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
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 + … + wd xd
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 + … + wd xd
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)
<matplotlib.axes._subplots.AxesSubplot at 0x172b9e28cf8>
# 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
24, w0:3.06937335, w1:2.82444853, loss:0.82515337
25, w0:3.09462895, w1:2.84335548, loss:0.73010724
26, w0:3.11778986, w1:2.86031549, loss:0.65140537
27, w0:3.13904648, w1:2.87550680, loss:0.58620385
28, w0:3.15857186, w1:2.88909135, loss:0.53215381
29, w0:3.17652325, w1:2.90121635, loss:0.48731526
30, w0:3.19304357, w1:2.91201557, loss:0.45008589
31, w0:3.20826270, w1:2.92161056, loss:0.41914232
32, w0:3.22229870, w1:2.93011181, loss:0.39339153
33, w0:3.23525885, w1:2.93761973, loss:0.37193074
34, w0:3.24724064, w1:2.94422555, loss:0.35401433
35, w0:3.25833268, w1:2.95001219, loss:0.33902647
36, w0:3.26861551, w1:2.95505501, loss:0.32645850
37, w0:3.27816232, w1:2.95942250, loss:0.31589033
38, w0:3.28703961, w1:2.96317688, loss:0.30697497
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40, w0:3.30302197, w1:2.96906741, loss:0.29300619
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854, w0:4.04254505, w1:2.28814070, loss:0.08820984
855, w0:4.04273961, w1:2.28793290, loss:0.08820174
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858, w0:4.04331978, w1:2.28731324, loss:0.08817776
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862, w0:4.04408524, w1:2.28649567, loss:0.08814645
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866, w0:4.04484153, w1:2.28568788, loss:0.08811588
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891, w0:4.04936763, w1:2.28085365, loss:0.08794064
892, w0:4.04954169, w1:2.28066775, loss:0.08793417
893, w0:4.04971523, w1:2.28048240, loss:0.08792773
894, w0:4.04988824, w1:2.28029760, loss:0.08792133
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896, w0:4.05023271, w1:2.27992969, loss:0.08790865
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898, w0:4.05057511, w1:2.27956397, loss:0.08789612
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900, w0:4.05091546, w1:2.27920046, loss:0.08788374
901, w0:4.05108486, w1:2.27901951, loss:0.08787760
902, w0:4.05125376, w1:2.27883912, loss:0.08787150
903, w0:4.05142215, w1:2.27865926, loss:0.08786544
904, w0:4.05159004, w1:2.27847995, loss:0.08785942
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906, w0:4.05192429, w1:2.27812294, loss:0.08784748
907, w0:4.05209067, w1:2.27794523, loss:0.08784156
908, w0:4.05225655, w1:2.27776806, loss:0.08783568
909, w0:4.05242193, w1:2.27759143, loss:0.08782983
910, w0:4.05258681, w1:2.27741532, loss:0.08782402
911, w0:4.05275119, w1:2.27723975, loss:0.08781825
912, w0:4.05291508, w1:2.27706470, loss:0.08781250
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915, w0:4.05340381, w1:2.27654270, loss:0.08779548
916, w0:4.05356574, w1:2.27636974, loss:0.08778988
917, w0:4.05372718, w1:2.27619731, loss:0.08778431
918, w0:4.05388814, w1:2.27602539, loss:0.08777877
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920, w0:4.05420861, w1:2.27568310, loss:0.08776779
921, w0:4.05436813, w1:2.27551273, loss:0.08776235
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925, w0:4.05500140, w1:2.27483635, loss:0.08774092
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930, w0:4.05578235, w1:2.27400223, loss:0.08771485
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933, w0:4.05624531, w1:2.27350776, loss:0.08769958
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937, w0:4.05685612, w1:2.27285536, loss:0.08767964
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939, w0:4.05715878, w1:2.27253209, loss:0.08766985
940, w0:4.05730943, w1:2.27237119, loss:0.08766500
941, w0:4.05745963, w1:2.27221077, loss:0.08766018
942, w0:4.05760937, w1:2.27205083, loss:0.08765538
943, w0:4.05775867, w1:2.27189137, loss:0.08765062
944, w0:4.05790751, w1:2.27173239, loss:0.08764588
945, w0:4.05805591, w1:2.27157389, loss:0.08764117
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954, w0:4.05937160, w1:2.27016863, loss:0.08760006
955, w0:4.05951560, w1:2.27001482, loss:0.08759563
956, w0:4.05965917, w1:2.26986148, loss:0.08759122
957, w0:4.05980230, w1:2.26970860, loss:0.08758684
958, w0:4.05994501, w1:2.26955618, loss:0.08758249
959, w0:4.06008729, w1:2.26940422, loss:0.08757816
960, w0:4.06022914, w1:2.26925271, loss:0.08757386
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967, w0:4.06121023, w1:2.26820483, loss:0.08754447
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969, w0:4.06148676, w1:2.26790947, loss:0.08753629
970, w0:4.06162441, w1:2.26776245, loss:0.08753224
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977, w0:4.06257642, w1:2.26674563, loss:0.08750457
978, w0:4.06271079, w1:2.26660211, loss:0.08750071
979, w0:4.06284475, w1:2.26645903, loss:0.08749687
980, w0:4.06297832, w1:2.26631637, loss:0.08749306
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987, w0:4.06390211, w1:2.26532969, loss:0.08746700
988, w0:4.06403250, w1:2.26519042, loss:0.08746336
989, w0:4.06416249, w1:2.26505158, loss:0.08745975
990, w0:4.06429210, w1:2.26491315, loss:0.08745616
991, w0:4.06442131, w1:2.26477514, loss:0.08745259
992, w0:4.06455014, w1:2.26463754, loss:0.08744904
993, w0:4.06467858, w1:2.26450035, loss:0.08744552
994, w0:4.06480664, w1:2.26436358, loss:0.08744201
995, w0:4.06493431, w1:2.26422722, loss:0.08743853
996, w0:4.06506160, w1:2.26409126, loss:0.08743506
997, w0:4.06518850, w1:2.26395572, loss:0.08743162
998, w0:4.06531502, w1:2.26382058, loss:0.08742820
999, w0:4.06544117, w1:2.26368585, loss:0.08742480
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)
<matplotlib.axes._subplots.AxesSubplot at 0x172ba5154a8>
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')
<matplotlib.axes._subplots.AxesSubplot at 0x24303391d30>
# 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)
<matplotlib.axes._subplots.AxesSubplot at 0x243076caf60>
# 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()
<matplotlib.legend.Legend at 0x1d666f9e780>
# 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()
<matplotlib.legend.Legend at 0x1d66760d9e8>
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()
<matplotlib.legend.Legend at 0x26d7642dcf8>
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
<matplotlib.legend.Legend at 0x26d7bfee668>
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()
<matplotlib.axes._subplots.AxesSubplot at 0x1dba88c7ef0>
# 判断每个字段与房价之间的关系
data.plot.scatter(x='RM', y='y')
<matplotlib.axes._subplots.AxesSubplot at 0x1dbabbafc88>
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
