Deming Regression
这里展示如何用TensorFlow求解线性戴明回归。
=+y=Ax+b
我们用iris数据集,特别是:
y = Sepal Length 且 x = Petal Width。
戴明回归Deming regression也称为total least squares, 其中我们最小化从预测线到实际点(x,y)的最短的距离。最小二乘线性回归最小化与预测线的垂直距离，戴明回归最小化与预测线的总的距离，这种回归线最小化y值和x值的误差。 我们从加载必要的库开始。
#List3-40
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()
#tf.set_random_seed(42)
np.random.seed(42)
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]) # Petal Width
y_vals = np.array(y\[0 for y in iris.data]) # Sepal Length
定义模型，损失函数，变量的梯度。
对于戴明损失,我们想要计算:
它告诉我们点(x,y)到预测线的最短距离, ⋅+A⋅x+b.
def model(x,w,b):
Declare model operations
model_output = tf.add(tf.matmul(x, w), b)
return model_output
def loss1(x,y,w,b):
Declare Deming loss function
deming_numerator = tf.abs(tf.subtract(tf.add(tf.matmul(x, w), b), y))
deming_denominator = tf.sqrt(tf.add(tf.square(w),1))
loss = tf.reduce_mean(tf.truediv(deming_numerator, deming_denominator))
return loss
def grad1(x,y,w,b):
with tf.GradientTape() as tape:
loss_1 = loss1(x,y,w,b)
return tape.gradient(loss_1,w,b)
Declare batch size
batch_size = 125
learning_rate = 0.25 # Will not converge with learning rate at 0.4
iterations = 50
Create variables for linear regression
w1 = tf.Variable(tf.random.normal(shape=1,1),tf.float32)
b1 = tf.Variable(tf.random.normal(shape=1,1),tf.float32)
optimizer = tf.optimizers.Adam(learning_rate)
Training loop
loss_vec = \[\]
for i in range(5000):
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])
x=tf.cast(rand_x,tf.float32)
y=tf.cast(rand_y,tf.float32)
grads1=grad1(x,y,w1,b1)
optimizer.apply_gradients(zip(grads1,w1,b1))
#sess.run(train_step, feed_dict={x_data: rand_x, y_target: rand_y})
temp_loss1 = loss1(x, y,w1,b1).numpy()
#sess.run(loss, feed_dict={x_data: rand_x, y_target: rand_y})
loss_vec.append(temp_loss1)
if (i+1)%25==0:
print('Step #' + str(i+1) + ' A = ' + str(w1.numpy()) + ' b = ' + str(b1.numpy()))
print('Loss = ' + str(temp_loss1))
Get the optimal coefficients
slope = w1.numpy()
y_intercept = b1.numpy()
Get best fit line
best_fit1 = \[\]
for i in x_vals:
best_fit1.append(slope*i+y_intercept)
Plot the result
plt.plot(x_vals, y_vals, 'o', label='Data Points')
plt.plot(x_vals, best_fit1, 'r-', label='Best fit line', linewidth=3)
plt.legend(loc='upper left')
plt.title('Sepal Length vs Petal Width')
plt.xlabel('Petal Width')
plt.ylabel('Sepal Length')
plt.show()
Plot loss over time
plt.plot(loss_vec, 'k-')
plt.title('L1 Loss per Generation')
plt.xlabel('Generation')
plt.ylabel('L1 Loss')
plt.show()
