TensorFlow实现iris数据集线性回归 TensorFlow实现iris数据集线性回归
lilongsy 人气:0本文将遍历批量数据点并让TensorFlow更新斜率和y截距。这次将使用Scikit Learn的内建iris数据集。特别地,我们将用数据点(x值代表花瓣宽度,y值代表花瓣长度)找到最优直线。选择这两种特征是因为它们具有线性关系,在后续结果中将会看到。本文将使用L2正则损失函数。
# 用TensorFlow实现线性回归算法 #---------------------------------- # # This function shows how to use TensorFlow to # solve linear regression. # y = Ax + b # # We will use the iris data, specifically: # y = Sepal Length # x = Petal Width import matplotlib.pyplot as plt import numpy as np import tensorflow as tf from sklearn import datasets from tensorflow.python.framework import ops ops.reset_default_graph() # Create graph sess = tf.Session() # Load the data # iris.data = [(Sepal Length, Sepal Width, Petal Length, Petal Width)] iris = datasets.load_iris() x_vals = np.array([x[3] for x in iris.data]) y_vals = np.array([y[0] for y in iris.data]) # 批量大小 batch_size = 25 # Initialize 占位符 x_data = tf.placeholder(shape=[None, 1], dtype=tf.float32) y_target = tf.placeholder(shape=[None, 1], dtype=tf.float32) # 模型变量 A = tf.Variable(tf.random_normal(shape=[1,1])) b = tf.Variable(tf.random_normal(shape=[1,1])) # 增加线性模型,y=Ax+b model_output = tf.add(tf.matmul(x_data, A), b) # 声明L2损失函数,其为批量损失的平均值。 loss = tf.reduce_mean(tf.square(y_target - model_output)) # 声明优化器 学习率设为0.05 my_opt = tf.train.GradientDescentOptimizer(0.05) train_step = my_opt.minimize(loss) # 初始化变量 init = tf.global_variables_initializer() sess.run(init) # 批量训练遍历迭代 # 迭代100次,每25次迭代输出变量值和损失值 loss_vec = [] for i in range(100): rand_index = np.random.choice(len(x_vals), size=batch_size) rand_x = np.transpose([x_vals[rand_index]]) rand_y = np.transpose([y_vals[rand_index]]) sess.run(train_step, feed_dict={x_data: rand_x, y_target: rand_y}) temp_loss = sess.run(loss, feed_dict={x_data: rand_x, y_target: rand_y}) loss_vec.append(temp_loss) if (i+1)%25==0: print('Step #' + str(i+1) + ' A = ' + str(sess.run(A)) + ' b = ' + str(sess.run(b))) print('Loss = ' + str(temp_loss)) # 抽取系数 [slope] = sess.run(A) [y_intercept] = sess.run(b) # 创建最佳拟合直线 best_fit = [] for i in x_vals: best_fit.append(slope*i+y_intercept) # 绘制两幅图 # 拟合的直线 plt.plot(x_vals, y_vals, 'o', label='Data Points') plt.plot(x_vals, best_fit, 'r-', label='Best fit line', linewidth=3) plt.legend(loc='upper left') plt.title('Sepal Length vs Pedal Width') plt.xlabel('Pedal Width') plt.ylabel('Sepal Length') plt.show() # Plot loss over time # 迭代100次的L2正则损失函数 plt.plot(loss_vec, 'k-') plt.title('L2 Loss per Generation') plt.xlabel('Generation') plt.ylabel('L2 Loss') plt.show()
结果:
Step #25 A = [[ 1.93474162]] b = [[ 3.11190438]] Loss = 1.21364 Step #50 A = [[ 1.48641717]] b = [[ 3.81004381]] Loss = 0.945256 Step #75 A = [[ 1.26089203]] b = [[ 4.221035]] Loss = 0.254756 Step #100 A = [[ 1.1693294]] b = [[ 4.47258472]] Loss = 0.281654
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