy', optimizer='sgd', metrics= 
    ['accuracy'])
    return model
seed = 1
np.random.seed(seed)
random.set_seed(seed)
model = KerasClassifier(build_fn=build_model, verbose=0, shuffle=False)
lambda_parameter = [0.01, 0.5, 1]
epochs = [50, 100]
batch_size = [20]
param_grid = dict(lambda_parameter=lambda_parameter, epochs=epochs, 
batch_size=batch_size)
grid_search = GridSearchCV(estimator=model, param_grid=param_grid, cv=5)
results_1 = grid_search.fit(X, y)
print(f"Best cross-validation score = {results_1.best_score_}")
print(f"Parameters for best cross-validation score = 
{results_1.best_params_}")
accuracy_means = results_1.cv_results_['mean_test_score']
accuracy_stds = results_1.cv_results_['std_test_score']
parameters = results_1.cv_results_['params']
for p in range(len(parameters)):
    print(f"Accuracy {accuracy_means[p]} for params {accuracy_stds[p]}, 
    {parameters[p]}

但是在切换到 Scikeras 之后，我总是得到一个 ValueError：

ValueError: Invalid parameter lambda_parameter for estimator 
KerasClassifier.
This issue can likely be resolved by setting this parameter in the 
KerasClassifier constructor: KerasClassifier(lambda_parameter=0.01)`
Check the list of available parameters with 
estimator.get_params().keys()`

我在 KerasClassifiet 中添加了 lambda_parameter=0.01 来解决问题

model = KerasClassifier(model=build_model, verbose=0, shuffle=False, 
lambda_parameter=0.01)

【讨论】：