TensorFlow实现iris数据集线性回归

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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