文章目录
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 | |
|---|---|---|
| 0 | 6.1101 | 17.5920 |
| 1 | 5.5277 | 9.1302 |
| 2 | 8.5186 | 13.6620 |
| 3 | 7.0032 | 11.8540 |
| 4 | 5.8598 | 6.8233 |
#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')
1.2 训练数据的准备
data.insert(0,"ones",1)
data
| ones | 人口 | 收益 | |
|---|---|---|---|
| 0 | 1 | 6.1101 | 17.59200 |
| 1 | 1 | 5.5277 | 9.13020 |
| 2 | 1 | 8.5186 | 13.66200 |
| 3 | 1 | 7.0032 | 11.85400 |
| 4 | 1 | 5.8598 | 6.82330 |
| ... | ... | ... | ... |
| 92 | 1 | 5.8707 | 7.20290 |
| 93 | 1 | 5.3054 | 1.98690 |
| 94 | 1 | 8.2934 | 0.14454 |
| 95 | 1 | 13.3940 | 9.05510 |
| 96 | 1 | 5.4369 | 0.61705 |
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()
1.7 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()
w
array([[-0.32791203],
[ 0.83460252]])
总结：
- 数据准备
- 初始化w
- 定义了假设函数
- 定义了损失函数或者代价函数
- 定义梯度下降算法
- 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 | |
|---|---|---|---|
| 0 | 2104 | 3 | 399900 |
| 1 | 1600 | 3 | 329900 |
| 2 | 2400 | 3 | 369000 |
| 3 | 1416 | 2 | 232000 |
| 4 | 3000 | 4 | 539900 |
data2=(data2-data2.mean())/data2.std()
data2.head()
| 面积 | 房间数 | 价格 | |
|---|---|---|---|
| 0 | 0.130010 | -0.223675 | 0.475747 |
| 1 | -0.504190 | -0.223675 | -0.084074 |
| 2 | 0.502476 | -0.223675 | 0.228626 |
| 3 | -0.735723 | -1.537767 | -0.867025 |
| 4 | 1.257476 | 1.090417 | 1.595389 |
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,
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0.2466270514254147,
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0.2423032852972262,
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0.23819315166076374,
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0.13071117243244995,
0.1307109602090767,
0.13071074980973368,
0.13071054121874362,
0.13071033442056404,
0.13071012939978588,
0.1307099261411327,
0.13070972462945923,
0.13070952484975046,
0.1307093267871204,
0.13070913042681098,
0.130708935754191,
0.13070874275475497,
0.1307085514141222,
0.13070836171803538,
0.13070817365235993,
0.1307079872030827,
0.13070780235631096,
0.1307076190982714,
0.1307074374153091,
0.13070725729388652,
0.13070707872058232,
0.1307069016820908,
0.13070672616522036,
0.13070655215689286,
0.13070637964414267,
0.13070620861411547,
0.1307060390540674,
0.13070587095136435,
0.1307057042934805,
0.1307055390679979,
0.13070537526260517,
0.13070521286509698,
0.13070505186337264,
0.13070489224543566,
0.1307047339993925,
0.13070457711345204,
0.13070442157592427,
0.13070426737521984,
0.1307041144998489,
0.1307039629384204,
0.13070381267964126,
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()
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=1∑m(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(Xw−y)2 ，其中 X X X为 m m m行 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为 n + 1 n+1 n+1行1列的矩阵(包含了 w 0 w_0 w0)， y y y为 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()
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)=21∑i=1m(hw(x(i))−y(i))2+λ∑j=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()
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)=21∑i=1m(hw(x(i))−y(i))2+λ∑j=1n∣wj∣，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()
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)=21∑i=1m(hw(x(i))−y(i))2+λ∑j=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()
