我有一个前馈 DNN 模型,具有多个层来执行二元分类。输出层为1个sigmoid单元,损失函数为binary_crossentropy。作为预测,我期望一个带有零/一的向量。为此,我对预测进行了总结并进行了解释。然后我使用 sklearn 评分函数来计算(f1score、rocauc、 precision、recall、mcc)。问题是我得到的预测向量与我假装的独热编码不匹配。尽管如果我使用 mse 损失函数,它会像假装的那样工作。
=>模型创建函数:
def create_DNN_model(self, verbose=True):
print("Creating DNN model")
fundamental_parameters = ['dropout', 'output_activation', 'optimization', 'learning_rate',
'units_in_input_layer',
'units_in_hidden_layers', 'nb_epoch', 'batch_size']
for param in fundamental_parameters:
if self.parameters[param] == None:
print("Parameter not set: " + param)
return
self.print_parameter_values()
model = Sequential()
# Input layer
model.add(Dense(self.parameters['units_in_input_layer'], input_dim=self.feature_number, activation='relu'))
model.add(BatchNormalization())
model.add(Dropout(self.parameters['dropout']))
# constructing all hidden layers
for layer in self.parameters['units_in_hidden_layers']:
model.add(Dense(layer, activation='relu'))
model.add(BatchNormalization())
model.add(Dropout(self.parameters['dropout']))
# constructing the final layer
model.add(Dense(1))
model.add(Activation(self.parameters['output_activation']))
if self.parameters['optimization'] == 'SGD':
optim = SGD()
optim.lr.set_value(self.parameters['learning_rate'])
elif self.parameters['optimization'] == 'RMSprop':
optim = RMSprop()
optim.lr.set_value(self.parameters['learning_rate'])
elif self.parameters['optimization'] == 'Adam':
optim = Adam()
elif self.parameters['optimization'] == 'Adadelta':
optim = Adadelta()
model.add(BatchNormalization())
model.compile(loss='binary_crossentropy', optimizer=optim, metrics=[matthews_correlation])
if self.verbose == 1: str(model.summary())
print("DNN model sucessfully created")
return model
=> 评估函数:
def evaluate_model(self, X_test, y_test):
print("Evaluating model with hold out test set.")
y_pred = self.model.predict(X_test)
y_pred = [float(np.round(x)) for x in y_pred]
y_pred = np.ravel(y_pred)
scores = dict()
scores['roc_auc'] = roc_auc_score(y_test, y_pred)
scores['accuracy'] = accuracy_score(y_test, y_pred)
scores['f1_score'] = f1_score(y_test, y_pred)
scores['mcc'] = matthews_corrcoef(y_test, y_pred)
scores['precision'] = precision_score(y_test, y_pred)
scores['recall'] = recall_score(y_test, y_pred)
scores['log_loss'] = log_loss(y_test, y_pred)
for metric, score in scores.items():
print(metric + ': ' + str(score))
return scores
=> 预测向量“y_pred”:
[-1. -1. 2. -0. 2. -1. -1. -1. 2. -1. -1. 2. -1. 2. -1. 2. -1. -1. 2. -1. 2. -1. -1. 2. -1. 2. 2. 2. -1. -1. 2. 2. 2. 2. -1. -1. 2. 2. 2. -1. 2. 2. -1. 2. -1. -1. -1. 1. -1. -1. -1.]
提前致谢。
最佳答案
您在输出层中使用线性激活(默认),而您应该采用 sigmoid。
关于python-3.x - Keras 二元分类将输出压缩为 0/1,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46697007/