yegeli

机器学习02-(损失函数loss、梯度下降、线性回归、评估训练、模型加载、岭回归、多项式回归)

机器学习-02

- 回归模型
- - 线性回归
  - - 评估训练结果误差（metrics）
    - 模型的保存和加载
  - 岭回归
  - 多项式回归
- 代码总结
- - 线性回归
  - - 绘制图像，观察w0、w1、loss的变化过程
    - 以等高线的方式绘制梯度下降的过程
    - 薪水预测
    - 评估误差
    - 把训练好的模型存入文件
  - 加载模型
  - - 封装预测模型对象，提供薪资预测服务
  - 岭回归
  - - 如何选择合适的超参数C？
  - 多项式回归
  - - 基于这组数据训练多项式回归模型
    - 案例：波士顿房屋价格数据分析与房价预测
    - - 训练回归模型，预测房屋价格

回归模型

线性回归

输入		输出
0.5      5.0
0.6      5.5
0.8      6.0
1.1      6.8
1.4      7.0
...
y = f(x)

预测函数：y = w₀+w₁x
x: 输入
y: 输出
w₀和w₁: 模型参数

所谓模型训练，就是根据已知的x和y，找到最佳的模型参数w₀ 和 w₁，尽可能精确地描述出输入和输出的关系。

5.0 = w₀ + w₁ × 0.5
5.5 = w₀ + w₁ × 0.6

单样本误差：

根据预测函数求出输入为x时的预测值：y’ = w₀ + w₁x，单样本误差为1/2(y’ - y)²。

总样本误差：

把所有单样本误差相加即是总样本误差：1/2 Σ(y’ - y)²

损失函数：

loss = 1/2 Σ(w₀ + w₁x - y)²

所以损失函数就是总样本误差关于模型参数的函数，该函数属于三维数学模型，即需要找到一组w₀ w₁使得loss取极小值。

案例：画图模拟梯度下降的过程

整理训练集数据，自定义梯度下降算法规则，求出w₀ ， w₁ ，绘制回归线。

import numpy as np
import matplotlib.pyplot as mp
train_x = np.array([0.5, 0.6, 0.8, 1.1, 1.4])
train_y = np.array([5.0, 5.5, 6.0, 6.8, 7.0])
test_x = np.array([0.45, 0.55, 1.0, 1.3, 1.5])
test_y = np.array([4.8, 5.3, 6.4, 6.9, 7.3])

times = 1000	# 定义梯度下降次数
lrate = 0.01	# 记录每次梯度下降参数变化率
epoches = []	# 记录每次梯度下降的索引
w0, w1, losses = [1], [1], []
for i in range(1, times + 1):
    epoches.append(i)
    loss = (((w0[-1] + w1[-1] * train_x) - train_y) ** 2).sum() / 2
    losses.append(loss)
    d0 = ((w0[-1] + w1[-1] * train_x) - train_y).sum()
    d1 = (((w0[-1] + w1[-1] * train_x) - train_y) * train_x).sum()
    print('{:4}> w0={:.8f}, w1={:.8f}, loss={:.8f}'.format(epoches[-1], w0[-1], w1[-1], losses[-1]))
    w0.append(w0[-1] - lrate * d0)
    w1.append(w1[-1] - lrate * d1)

pred_test_y = w0[-1] + w1[-1] * test_x
mp.figure('Linear Regression', facecolor='lightgray')
mp.title('Linear Regression', fontsize=20)
mp.xlabel('x', fontsize=14)
mp.ylabel('y', fontsize=14)
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.scatter(train_x, train_y, marker='s', c='dodgerblue', alpha=0.5, s=80, label='Training')
mp.scatter(test_x, test_y, marker='D', c='orangered', alpha=0.5, s=60, label='Testing')
mp.scatter(test_x, pred_test_y, c='orangered', alpha=0.5, s=80, label='Predicted')
mp.plot(test_x, pred_test_y, '--', c='limegreen', label='Regression', linewidth=1)
mp.legend()
mp.show()

绘制随着每次梯度下降，w₀，w₁，loss的变化曲线。

w0 = w0[:-1]
w1 = w1[:-1]

mp.figure('Training Progress', facecolor='lightgray')
mp.subplot(311)
mp.title('Training Progress', fontsize=20)
mp.ylabel('w0', fontsize=14)
mp.gca().xaxis.set_major_locator(mp.MultipleLocator(100))
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.plot(epoches, w0, c='dodgerblue', label='w0')
mp.legend()
mp.subplot(312)
mp.ylabel('w1', fontsize=14)
mp.gca().xaxis.set_major_locator(mp.MultipleLocator(100))
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.plot(epoches, w1, c='limegreen', label='w1')
mp.legend()

mp.subplot(313)
mp.xlabel('epoch', fontsize=14)
mp.ylabel('loss', fontsize=14)
mp.gca().xaxis.set_major_locator(mp.MultipleLocator(100))
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.plot(epoches, losses, c='orangered', label='loss')
mp.legend()

基于三维曲面绘制梯度下降过程中的每一个点。

import mpl_toolkits.mplot3d as axes3d

grid_w0, grid_w1 = np.meshgrid(
    np.linspace(0, 9, 500),
    np.linspace(0, 3.5, 500))

grid_loss = np.zeros_like(grid_w0)
for x, y in zip(train_x, train_y):
    grid_loss += ((grid_w0 + x*grid_w1 - y) ** 2) / 2

mp.figure('Loss Function')
ax = mp.gca(projection='3d')
mp.title('Loss Function', fontsize=20)
ax.set_xlabel('w0', fontsize=14)
ax.set_ylabel('w1', fontsize=14)
ax.set_zlabel('loss', fontsize=14)
ax.plot_surface(grid_w0, grid_w1, grid_loss, rstride=10, cstride=10, cmap='jet')
ax.plot(w0, w1, losses, 'o-', c='orangered', label='BGD')
mp.legend()

以等高线的方式绘制梯度下降的过程。

mp.figure('Batch Gradient Descent', facecolor='lightgray')
mp.title('Batch Gradient Descent', fontsize=20)
mp.xlabel('x', fontsize=14)
mp.ylabel('y', fontsize=14)
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.contourf(grid_w0, grid_w1, grid_loss, 10, cmap='jet')
cntr = mp.contour(grid_w0, grid_w1, grid_loss, 10,
                  colors='black', linewidths=0.5)
mp.clabel(cntr, inline_spacing=0.1, fmt='%.2f',
          fontsize=8)
mp.plot(w0, w1, 'o-', c='orangered', label='BGD')
mp.legend()
mp.show()

线性回归相关API：

import sklearn.linear_model as lm
# 创建模型
model = lm.LinearRegression()
# 训练模型
# 输入为一个二维数组表示的样本矩阵
# 输出为每个样本最终的结果
model.fit(输入, 输出) # 通过梯度下降法计算模型参数
# 预测输出  
# 输入array是一个二维数组，每一行是一个样本，每一列是一个特征。
result = model.predict(array)


浓度		深度		温度			腐蚀速率
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002

0.002	 200      -2              ?
0.003	 300      -4              ?

案例：基于线性回归训练single.txt中的训练样本，使用模型预测测试样本。

import numpy as np
import sklearn.linear_model as lm
import matplotlib.pyplot as mp
# 采集数据
x, y = np.loadtxt('../data/single.txt', delimiter=',', usecols=(0,1), unpack=True)
x = x.reshape(-1, 1)
# 创建模型
model = lm.LinearRegression()  # 线性回归
# 训练模型
model.fit(x, y)
# 根据输入预测输出
pred_y = model.predict(x)
mp.figure('Linear Regression', facecolor='lightgray')
mp.title('Linear Regression', fontsize=20)
mp.xlabel('x', fontsize=14)
mp.ylabel('y', fontsize=14)
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.scatter(x, y, c='dodgerblue', alpha=0.75, s=60, label='Sample')
mp.plot(x, pred_y, c='orangered', label='Regression')
mp.legend()
mp.show()

评估训练结果误差（metrics）

线性回归模型训练完毕后，可以利用测试集评估训练结果误差。sklearn.metrics提供了计算模型误差的几个常用算法：

import sklearn.metrics as sm

# 平均绝对值误差：1/m∑|实际输出-预测输出|
sm.mean_absolute_error(y, pred_y)
# 平均平方误差：SQRT(1/mΣ(实际输出-预测输出)^2)
sm.mean_squared_error(y, pred_y)
# 中位绝对值误差：MEDIAN(|实际输出-预测输出|)
sm.median_absolute_error(y, pred_y)
# R2得分，(0,1]区间的分值。分数越高，误差越小。
sm.r2_score(y, pred_y)

案例：在上一个案例中使用sm评估模型误差。

# 平均绝对值误差：1/m∑|实际输出-预测输出|
print(sm.mean_absolute_error(y, pred_y))
# 平均平方误差：SQRT(1/mΣ(实际输出-预测输 出)^2)
print(sm.mean_squared_error(y, pred_y))
# 中位绝对值误差：MEDIAN(|实际输出-预测输出|)
print(sm.median_absolute_error(y, pred_y))
# R2得分，(0,1]区间的分值。分数越高，误差越小。
print(sm.r2_score(y, pred_y))

模型的保存和加载

模型训练是一个耗时的过程，一个优秀的机器学习是非常宝贵的。可以模型保存到磁盘中，也可以在需要使用的时候从磁盘中重新加载模型即可。不需要重新训练。

模型保存和加载相关API：

import pickle
pickle.dump(内存对象, 磁盘文件) # 保存模型
model = pickle.load(磁盘文件) # 加载模型

案例：把训练好的模型保存到磁盘中。

# 将训练好的模型对象保存到磁盘文件中
with open('../../data/linear.pkl', 'wb') as f:
    pickle.dump(model, f)

# 从磁盘文件中加载模型对象
with open('../../data/linear.pkl', 'rb') as f:
    model = pickle.load(f)
# 根据输入预测输出
pred_y = model.predict(x)

岭回归

普通线性回归模型使用基于梯度下降的最小二乘法，在最小化损失函数的前提下，寻找最优模型参数，于此过程中，包括少数异常样本在内的全部训练数据都会对最终模型参数造成程度相等的影响，异常值对模型所带来影响无法在训练过程中被识别出来。为此，岭回归在模型迭代过程所依据的损失函数中增加了正则项，以限制模型参数对异常样本的匹配程度，进而提高模型面对多数正常样本的拟合精度。

import sklearn.linear_model as lm
# 创建模型
model = lm.Ridge(正则强度，fit_intercept=是否训练截距, max_iter=最大迭代次数)
# 训练模型
# 输入为一个二维数组表示的样本矩阵
# 输出为每个样本最终的结果
model.fit(输入, 输出)
# 预测输出  
# 输入array是一个二维数组，每一行是一个样本，每一列是一个特征。
result = model.predict(array)

案例：加载abnormal.txt文件中的数据，基于岭回归算法训练回归模型。

import numpy as np
import sklearn.linear_model as lm
import matplotlib.pyplot as mp
# 采集数据
x, y = np.loadtxt('../data/single.txt', delimiter=',', usecols=(0,1), unpack=True)
x = x.reshape(-1, 1)
# 创建线性回归模型
model = lm.LinearRegression() 
# 训练模型
model.fit(x, y)
# 根据输入预测输出
pred_y1 = model.predict(x)
# 创建岭回归模型
model = lm.Ridge(150, fit_intercept=True, max_iter=10000) 
# 训练模型
model.fit(x, y)
# 根据输入预测输出
pred_y2 = model.predict(x)

mp.figure('Linear & Ridge', facecolor='lightgray')
mp.title('Linear & Ridge', fontsize=20)
mp.xlabel('x', fontsize=14)
mp.ylabel('y', fontsize=14)
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.scatter(x, y, c='dodgerblue', alpha=0.75,
           s=60, label='Sample')
sorted_indices = x.T[0].argsort()
mp.plot(x[sorted_indices], pred_y1[sorted_indices],
        c='orangered', label='Linear')
mp.plot(x[sorted_indices], pred_y2[sorted_indices],
        c='limegreen', label='Ridge')
mp.legend()
mp.show()

多项式回归

浓度		深度		温度			腐蚀速率
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002
0.001	 100	  -1			0.0002

0.002	 200      -2              ?
0.003	 300      -4              ?

若希望回归模型更好的拟合训练样本数据，可以使用多项式回归器。

一元多项式回归

y=w₀ + w₁ x + w₂ x² + w₃ x³ + … + w_d x^d

将高次项看做对一次项特征的扩展得到：

y=w₀ + w₁ x₁ + w₂ x₂ + w₃ x₃ + … + w_d x_d

那么一元多项式回归即可以看做为多元线性回归，可以使用LinearRegression模型对样本数据进行模型训练。

所以一元多项式回归的实现需要两个步骤：

将一元多项式回归问题转换为多元线性回归问题（只需给出多项式最高次数即可）。
将1步骤得到多项式的结果中 w₁ w₂ … 当做样本特征，交给线性回归器训练多元线性模型。

使用sklearn提供的数据管线实现两个步骤的顺序执行：

import sklearn.pipeline as pl
import sklearn.preprocessing as sp
import sklearn.linear_model as lm

model = pl.make_pipeline(
    sp.PolynomialFeatures(10),  # 多项式特征扩展器
    lm.LinearRegression())      # 线性回归器

案例：

import numpy as np
import sklearn.pipeline as pl
import sklearn.preprocessing as sp
import sklearn.linear_model as lm
import sklearn.metrics as sm
import matplotlib.pyplot as mp
# 采集数据
x, y = np.loadtxt('../data/single.txt', delimiter=',', usecols=(0,1), unpack=True)
x = x.reshape(-1, 1)
# 创建模型(管线)
model = pl.make_pipeline(
    sp.PolynomialFeatures(10),  # 多项式特征扩展器
    lm.LinearRegression())      # 线性回归器
# 训练模型
model.fit(x, y)
# 根据输入预测输出
pred_y = model.predict(x)
test_x = np.linspace(x.min(), x.max(), 1000).reshape(-1, 1)
pred_test_y = model.predict(test_x)
mp.figure('Polynomial Regression', facecolor='lightgray')
mp.title('Polynomial Regression', fontsize=20)
mp.xlabel('x', fontsize=14)
mp.ylabel('y', fontsize=14)
mp.tick_params(labelsize=10)
mp.grid(linestyle=':')
mp.scatter(x, y, c='dodgerblue', alpha=0.75, s=60, label='Sample')
mp.plot(test_x, pred_test_y, c='orangered', label='Regression')
mp.legend()
mp.show()

过于简单的模型，无论对于训练数据还是测试数据都无法给出足够高的预测精度，这种现象叫做欠拟合。

过于复杂的模型，对于训练数据可以得到较高的预测精度，但对于测试数据通常精度较低，这种现象叫做过拟合。

一个性能可以接受的学习模型应该对训练数据和测试数据都有接近的预测精度，而且精度不能太低。

训练集R2   测试集R2
0.3        0.4        欠拟合：过于简单，无法反映数据的规则
0.9        0.2        过拟合：过于复杂，太特殊，缺乏一般性
0.7        0.6        可接受：复杂度适中，既反映数据的规则，同时又不失一般性

案例：预测波士顿地区房屋价格。

读取数据，打断原始数据集。划分训练集和测试集。

import sklearn.datasets as sd
import sklearn.utils as su
# 加载波士顿地区房价数据集
boston = sd.load_boston()
print(boston.feature_names)
# |CRIM|ZN|INDUS|CHAS|NOX|RM|AGE|DIS|RAD|TAX|PTRATIO|B|LSTAT|
# 犯罪率|住宅用地比例|商业用地比例|是否靠河|空气质量|房间数|年限|距中心区距离|路网密度|房产税|师生比|黑人比例|低地位人口比例|
# 打乱原始数据集的输入和输出
x, y = su.shuffle(boston.data, boston.target, random_state=7)
# 划分训练集和测试集
train_size = int(len(x) * 0.8)
train_x, test_x, train_y, test_y = \
    x[:train_size], x[train_size:], \
    y[:train_size], y[train_size:]

基于岭回归与多项式回归训练模型并测试模型性能。

代码查看代码总结中的波士顿房屋价格数据分析与房价预测案例

代码总结

线性回归

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

x = np.array([0.5, 0.6, 0.8, 1.1, 1.4])
y = np.array([5.0, 5.5, 6.0, 6.8, 7.0])
data = pd.DataFrame({
     'x':x, 'y':y})
data.plot.scatter(x='x', y='y', s=80)

# 基于线性回归，梯度下降理论找到最优秀的直线拟合当前这组样本数据
w0, w1 = 1, 1  # 模型参数
lrate = 0.01   # 学习率
times = 1000   # 迭代次数

# 整理一些数组，用于后期画图
epoches, w0s, w1s, losses= [], [], [], []
for i in range(times):
    # 计算当前w0与w1的状态下的loss函数值
    loss = ((w0 + w1*x - y)**2).sum()/2
    epoches.append(i+1)
    w0s.append(w0)
    w1s.append(w1)
    losses.append(loss)
    # 输出模型参数的变化过程
    print('{:4}, w0:{:.8f}, w1:{:.8f}, loss:{:.8f}'.format(i+1, w0, w1, loss))
    # 更新w0与w1 （需要求出w0与w1方向上的偏导数，带入更新公式）
    d0 = (w0 + w1*x - y).sum()
    d1 = (x*(w0 + w1*x - y)).sum()
    w0 = w0 - lrate * d0
    w1 = w1 - lrate * d1

   1, w0:1.00000000, w1:1.00000000, loss:44.17500000
   2, w0:1.20900000, w1:1.19060000, loss:36.53882794
   3, w0:1.39916360, w1:1.36357948, loss:30.23168666
   4, w0:1.57220792, w1:1.52054607, loss:25.02222743
   5, w0:1.72969350, w1:1.66296078, loss:20.71937337
   6, w0:1.87303855, w1:1.79215140, loss:17.16530917
   7, w0:2.00353196, w1:1.90932461, loss:14.22969110
   8, w0:2.12234508, w1:2.01557706, loss:11.80486494
   9, w0:2.23054244, w1:2.11190537, loss:9.80191627
  10, w0:2.32909148, w1:2.19921529, loss:8.14740839
  11, w0:2.41887143, w1:2.27832995, loss:6.78068803
  12, w0:2.50068134, w1:2.34999742, loss:5.65166010
  13, w0:2.57524739, w1:2.41489755, loss:4.71894976
  14, w0:2.64322953, w1:2.47364820, loss:3.94838447
  15, w0:2.70522753, w1:2.52681085, loss:3.31174023
  16, w0:2.76178648, w1:2.57489580, loss:2.78570611
  17, w0:2.81340174, w1:2.61836680, loss:2.35102901
  18, w0:2.86052351, w1:2.65764531, loss:1.99180729
  19, w0:2.90356094, w1:2.69311435, loss:1.69490738
  20, w0:2.94288586, w1:2.72512202, loss:1.44948190
  21, w0:2.97883620, w1:2.75398465, loss:1.24657173
  22, w0:3.01171907, w1:2.77998973, loss:1.07877728
  23, w0:3.04181357, w1:2.80339855, loss:0.93998705
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1000, w0:4.06556693, w1:2.26355153, loss:0.08742142

# 可视化
# 通过求得的w0与w1，得到拟合直线
pred_y = w0 + w1 * x
data['pred_y'] = pred_y

ax = data.plot(x='x', y='pred_y', color='orangered')
data.plot.scatter(x='x', y='y', s=80, ax=ax)

绘制图像，观察w0、w1、loss的变化过程

plt.subplot(3,1,1)
plt.grid(linestyle=':')
plt.ylabel('w0')
plt.plot(epoches, w0s, color='dodgerblue', label='w0')
plt.subplot(3,1,2)
plt.grid(linestyle=':')
plt.ylabel('w1')
plt.plot(epoches, w1s, color='dodgerblue', label='w1')
plt.subplot(3,1,3)
plt.grid(linestyle=':')
plt.ylabel('loss')
plt.plot(epoches, losses, color='orangered', label='loss')
plt.tight_layout()

以等高线的方式绘制梯度下降的过程

import mpl_toolkits.mplot3d as axes3d

grid_w0, grid_w1 = np.meshgrid(
    np.linspace(0, 9, 500),
    np.linspace(0, 3.5, 500))

grid_loss = np.zeros_like(grid_w0)
for xs, ys in zip(x, y):
    grid_loss += ((grid_w0 + xs*grid_w1 - ys) ** 2) / 2

plt.figure('Batch Gradient Descent', facecolor='lightgray')
plt.title('Batch Gradient Descent', fontsize=20)
plt.xlabel('w0', fontsize=14)
plt.ylabel('w1', fontsize=14)
plt.tick_params(labelsize=10)
plt.grid(linestyle=':')
plt.contourf(grid_w0, grid_w1, grid_loss, 10, cmap='jet')
cntr = plt.contour(grid_w0, grid_w1, grid_loss, 10,
                  colors='black', linewidths=0.5)
plt.clabel(cntr, inline_spacing=0.1, fmt='%.2f',
          fontsize=8)
plt.plot(w0s, w1s, 'o-', c='orangered', label='BGD')
plt.legend()
plt.show()

薪水预测

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# 加载数据
data = pd.read_csv('../data/Salary_Data.csv')
data.plot.scatter(x='YearsExperience', y='Salary')

# 训练线性回归模型
import sklearn.linear_model as lm
model = lm.LinearRegression()
x, y = data.loc[:, :'YearsExperience'], data['Salary']
model.fit(x, y)

# 通过训练好的模型，绘制回归线
pred_y = model.predict(x)  # 计算30个样本的预测输出
data['pred_y'] = pred_y

ax = data.plot.scatter(x='YearsExperience', y='Salary')
data.plot(x='YearsExperience', y='pred_y', color='orangered', ax=ax)

# 计算15年工作经验、18年工作经验的薪水预测结果
# test_x = [15, 18]
test_x = [[15], [18]]
model.predict(test_x)

array([167541.63502049, 195891.52198486])

评估误差

把训练集中的30个样本当做测试样本，输出预测结果并评估误差

import sklearn.metrics as sm
print(sm.mean_absolute_error(y, pred_y))
# print(sm.mean_squared_error(y, pred_y))
print(sm.median_absolute_error(y, pred_y))

4644.2012894435375
4017.9270292179935

可以根据误差的大小来判断当前模型是否达到业务标准，如果达到要求，则可以筹备上线。如果没有达到，找原因，优化模型、样本。

sm.r2_score(y, pred_y)

0.9569566641435086

把训练好的模型存入文件

import pickle
with open('salary_model.pkl', 'wb') as f:
    pickle.dump(model, f)
print('dump success!')

dump success!

加载模型

import numpy as np
import pickle
import sklearn.linear_model as lm

# 加载模型
with open('salary_model.pkl', 'rb') as f:
    model = pickle.load(f)

model.predict([[15.5]])

array([172266.61618122])

封装预测模型对象，提供薪资预测服务

class SalaryPredictionModel():
    
    def __init__(self):
        with open('salary_model.pkl', 'rb') as f:
            self.model = pickle.load(f)
        
    def predict(self, exps):
        """
        预测薪水
        exps：接收一个array（存储每个人的工作年限）
        """
        exps = np.array(exps).reshape(-1, 1)  
        return self.model.predict(exps)

# 使用封装好的对象，实现预测业务
model = SalaryPredictionModel()
model.predict([5, 6, 7, 8, 9, 10, 14, 13.5, 5.3, 1.2])

array([ 73042.01180594,  82491.9741274 ,  91941.93644885, 101391.89877031,
       110841.86109176, 120291.82341322, 158091.67269904, 153366.69153831,
        75877.00050238,  37132.15498441])

岭回归

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# 加载数据
data = pd.read_csv('../data/Salary_Data2.csv')
data.head(3)

	YearsExperience	Salary
0	1.1	39343
1	1.3	46205
2	1.5	37731

# 训练普通的线性回归模型，观察
import sklearn.linear_model as lm

x, y = data.loc[:, :'YearsExperience'], data['Salary']
model = lm.LinearRegression()
model.fit(x, y)
pred_y = model.predict(x)
plt.grid(linestyle=':')
plt.scatter(data['YearsExperience'], data['Salary'], s=60, label='points')
plt.plot(data['YearsExperience'], pred_y, color='orangered', label='regression')
plt.legend()

# 训练一个岭回归模型，这样可以更好的拟合这些普通样本
model = lm.Ridge(100)
model.fit(x, y)
pred_y = model.predict(x)
plt.grid(linestyle=':')
plt.scatter(data['YearsExperience'], data['Salary'], s=60, label='points')
plt.plot(data['YearsExperience'], pred_y, color='orangered', label='regression')
plt.legend()

如何选择合适的超参数C？

import sklearn.metrics as sm
# 找一组测试数据，针对不同的C训练不同的模型，比较每个模型r2得分即可。
test_data = data.iloc[0:-3:3]
test_x, test_y = test_data.loc[:, :'YearsExperience'], test_data['Salary']
pred_test_y = model.predict(test_x)
sm.r2_score(test_y, pred_test_y)

0.9196733030163875

多项式回归

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# 加载数据
data = pd.read_csv('../data/Salary_Data.csv')
data.head(3)

	YearsExperience	Salary
0	1.1	39343
1	1.3	46205
2	1.5	37731

# 训练普通的线性回归模型，观察
import sklearn.linear_model as lm

x, y = data.loc[:, :'YearsExperience'], data['Salary']
model = lm.LinearRegression()
model.fit(x, y)
pred_y = model.predict(x)
plt.grid(linestyle=':')
plt.scatter(data['YearsExperience'], data['Salary'], s=60, label='points')
plt.plot(data['YearsExperience'], pred_y, color='orangered', label='regression')
plt.legend()

基于这组数据训练多项式回归模型

import sklearn.pipeline as pl
import sklearn.preprocessing as sp

# 通过数据管线，链接两个操作，特征扩展->回归模型
model = pl.make_pipeline(
    sp.PolynomialFeatures(10), lm.LinearRegression())
model.fit(x, y)
# 输出模型的r2得分
pred_y = model.predict(x)
print(sm.r2_score(y, pred_y))

# 从x的最小值到最大值拆出200个x值，预测得到200个y值，绘制模型曲线。
test_x = np.linspace(x.min(), x.max(), 200)
pred_test_y = model.predict(test_x.reshape(-1, 1))
# 可视化
plt.grid(linestyle=':')
plt.scatter(data['YearsExperience'], data['Salary'], s=60, label='points')
plt.plot(test_x, pred_test_y, color='orangered', label='poly regression')
plt.legend()

0.980983738515142

案例：波士顿房屋价格数据分析与房价预测

import numpy as np
import matplotlib.pyplot as plt
import sklearn.datasets as sd
import sklearn.utils as su
import pandas as pd

# 加载数据集
boston = sd.load_boston()
# print(boston.DESCR)
x, y, header = boston.data, boston.target, boston.feature_names

# 针对当前数据集，做简单的数据分析
data = pd.DataFrame(x, columns=header)
data['y'] = y
data.describe()

	CRIM	ZN	INDUS	CHAS	NOX	RM	AGE	DIS	RAD	TAX	PTRATIO	B	LSTAT	y
count	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000	506.000000
mean	3.593761	11.363636	11.136779	0.069170	0.554695	6.284634	68.574901	3.795043	9.549407	408.237154	18.455534	356.674032	12.653063	22.532806
std	8.596783	23.322453	6.860353	0.253994	0.115878	0.702617	28.148861	2.105710	8.707259	168.537116	2.164946	91.294864	7.141062	9.197104
min	0.006320	0.000000	0.460000	0.000000	0.385000	3.561000	2.900000	1.129600	1.000000	187.000000	12.600000	0.320000	1.730000	5.000000
25%	0.082045	0.000000	5.190000	0.000000	0.449000	5.885500	45.025000	2.100175	4.000000	279.000000	17.400000	375.377500	6.950000	17.025000
50%	0.256510	0.000000	9.690000	0.000000	0.538000	6.208500	77.500000	3.207450	5.000000	330.000000	19.050000	391.440000	11.360000	21.200000
75%	3.647423	12.500000	18.100000	0.000000	0.624000	6.623500	94.075000	5.188425	24.000000	666.000000	20.200000	396.225000	16.955000	25.000000
max	88.976200	100.000000	27.740000	1.000000	0.871000	8.780000	100.000000	12.126500	24.000000	711.000000	22.000000	396.900000	37.970000	50.000000

data.pivot_table(index='CHAS', values='y')

	y
CHAS
0.0	22.093843
1.0	28.440000

data['DIS'].plot.box()

# 判断每个字段与房价之间的关系
data.plot.scatter(x='RM', y='y')

训练回归模型，预测房屋价格

整理数据集（输入集、输出集）
打乱数据集，拆分测试集、训练集。
选择模型，使用训练集训练模型，用测试集测试。

# 整理数据集（输入集、输出集）
x, y = data.iloc[:, :-1], data['y']
# 打乱数据集。  su:  sklearn.utils
# random_state：随机种子。 若两次随机操作使用的随机种子相同，则随机结果也相同。
x, y = su.shuffle(x, y, random_state=7)
# 拆分测试集、训练集。
train_size = int(len(x) * 0.9)
train_x, test_x, train_y, test_y = \
    x[:train_size], x[train_size:], y[:train_size], y[train_size:]
train_x.shape, test_x.shape, train_y.shape, test_y.shape

((455, 13), (51, 13), (455,), (51,))

# 选择线性模型，使用训练集训练模型，用测试集测试。
import sklearn.linear_model as lm
import sklearn.metrics as sm

model = lm.LinearRegression()
model.fit(train_x, train_y)  # 针对训练集数据进行训练
pred_test_y = model.predict(test_x)  # 针对测试集数据进行测试
print(sm.r2_score(test_y, pred_test_y))
print(sm.mean_absolute_error(test_y, pred_test_y))

0.8188356183218533
2.405641089772746

# 选择岭回归模型，使用训练集训练模型，用测试集测试
model = lm.Ridge(3)
model.fit(train_x,train_y)# 针对训练集数据进行训练
pred_test_y = model.predict(test_x)# 针对测试集数据进行测试
print(sm.r2_score(test_y,pred_test_y)) 
print(sm.mean_absolute_error(test_y,pred_test_y))

0.8106201351332835
2.512282622308928

# 选择多项式回归模型，使用训练集训练模型，用测试集测试
import sklearn.pipeline as pl
import sklearn.preprocessing as sp
model = pl.make_pipeline(sp.PolynomialFeatures(2),lm.Ridge(50))
model.fit(train_x,train_y)# 针对训练集数据进行训练
pred_test_y = model.predict(test_x) # 针对测试集数据进行测试
print(sm.r2_score(test_y,pred_test_y))
print(sm.mean_absolute_error(test_y,pred_test_y))

0.8936132765852787
2.008395111436512

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做了那么多年开发，自学了很多门编程语言，我很明白学习资源对于学一门新语言的重要性，这些年也收藏了不少的Python干货，对我来说这些东西确实已经用不到了，但对于准备自学Python的人来说，或许它就是一个宝藏，可以给你省去很多的时间和精力。别在网上瞎学了，我最近也做了一些资源的更新，只要你是我的粉丝，这期福利你都可拿走。我先来介绍一下这些东西怎么用，文末抱走。（1）Python所有方向的学习路线（
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1.基本概念机器学习：发明算法将数据转化为智能行为数据挖掘VS机器学习：前者侧重寻找有价值的信息，后者侧重执行已知的任务。后者是前者的先期准备过程：数据——>抽象化——>一般化。或者：收集数据——推理数据——归纳数据——发现规律抽象化：训练：用一个特定模型来拟合数据集的过程用方程来拟合观测的数据：观测现象——数据呈现——模型建立。通过不同的格式来把信息概念化一般化：一般化：将抽象化的知识转换成可用
Python前沿技术：机器学习与人工智能 4.0啊 Python 人工智能 python 机器学习
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Python 初学者入门必知： Anaconda是什么？有什么作用？怎么使用？懒大王爱吃狼 Python基础 python 开发语言 python基础 python学习 anaconda anaconda安装 python教程
初学者在学习Python时，经常看到的一个名字是Anaconda。究竟什么是Anaconda，为什么它如此受欢迎？在这篇文章中，我们将探讨Anaconda，了解Anaconda的从安装到使用的。Anaconda是一个免费开源的Python和R编程发行版，包含上千个适用于数据科学和机器学习的包。同时，配备了Spyder和Jupyternotebook等工具，初学者可以使用它们来学习Python，使用
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一切皆是映射：AI的去中心化：区块链技术的融合作者：禅与计算机程序设计艺术/ZenandtheArtofComputerProgramming关键词：AI，区块链，去中心化，智能合约，共识机制，数据安全，隐私保护，分布式账本技术，机器学习，数据隐私1.背景介绍1.1问题的由来随着人工智能（AI）技术的快速发展，其在各个领域的应用越来越广泛，从自动驾驶、智能医疗到金融服务，AI正在改变着我们的生活。
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