# 使用 Keras 的简单 RNN 通过添加具有内部神经元数量和输入张量形状的 SimpleRNN 层，可以在 Keras 中轻松构建 RNN 模型，不包括样本维数。以下代码创建，编译和拟合 SimpleRNN： ```py # create and fit the SimpleRNN model model = Sequential() model.add(SimpleRNN(units=4, input_shape=(X_train.shape[1], X_train.shape[2]))) model.add(Dense(1)) model.compile(loss='mean_squared_error', optimizer='adam') model.fit(X_train, Y_train, epochs=20, batch_size=1) ``` 由于我们的数据集很小，我们使用`batch_size`为 1 并训练 20 次迭代，但对于较大的数据集，您需要调整这些和其他超参数的值。 该模型的结构如下： ```py _________________________________________________________________ Layer (type) Output Shape Param # ================================================================= simple_rnn_1 (SimpleRNN) (None, 4) 24 _________________________________________________________________ dense_1 (Dense) (None, 1) 5 ================================================================= Total params: 29 Trainable params: 29 Non-trainable params: 0 ``` 训练的结果如下： ```py Epoch 1/20 95/95 [==============================] - 0s - loss: 0.0161 Epoch 2/20 95/95 [==============================] - 0s - loss: 0.0074 Epoch 3/20 95/95 [==============================] - 0s - loss: 0.0063 Epoch 4/20 95/95 [==============================] - 0s - loss: 0.0051 -- epoch 5 to 14 removed for the sake of brevity -- Epoch 14/20 95/95 [==============================] - 0s - loss: 0.0021 Epoch 15/20 95/95 [==============================] - 0s - loss: 0.0020 Epoch 16/20 95/95 [==============================] - 0s - loss: 0.0020 Epoch 17/20 95/95 [==============================] - 0s - loss: 0.0020 Epoch 18/20 95/95 [==============================] - 0s - loss: 0.0020 Epoch 19/20 95/95 [==============================] - 0s - loss: 0.0020 Epoch 20/20 95/95 [==============================] - 0s - loss: 0.0020 ``` 损失从 0.0161 开始，平稳在 0.0020。让我们做出预测并重新调整预测和原件。我们使用 Keras 提供的函数来计算均方根误差： ```py from keras.losses import mean_squared_error as k_mse from keras.backend import sqrt as k_sqrt import keras.backend as K # make predictions y_train_pred = model.predict(X_train) y_test_pred = model.predict(X_test) # invert predictions y_train_pred = scaler.inverse_transform(y_train_pred) y_test_pred = scaler.inverse_transform(y_test_pred) #invert originals y_train_orig = scaler.inverse_transform(Y_train) y_test_orig = scaler.inverse_transform(Y_test) # calculate root mean squared error trainScore = k_sqrt(k_mse(y_train_orig[:,0], y_train_pred[:,0]) ).eval(session=K.get_session()) print('Train Score: {0:.2f} RMSE'.format(trainScore)) testScore = k_sqrt(k_mse(y_test_orig[:,0], y_test_pred[:,0]) ).eval(session=K.get_session()) print('Test Score: {0:.2f} RMSE'.format(testScore)) ``` 我们得到以下结果： ```py Train Score: 23.27 RMSE Test Score: 54.13 RMSE ```  我们可以看到，这不像我们在 TensorFlow 部分得到的那样完美;但是，这种差异是因为超参数值。我们留给您尝试不同的超参数值来调整此 Keras 模型以获得更好的结果。