TensorFlow实现iris数据集线性回归_Python

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