numpy 实现线性回归
import numpy as np
from matplotlib import pyplot
x_train = np.array([[3.3], [4.4], [5.5], [6.71], [6.93], [4.168],
[9.779], [6.182], [7.59], [2.167], [7.042],
[10.791], [5.313], [7.997], [3.1]], dtype=np.float32)
y_train = np.array([[1.7], [2.76], [2.09], [3.19], [1.694], [1.573],
[3.366], [2.596], [2.53], [1.221], [2.827],
[3.465], [1.65], [2.904], [1.3]], dtype=np.float32)
import numpy as np
import matplotlib.pyplot as plt
def compute_cost(theta, X, y):
"""
:param theta:权重参数
:param X:
:param y:
:return:
"""
return np.dot(np.transpose(np.dot(X, theta) - y), (np.dot(X, theta) - y)) / 2 * len(X)
def scatter_plot(X, y):
"""
散点图展示
:param X:
:param y:
:param path: save_file_path
:return:
"""
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.scatter(X, y)
plt.show()
def func_plot(X, y):
"""
函数直线绘图
:param X:
:param y:
:param path:
:return:
"""
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.plot(X, y)
plt.show()
def plotJ_iters(J, iters):
x = np.arange(1, iters + 1)
plt.plot(x, J)
plt.xlabel('iter')
plt.ylabel('cost')
plt.title('iter-cost')
plt.show()
def gradient_descent(X, y, theta, alpha, iter_num):
"""
:param X: 特征值
:param y: 结果值
:param theta: 学习参数
:param alpha: 学习速度
:param iter_num: 迭代次数
:return:
"""
m = X.shape[0] # 数据集数量
n = X.shape[1] # 特征数量
J_history = np.zeros(shape=(iter_num, 1))
for i in range(iter_num):
h = np.dot(X, theta) # 预测值 theta shape:(n, 1) X shape:(m, n) h shape:(m,1)
theta = theta - alpha * 1 / m * np.dot(np.transpose(X), h - y)
if (i % 2 == 0):
fig = plt.figure()
ax = fig.add_subplot(1,1,1)
ax.scatter(X[:,0], y)
ax.plot(X[:,0], h, 'red')
plt.show()
J_history[i] = compute_cost(theta, X, y)
return theta, J_history
def linear_regression():
X = x_train
y = y_train
# 增加噪音参数
noise = np.ones(shape=(len(X), 1), dtype=np.float64)
X = np.column_stack((X, noise))
# y = y.reshape(-1, 1)
theta = np.zeros(shape=(X.shape[1], 1))
theta, J_history = gradient_descent(X, y, theta, 0.01, iter_num=10)
# plotJ_iters(J_history, 50)
print('theta :', theta.reshape(1, X.shape[1]))
linear_regression()
pytorch 实现线性模型
# pytorch 实现 线性模型
import torch
from torch.autograd import Variable
from torch import nn
from torch import optim
x_train = np.array([[3.3], [4.4], [5.5], [6.71], [6.93], [4.168],
[9.779], [6.182], [7.59], [2.167], [7.042],
[10.791], [5.313], [7.997], [3.1]], dtype=np.float32)
y_train = np.array([[1.7], [2.76], [2.09], [3.19], [1.694], [1.573],
[3.366], [2.596], [2.53], [1.221], [2.827],
[3.465], [1.65], [2.904], [1.3]], dtype=np.float32)
# 将narray转化为张量
x_train = torch.from_numpy(x_train)
y_train = torch.from_numpy(y_train)
# 定义线性模型
class LinearRegression(nn.Module):
def __init__(self):
super(LinearRegression, self).__init__()
self.linear = nn.Linear(1, 1)
def forward(self, x):
out = self.linear(x)
return out
linear_model = LinearRegression()
# 定义 loss函数(最小二乘) 梯度下降
criterion = nn.MSELoss()
optimizer = optim.SGD(linear_model.parameters(), lr = 0.001) # 学习率 0.001
# 训练
epochs = 1000
for epoch in range(epochs):
inputs = Variable(x_train)
target = Variable(y_train)
# 向前传播
out = linear_model(inputs)
loss = criterion(out, target)
# 反向传播
optimizer.zero_grad() # 梯度清零
loss.backward() # 反向传播
optimizer.step() # 更新参数
if (epoch + 1) % 20 == 0:
# print('Epoch[{}/{}], loss: {:6f}'.format(epoch, epochs, loss.data[0]))
print('Epoch[{}/{}], loss: {:.6f}'.format(epoch + 1, epochs, loss.data))
linear_model.eval()
predict = linear_model(Variable(x_train))
predict = predict.data.numpy()
pyplot.plot(x_train.numpy(), y_train.numpy(), 'ro', label='Original data')
pyplot.plot(x_train.numpy(), predict, label='Fitting Line')
# 显示图例
pyplot.legend()
pyplot.show()
# 保存模型
torch.save(linear_model.state_dict(), './linear.pth')
