# 套索正则化 我们将 lasso 参数定义为值 0.8： ```py lasso_param = tf.Variable(0.8, dtype=tf.float32) lasso_loss = tf.reduce_mean(tf.abs(w)) * lasso_param ``` 将套索参数设置为零意味着没有正则化，因为该项变为零。正则化项的值越高，惩罚越高。以下是套索正则化回归的完整代码，用于训练模型以预测波士顿房屋定价： 下面的代码假定训练和测试数据集已按照前面的示例进行拆分。 ```py num_outputs = y_train.shape[1] num_inputs = X_train.shape[1] x_tensor = tf.placeholder(dtype=tf.float32, shape=[None, num_inputs], name='x') y_tensor = tf.placeholder(dtype=tf.float32, shape=[None, num_outputs], name='y') w = tf.Variable(tf.zeros([num_inputs, num_outputs]), dtype=tf.float32, name='w') b = tf.Variable(tf.zeros([num_outputs]), dtype=tf.float32, name='b') model = tf.matmul(x_tensor, w) + b lasso_param = tf.Variable(0.8, dtype=tf.float32) lasso_loss = tf.reduce_mean(tf.abs(w)) * lasso_param loss = tf.reduce_mean(tf.square(model - y_tensor)) + lasso_loss learning_rate = 0.001 optimizer = tf.train.GradientDescentOptimizer(learning_rate).minimize(loss) mse = tf.reduce_mean(tf.square(model - y_tensor)) y_mean = tf.reduce_mean(y_tensor) total_error = tf.reduce_sum(tf.square(y_tensor - y_mean)) unexplained_error = tf.reduce_sum(tf.square(y_tensor - model)) rs = 1 - tf.div(unexplained_error, total_error) num_epochs = 1500 loss_epochs = np.empty(shape=[num_epochs],dtype=np.float32) mse_epochs = np.empty(shape=[num_epochs],dtype=np.float32) rs_epochs = np.empty(shape=[num_epochs],dtype=np.float32) mse_score = 0.0 rs_score = 0.0 num_epochs = 1500 loss_epochs = np.empty(shape=[num_epochs], dtype=np.float32) mse_epochs = np.empty(shape=[num_epochs], dtype=np.float32) rs_epochs = np.empty(shape=[num_epochs], dtype=np.float32) mse_score = 0.0 rs_score = 0.0 with tf.Session() as tfs: tfs.run(tf.global_variables_initializer()) for epoch in range(num_epochs): feed_dict = {x_tensor: X_train, y_tensor: y_train} loss_val,_ = tfs.run([loss,optimizer], feed_dict) loss_epochs[epoch] = loss_val feed_dict = {x_tensor: X_test, y_tensor: y_test} mse_score,rs_score = tfs.run([mse,rs], feed_dict) mse_epochs[epoch] = mse_score rs_epochs[epoch] = rs_score print('For test data : MSE = {0:.8f}, R2 = {1:.8f} '.format( mse_score, rs_score)) ``` 我们得到以下输出： ```py For test data : MSE = 30.48978233, R2 = 0.64166653 ``` 让我们使用以下代码绘制 MSE 和 r 平方的值： ```py plt.figure(figsize=(14,8)) plt.axis([0,num_epochs,0,np.max([loss_epochs,mse_epochs])]) plt.plot(loss_epochs, label='Loss on X_train') plt.plot(mse_epochs, label='MSE on X_test') plt.title('Loss in Iterations') plt.xlabel('# Epoch') plt.ylabel('Loss or MSE') plt.legend() plt.show() plt.figure(figsize=(14,8)) plt.axis([0,num_epochs,np.min(rs_epochs),np.max(rs_epochs)]) plt.title('R-squared in Iterations') plt.plot(rs_epochs, label='R2 on X_test') plt.xlabel('# Epoch') plt.ylabel('R2') plt.legend() plt.show() ``` 我们得到以下损失绘图：  迭代中 R 平方的图如下：  让我们用岭回归重复相同的例子。