【Python机器学习】实验02 线性回归

news2024/11/18 8:37:53

文章目录

  • 线性回归
    • 1. 单变量的线性回归
      • 1.1 数据读取
      • 1.2 训练数据的准备
      • 1.3 假设函数定义--假设函数是为了去预测
      • 1.4 损失函数的定义
      • 1.5 利用梯度下降算法来优化参数w
      • 1.6 可视化误差曲线
      • 1.7 可视化回归线/回归平面
    • 1.2 单变量的线性回归--基于sklearn试试?
    • 1.3 多变量线性回归
      • 实验要求1 准备训练数据
      • 实验要求2 调用前面的梯度下降算法
      • 实验要求3 绘制误差曲线
    • 1.4 最小二乘法求参数
    • 1.5 来点正则化?
      • 1.5.1 普通的线性回归
      • 1.5.2 岭回归
      • 1.5.3 Lasso回归
      • 实验要求4 手写代码实现单变量的L2正则化

线性回归

1. 单变量的线性回归

import pandas as pd 
import numpy as np
import matplotlib.pyplot as plt
plt.rcParams['font.sans-serif']=['SimHei'] #用来正常显示中文标签
plt.rcParams['axes.unicode_minus']=False #用来正常显示负号

1.1 数据读取

data=pd.read_csv("data/regress_data1.csv")
data.head()
人口收益
06.110117.5920
15.52779.1302
28.518613.6620
37.003211.8540
45.85986.8233
#可视化人口与收益之间的关系
data.plot(kind="scatter",x="人口",y="收益")
plt.xlabel("人口",fontsize=10)
plt.ylabel("收益",fontsize=10)
plt.title("人口与收益之间的关系")
Text(0.5, 1.0, '人口与收益之间的关系')

0

1.2 训练数据的准备

data.insert(0,"ones",1)
data
ones人口收益
016.110117.59200
115.52779.13020
218.518613.66200
317.003211.85400
415.85986.82330
............
9215.87077.20290
9315.30541.98690
9418.29340.14454
95113.39409.05510
9615.43690.61705

97 rows × 3 columns

col_num=data.shape[1]
m=data.shape[0]
#训练集中的特征
X=data.iloc[:,:col_num-1]
#训练集中的标签
y=data.iloc[:,col_num-1]
X=X.values
y=y.values
X.shape,y.shape
((97, 2), (97,))
y=y.reshape((m,1))
y.shape
(97, 1)
#初始化权重向量
w=np.zeros((col_num-1,1))
w.shape
(2, 1)

1.3 假设函数定义–假设函数是为了去预测

#估计yhat
def h(X,w):
    #X的维度m,col_num-1, w的维度col_num-1,1
    temp=X@w
    return temp

1.4 损失函数的定义

#定义MSE损失,均方损失函数
def cost(X,y,w):
    temp=h(X,w)
    cost=np.sum(np.square(temp-y))/(2*m)
    return cost
def computeCost(X,y,w):
    inner = np.power(((X @ w) - y), 2)# (m,n) @ (n, 1) -> (n, 1)
#     return np.sum(inner)/(2 * len(X))
    return np.sum(inner) / (2*m)
cost(X,y,w)
32.072733877455676
error=h(X,w)-y
error.shape
(97, 1)
x1=np.array([1,2]).reshape(2,1)
x2=np.array([3,4]).reshape(2,1)
np.multiply(x1,x2)
array([[3],
       [8]])
X[:,1].shape
(97,)
X.shape,w.shape,y.shape
((97, 2), (2, 1), (97, 1))
h(X,w)-y
array([[-17.592  ],
       [ -9.1302 ],
       [-13.662  ],
       [-11.854  ],
       [ -6.8233 ],
       [-11.886  ],
       [ -4.3483 ],
       [-12.     ],
       [ -6.5987 ],
       [ -3.8166 ],
       [ -3.2522 ],
       [-15.505  ],
       [ -3.1551 ],
       [ -7.2258 ],
       [ -0.71618],
       [ -3.5129 ],
       [ -5.3048 ],
       [ -0.56077],
       [ -3.6518 ],
       [ -5.3893 ],
       [ -3.1386 ],
       [-21.767  ],
       [ -4.263  ],
       [ -5.1875 ],
       [ -3.0825 ],
       [-22.638  ],
       [-13.501  ],
       [ -7.0467 ],
       [-14.692  ],
       [-24.147  ],
       [  1.22   ],
       [ -5.9966 ],
       [-12.134  ],
       [ -1.8495 ],
       [ -6.5426 ],
       [ -4.5623 ],
       [ -4.1164 ],
       [ -3.3928 ],
       [-10.117  ],
       [ -5.4974 ],
       [ -0.55657],
       [ -3.9115 ],
       [ -5.3854 ],
       [ -2.4406 ],
       [ -6.7318 ],
       [ -1.0463 ],
       [ -5.1337 ],
       [ -1.844  ],
       [ -8.0043 ],
       [ -1.0179 ],
       [ -6.7504 ],
       [ -1.8396 ],
       [ -4.2885 ],
       [ -4.9981 ],
       [ -1.4233 ],
       [  1.4211 ],
       [ -2.4756 ],
       [ -4.6042 ],
       [ -3.9624 ],
       [ -5.4141 ],
       [ -5.1694 ],
       [  0.74279],
       [-17.929  ],
       [-12.054  ],
       [-17.054  ],
       [ -4.8852 ],
       [ -5.7442 ],
       [ -7.7754 ],
       [ -1.0173 ],
       [-20.992  ],
       [ -6.6799 ],
       [ -4.0259 ],
       [ -1.2784 ],
       [ -3.3411 ],
       [  2.6807 ],
       [ -0.29678],
       [ -3.8845 ],
       [ -5.7014 ],
       [ -6.7526 ],
       [ -2.0576 ],
       [ -0.47953],
       [ -0.20421],
       [ -0.67861],
       [ -7.5435 ],
       [ -5.3436 ],
       [ -4.2415 ],
       [ -6.7981 ],
       [ -0.92695],
       [ -0.152  ],
       [ -2.8214 ],
       [ -1.8451 ],
       [ -4.2959 ],
       [ -7.2029 ],
       [ -1.9869 ],
       [ -0.14454],
       [ -9.0551 ],
       [ -0.61705]])
np.multiply((h(X,w)-y).ravel(),X[:,1]).shape
(97,)

1.5 利用梯度下降算法来优化参数w

#超参数为I,学习率alpha,对所有样本
def gradient_descent(X,y,w,iter_num,alpha):
    temp=np.zeros((col_num-1,1))
    cost_lst=[] 
    for i in range(iter_num):
        error=h(X,w)-y
        for j in range(col_num-1):
            incre=np.multiply(error.ravel(),X[:,j].ravel())
            temp[j,0]=w[j,0]-((alpha/m)*np.sum(incre))
        w=temp
        cost_lst.append(cost(X,y,w))
    return w,cost_lst        
iter_num=200
alpha=0.003
w=np.zeros((col_num-1,1))
w,cost_lst=gradient_descent(X,y,w,iter_num,alpha)
w
array([[-0.32791203],
       [ 0.83460252]])
cost
<function __main__.cost(X, y, w)>

1.6 可视化误差曲线

plt.plot(range(iter_num),cost_lst,"r-+")
plt.xlabel("迭代次数")
plt.ylabel("误差")
plt.show()

1

1.7 可视化回归线/回归平面

x=np.linspace(data["人口"].min(),data["人口"].max(),50)
y1=w[0,0]*1+w[1,0]*x
plt.plot(x,y1,"r-+",label="预测线")
plt.scatter(data["人口"],data["收益"], label='训练数据')
plt.xlabel("人口",fontsize=10)
plt.ylabel("收益",fontsize=10)
plt.title("人口与收益之间的关系")
plt.show()

2

w
array([[-0.32791203],
       [ 0.83460252]])

总结:

  1. 数据准备
  2. 初始化w
  3. 定义了假设函数
  4. 定义了损失函数或者代价函数
  5. 定义梯度下降算法
  6. 可视化分析

1.2 单变量的线性回归–基于sklearn试试?

X.shape,y.shape
((97, 2), (97, 1))
import sklearn
from sklearn import linear_model
reg=linear_model.LinearRegression()
reg.fit(X,y)
reg.coef_
array([[0.        , 1.19303364]])
w
array([[-0.32791203],
       [ 0.83460252]])
reg.intercept_
array([-3.89578088])
reg.get_params()
{'copy_X': True,
 'fit_intercept': True,
 'n_jobs': None,
 'normalize': 'deprecated',
 'positive': False}
reg.predict(X)-y
array([[-14.19822601],
       [ -6.4312488 ],
       [ -7.39480448],
       [ -7.39472766],
       [ -3.72814233],
       [ -5.78069914],
       [  0.67551586],
       [ -5.66181898],
       [ -2.75622606],
       [ -1.68207302],
       [ -0.33492365],
       [ -2.50265234],
       [ -0.21002596],
       [ -1.09007678],
       [  2.117584  ],
       [ -0.99087569],
       [ -1.60644452],
       [  1.66383102],
       [  0.12314824],
       [ -0.84937859],
       [  0.34942365],
       [ -1.47998891],
       [ -1.60890687],
       [ -1.53603074],
       [ -0.33916795],
       [ -3.93175849],
       [ -2.09254529],
       [  2.12958876],
       [ -2.86836958],
       [ -1.55385488],
       [  3.59050903],
       [ -2.03100498],
       [ -4.99636713],
       [  1.28383475],
       [ -0.64226232],
       [  1.00673223],
       [  1.6465002 ],
       [ -0.60007636],
       [  1.30099898],
       [ -1.81336092],
       [  1.99826273],
       [  0.40377318],
       [  4.68685703],
       [  0.55183747],
       [ -1.29245052],
       [  3.52022606],
       [ -2.9805617 ],
       [  1.18148451],
       [  2.05841276],
       [  1.69763436],
       [ -1.65046859],
       [  0.59688379],
       [  0.67268159],
       [  0.17687322],
       [  2.23616258],
       [  5.11170076],
       [  1.11395081],
       [ -1.77162904],
       [  3.24920096],
       [  1.96858198],
       [  1.46381825],
       [  3.02608828],
       [  3.56178204],
       [  1.83596469],
       [  1.66894398],
       [ -0.16942543],
       [  0.2563525 ],
       [  0.5407115 ],
       [  1.64788834],
       [ -0.62028352],
       [  1.51690814],
       [  0.82862438],
       [  1.9914178 ],
       [  1.38386093],
       [  4.78217995],
       [  3.61930412],
       [  1.21352255],
       [ -3.58846693],
       [  1.60884678],
       [  0.14027707],
       [  2.45981748],
       [  2.08994488],
       [  3.00817305],
       [  0.21510688],
       [ -1.46569296],
       [  2.02402528],
       [  0.25840658],
       [  2.33785705],
       [  2.53824205],
       [ -0.68114646],
       [  1.06859725],
       [  0.91903985],
       [ -4.09473826],
       [  0.44683982],
       [  5.85398435],
       [  3.02861175],
       [  1.97357374]])
reg.score(X,y)
0.7020315537841397

1.3 多变量线性回归

path = 'data/regress_data2.csv'
data2 = pd.read_csv(path)
data2.head()
面积房间数价格
021043399900
116003329900
224003369000
314162232000
430004539900
data2=(data2-data2.mean())/data2.std()
data2.head()
面积房间数价格
00.130010-0.2236750.475747
1-0.504190-0.223675-0.084074
20.502476-0.2236750.228626
3-0.735723-1.537767-0.867025
41.2574761.0904171.595389

实验要求1 准备训练数据

data2.insert(0,"ones",1)
col_num2=data2.shape[1]
m2=data2.shape[0]
X2=data2.iloc[:,:-1].values
y2=data2.iloc[:,-1].values.reshape((data2.shape[0],1))
w2=np.zeros((X2.shape[1],1))
X2.shape,y2.shape,w2.shape
((47, 3), (47, 1), (3, 1))

实验要求2 调用前面的梯度下降算法

#定义MSE损失,均方损失函数
def cost2(X,y,w):
    temp=h(X,w)
    cost=np.sum(np.square(temp-y))/(2*m2)
    return cost
#超参数为I,学习率alpha,对所有样本
def gradient_descent(X,y,w,iter_num,alpha):
    temp=np.zeros((col_num2-1,1))
    cost_lst=[] 
    for i in range(iter_num):
        error=h(X,w)-y
        for j in range(col_num2-1):
            incre=np.multiply(error.ravel(),X[:,j].ravel())
            temp[j,0]=w[j,0]-((alpha/m2)*np.sum(incre))
        w=temp
        cost_lst.append(cost2(X,y,w))
    return w,cost_lst        
iter_num2=1000
alpha2=0.01
w2,cost_lst2=gradient_descent(X2,y2,w2,iter_num2,alpha2)
w2
array([[-1.03191687e-16],
       [ 8.78503652e-01],
       [-4.69166570e-02]])
cost_lst2
[0.4805491041076719,
 0.47198587701203876,
 0.46366461618706284,
 0.4555781400525299,
 0.44771948335326117,
 0.4400818906150644,
 0.43265880979889004,
 0.42544388614718714,
 0.41843095621663473,
 0.4116140420916035,
 0.4049873457728717,
 0.39854524373628347,
 0.3922822816562035,
 0.38619316928877434,
 0.3802727755101314,
 0.3745161235048873,
 0.36891838610032585,
 0.36347488124189714,
 0.3581810676057273,
 0.353032540343996,
 0.34802502695915444,
 0.3431543833030803,
 0.33841658969738386,
 0.3338077471711977,
 0.3293240738128865,
 0.32496190123222957,
 0.32071767112972566,
 0.3165879319697778,
 0.3125693357546089,
 0.3086586348958572,
 0.3048526791808924,
 0.301148412830983,
 0.29754287164853055,
 0.29403318025067643,
 0.290616549386659,
 0.2872902733363892,
 0.2840517273877804,
 0.2808983653904495,
 0.2778277173834725,
 0.2748373872949541,
 0.27192505071123485,
 0.2690884527136251,
 0.26632540578062175,
 0.26363378775362334,
 0.2610115398642199,
 0.2584566648211922,
 0.25596722495541263,
 0.2535413404208921,
 0.2511771874502738,
 0.24887299666312288,
 0.2466270514254147,
 0.2444376862586688,
 0.2423032852972262,
 0.24022228079221122,
 0.23819315166076374,
 0.23621442207916982,
 0.23428466011856308,
 0.23240247642190492,
 0.2305665229209955,
 0.22877549159230046,
 0.22702811325042083,
 0.2253231563780625,
 0.2236594259914031,
 0.22203576253978147,
 0.22045104083867165,
 0.21890416903493287,
 0.2173940876033578,
 0.2159197683735719,
 0.21448021358636368,
 0.21307445497855435,
 0.21170155289554488,
 0.21036059543069827,
 0.20905069759074849,
 0.2077710004864442,
 0.20652067054766643,
 0.20529889876227617,
 0.20410489993797507,
 0.20293791198648126,
 0.20179719522934592,
 0.20068203172475318,
 0.1995917246146703,
 0.19852559749172968,
 0.1974829937852473,
 0.19646327616579626,
 0.19546582596777473,
 0.1944900426294234,
 0.1935353431497636,
 0.19260116156194368,
 0.1916869484224977,
 0.1907921703160338,
 0.189916309374886,
 0.18905886281327441,
 0.18821934247553765,
 0.1873972743980089,
 0.1865921983841233,
 0.185803667592357,
 0.1850312481366081,
 0.1842745186986435,
 0.18353307015224665,
 0.18280650519871142,
 0.18209443801333897,
 0.18139649390260434,
 0.1807123089716706,
 0.18004152980193594,
 0.17938381313831125,
 0.17873882558593354,
 0.17810624331602853,
 0.17748575178064738,
 0.1768770454360068,
 0.17627982747417442,
 0.17569380956284517,
 0.17511871159296452,
 0.17455426143396124,
 0.17400019469635858,
 0.17345625450154267,
 0.17292219125846853,
 0.17239776244709656,
 0.1718827324083545,
 0.1713768721404272,
 0.170879959101184,
 0.17039177701655678,
 0.16991211569468956,
 0.16944077084568412,
 0.16897754390677372,
 0.16852224187275908,
 0.16807467713154944,
 0.16763466730465196,
 0.16720203509246181,
 0.16677660812420697,
 0.16635821881240656,
 0.1659467042117072,
 0.165541905881964,
 0.16514366975543857,
 0.16475184600798926,
 0.16436628893413294,
 0.16398685682586134,
 0.1636134118550983,
 0.16324581995968843,
 0.1628839507328093,
 0.16252767731570503,
 0.16217687629363958,
 0.16183142759497396,
 0.16149121439327116,
 0.1611561230123392,
 0.1608260428341213,
 0.1605008662093497,
 0.16018048837087778,
 0.15986480734960967,
 0.15955372389294997,
 0.1592471413856969,
 0.1589449657733043,
 0.1586471054874422,
 0.1583534713737859,
 0.1580639766219659,
 0.15777853669761477,
 0.15749706927644566,
 0.15721949418030304,
 0.1569457333151249,
 0.15667571061075897,
 0.15640935196257785,
 0.15614658517483707,
 0.15588733990572493,
 0.1556315476140529,
 0.15537914150753623,
 0.15513005649261752,
 0.15488422912578662,
 0.15464159756635176,
 0.15440210153061742,
 0.15416568224742783,
 0.15393228241503384,
 0.15370184615924348,
 0.15347431899281805,
 0.15324964777607533,
 0.15302778067866443,
 0.15280866714247615,
 0.1525922578456555,
 0.15237850466768188,
 0.15216736065548658,
 0.15195877999057447,
 0.1517527179571211,
 0.1515491309110149,
 0.15134797624981636,
 0.15114921238360698,
 0.1509527987067004,
 0.15075869557019017,
 0.15056686425530957,
 0.15037726694757766,
 0.15018986671170936,
 0.15000462746726564,
 0.1498215139650219,
 0.1496404917640332,
 0.14946152720937447,
 0.14928458741053677,
 0.14910964022045875,
 0.14893665421517455,
 0.14876559867406025,
 0.14859644356065968,
 0.1484291595040733,
 0.1482637177808928,
 0.14810009029766483,
 0.1479382495738683,
 0.14777816872538996,
 0.1476198214484827,
 0.1474631820041929,
 0.1473082252032421,
 0.14715492639134978,
 0.14700326143498368,
 0.14685320670752527,
 0.14670473907583773,
 0.14655783588722407,
 0.14641247495676463,
 0.1462686345550211,
 0.14612629339609778,
 0.14598543062604827,
 0.14584602581161768,
 0.14570805892930994,
 0.14557151035477142,
 0.14543636085247996,
 0.1453025915657318,
 0.14517018400691611,
 0.1450391200480697,
 0.14490938191170247,
 0.14478095216188633,
 0.14465381369559951,
 0.14452794973431848,
 0.1444033438158502,
 0.14427997978639776,
 0.14415784179285188,
 0.14403691427530232,
 0.14391718195976197,
 0.14379862985109793,
 0.14368124322616269,
 0.14356500762711993,
 0.14344990885495965,
 0.14333593296319558,
 0.1432230662517408,
 0.1431112952609561,
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 0.13072046350250424,
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 0.13070366371231526,
 0.13070351602534264,
 0.13070336960771892]

实验要求3 绘制误差曲线

plt.plot(range(iter_num2),cost_lst2,"r-+")
plt.xlabel("迭代次数")
plt.ylabel("误差")
plt.show()

3

1.4 最小二乘法求参数

最小二乘法的需要求解最优参数 w ∗ w^{*} w

已知:目标函数

J ( w ) = 1 2 m ∑ i = 1 m ( h ( x ( i ) ) − y ( i ) ) 2 J\left( w \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {h}\left( {x^{(i)}} \right)-{y^{(i)}} \right)}^{2}}} J(w)=2m1i=1m(h(x(i))y(i))2

其中: h ( x ) = w T X = w 0 x 0 + w 1 x 1 + w 2 x 2 + . . . + w n x n {h}\left( x \right)={w^{T}}X={w_{0}}{x_{0}}+{w_{1}}{x_{1}}+{w_{2}}{x_{2}}+...+{w_{n}}{x_{n}} h(x)=wTX=w0x0+w1x1+w2x2+...+wnxn

将向量表达形式转为矩阵表达形式,则有 J ( w ) = 1 2 ( X w − y ) 2 J(w )=\frac{1}{2}{{\left( Xw -y\right)}^{2}} J(w)=21(Xwy)2 ,其中 X X X m m m n + 1 n+1 n+1列的矩阵( m m m为样本个数, n n n为特征个数), w w w n + 1 n+1 n+1行1列的矩阵(包含了 w 0 w_0 w0), y y y m m m行1列的矩阵,则可以求得最优参数 w ∗ = ( X T X ) − 1 X T y w^{*} ={{\left( {X^{T}}X \right)}^{-1}}{X^{T}}y w=(XTX)1XTy

梯度下降与最小二乘法的比较:

梯度下降:需要选择学习率 α \alpha α,需要多次迭代,当特征数量 n n n大时也能较好适用,适用于各种类型的模型

最小二乘法:不需要选择学习率 α \alpha α,一次计算得出,需要计算 ( X T X ) − 1 {{\left( {{X}^{T}}X \right)}^{-1}} (XTX)1,如果特征数量 n n n较大则运算代价大,因为矩阵逆的计算时间复杂度为 O ( n 3 ) O(n^3) O(n3),通常来说当 n n n小于10000 时还是可以接受的,只适用于线性模型,不适合逻辑回归模型等其他模型

def lsm(X,y):
    w=np.linalg.inv(X.T@X)@X.T@y
    return w
def lsm_v(X,y):
    w=np.linalg.inv(np.dot(X.T,X))
    w=np.dot(w,X.T)
    w=np.dot(w,y)
    return w
lsm(X,y)
array([[-3.89578088],
       [ 1.19303364]])
lsm_v(X,y)
array([[-3.89578088],
       [ 1.19303364]])

1.5 来点正则化?

1.5.1 普通的线性回归

from sklearn import linear_model
reg=linear_model.LinearRegression()
reg.fit(X,y)
LinearRegression()
#回到单变量的线性回归中来
x=X
y_1=reg.predict(x)
plt.plot(x,y_1,"r-+",label="预测线")
plt.scatter(data["人口"],data["收益"], label='训练数据')
plt.xlim(4.7,10)
plt.xlabel("人口",fontsize=10)
plt.ylabel("收益",fontsize=10)
plt.title("人口与收益之间的关系")
plt.show()

4

reg.coef_,reg.intercept_,reg.score(X,y)
(array([[0.        , 1.19303364]]), array([-3.89578088]), 0.7020315537841397)

1.5.2 岭回归

J ( w ) = 1 2 ∑ i = 1 m ( h w ( x ( i ) ) − y ( i ) ) 2 + λ ∑ j = 1 n w j 2 J ( { w } ) = \frac { 1 } { 2 } \sum _ { i = 1 } ^ { m } ( h _ { w} ( x ^ { ( i ) } ) - y ^ { ( i ) } ) ^ { 2 } + \lambda \sum _ { j = 1 } ^ { n } w_ { j } ^ { 2 } J(w)=21i=1m(hw(x(i))y(i))2+λj=1nwj2,此时称作Ridge Regression

from sklearn import linear_model
reg_rigde=linear_model.Ridge()
reg_rigde.fit(X,y)
Ridge()
#回到单变量的线性回归中来,Ridge
x=X
y_1=reg_rigde.predict(x)
plt.plot(x,y_1,"r-+",label="预测线")
plt.scatter(data["人口"],data["收益"], label='训练数据')
plt.xlim(4.7,10)
plt.xlabel("人口",fontsize=10)
plt.ylabel("收益",fontsize=10)
plt.title("人口与收益之间的关系")
plt.show()

5

reg_rigde.coef_,reg_rigde.intercept_,reg_rigde.score(X,y)
(array([[0.       , 1.1922044]]), array([-3.88901439]), 0.7020312146131912)

1.5.3 Lasso回归

J ( w ) = 1 2 ∑ i = 1 m ( h w ( x ( i ) ) − y ( i ) ) 2 + λ ∑ j = 1 n ∣ w j ∣ J ( {w } ) = \frac { 1 } { 2 } \sum _ { i = 1 } ^ { m } ( h _ { w} ( x ^ { ( i ) } ) - y ^ { ( i ) } ) ^ { 2 } + \lambda \sum _ { j = 1 } ^ { n } | w _ { j } | J(w)=21i=1m(hw(x(i))y(i))2+λj=1nwj,此时称作Lasso Regression

from sklearn import linear_model
reg_lasso=linear_model.Lasso()
reg_lasso.fit(X,y)
Lasso()
#回到单变量的线性回归中来,Lasso
x=X
y_1=reg_lasso.predict(x)
plt.plot(x,y_1,"r-+",label="预测线")
plt.scatter(data["人口"],data["收益"], label='训练数据')
plt.xlim(4.7,10)
plt.xlabel("人口",fontsize=10)
plt.ylabel("收益",fontsize=10)
plt.title("人口与收益之间的关系")
plt.show()

6

reg_lasso.coef_,reg_lasso.intercept_,reg_lasso.score(X,y)
(array([0.        , 1.12556458]), array([-3.34524677]), 0.6997863246152711)

实验要求4 手写代码实现单变量的L2正则化

J ( w ) = 1 2 ∑ i = 1 m ( h w ( x ( i ) ) − y ( i ) ) 2 + λ ∑ j = 1 n w j 2 J ( { w } ) = \frac { 1 } { 2 } \sum _ { i = 1 } ^ { m } ( h _ { w} ( x ^ { ( i ) } ) - y ^ { ( i ) } ) ^ { 2 } + \lambda \sum _ { j = 1 } ^ { n } w_ { j } ^ { 2 } J(w)=21i=1m(hw(x(i))y(i))2+λj=1nwj2,此时称作Ridge Regression

#超参数为I,学习率alpha,对所有样本
def gradient_descent_l2(X,y,w,iter_num,alpha,lambd):
    temp=np.zeros((col_num-1,1))
    cost_lst=[] 
    for i in range(iter_num):
        error=h(X,w)-y
        for j in range(col_num-1):
            incre=np.multiply(error.ravel(),X[:,j].ravel())
            temp[j,0]=w[j,0]-((alpha/m)*(np.sum(incre)+2*lambd*w[j,0]))
        w=temp
        cost_lst.append(cost(X,y,w))      
    return w,cost_lst  
iter_num=200
alpha=0.001
lambd=2
w=np.zeros((col_num-1,1))
w,cost_lst=gradient_descent_l2(X,y,w,iter_num,alpha,lambd)
plt.plot(range(iter_num),cost_lst,"r-+")
plt.xlabel("迭代次数")
plt.ylabel("误差")
plt.show()

7

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