吴恩达机器学习中文版课后题（中文题目+数据集+python版答案）week2 逻辑回归
题目一：
你将建立一个逻辑回归模型来预测一个学生是否被大学录取。假设你是一所大学系的管理员，你想根据两次考试的成绩来决定每个申请人的录取机会。您有以前申请者的历史数据，可以用作逻辑回归的训练集。对于每个培训示例，您都有申请人在两次考试中的分数和录取决定。你的任务是建立一个分类模型，根据这两次考试的分数来估计申请人的录取概率。
数据集：
34.62365962451697,78.0246928153624,0
30.28671076822607,43.89499752400101,0
35.84740876993872,72.90219802708364,0
60.18259938620976,86.30855209546826,1
79.0327360507101,75.3443764369103,1
45.08327747668339,56.3163717815305,0
61.10666453684766,96.51142588489624,1
75.02474556738889,46.55401354116538,1
76.09878670226257,87.42056971926803,1
84.43281996120035,43.53339331072109,1
95.86155507093572,38.22527805795094,0
75.01365838958247,30.60326323428011,0
82.30705337399482,76.48196330235604,1
69.36458875970939,97.71869196188608,1
39.53833914367223,76.03681085115882,0
53.9710521485623,89.20735013750205,1
69.07014406283025,52.74046973016765,1
67.94685547711617,46.67857410673128,0
70.66150955499435,92.92713789364831,1
76.97878372747498,47.57596364975532,1
67.37202754570876,42.83843832029179,0
89.67677575072079,65.79936592745237,1
50.534788289883,48.85581152764205,0
34.21206097786789,44.20952859866288,0
77.9240914545704,68.9723599933059,1
62.27101367004632,69.95445795447587,1
80.1901807509566,44.82162893218353,1
93.114388797442,38.80067033713209,0
61.83020602312595,50.25610789244621,0
38.78580379679423,64.99568095539578,0
61.379289447425,72.80788731317097,1
85.40451939411645,57.05198397627122,1
52.10797973193984,63.12762376881715,0
52.04540476831827,69.43286012045222,1
40.23689373545111,71.16774802184875,0
54.63510555424817,52.21388588061123,0
33.91550010906887,98.86943574220611,0
64.17698887494485,80.90806058670817,1
74.78925295941542,41.57341522824434,0
34.1836400264419,75.2377203360134,0
83.90239366249155,56.30804621605327,1
51.54772026906181,46.85629026349976,0
94.44336776917852,65.56892160559052,1
82.36875375713919,40.61825515970618,0
51.04775177128865,45.82270145776001,0
62.22267576120188,52.06099194836679,0
77.19303492601364,70.45820000180959,1
97.77159928000232,86.7278223300282,1
62.07306379667647,96.76882412413983,1
91.56497449807442,88.69629254546599,1
79.94481794066932,74.16311935043758,1
99.2725269292572,60.99903099844988,1
90.54671411399852,43.39060180650027,1
34.52451385320009,60.39634245837173,0
50.2864961189907,49.80453881323059,0
49.58667721632031,59.80895099453265,0
97.64563396007767,68.86157272420604,1
32.57720016809309,95.59854761387875,0
74.24869136721598,69.82457122657193,1
71.79646205863379,78.45356224515052,1
75.3956114656803,85.75993667331619,1
35.28611281526193,47.02051394723416,0
56.25381749711624,39.26147251058019,0
30.05882244669796,49.59297386723685,0
44.66826172480893,66.45008614558913,0
66.56089447242954,41.09209807936973,0
40.45755098375164,97.53518548909936,1
49.07256321908844,51.88321182073966,0
80.27957401466998,92.11606081344084,1
66.74671856944039,60.99139402740988,1
32.72283304060323,43.30717306430063,0
64.0393204150601,78.03168802018232,1
72.34649422579923,96.22759296761404,1
60.45788573918959,73.09499809758037,1
58.84095621726802,75.85844831279042,1
99.82785779692128,72.36925193383885,1
47.26426910848174,88.47586499559782,1
50.45815980285988,75.80985952982456,1
60.45555629271532,42.50840943572217,0
82.22666157785568,42.71987853716458,0
88.9138964166533,69.80378889835472,1
94.83450672430196,45.69430680250754,1
67.31925746917527,66.58935317747915,1
57.23870631569862,59.51428198012956,1
80.36675600171273,90.96014789746954,1
68.46852178591112,85.59430710452014,1
42.0754545384731,78.84478600148043,0
75.47770200533905,90.42453899753964,1
78.63542434898018,96.64742716885644,1
52.34800398794107,60.76950525602592,0
94.09433112516793,77.15910509073893,1
90.44855097096364,87.50879176484702,1
55.48216114069585,35.57070347228866,0
74.49269241843041,84.84513684930135,1
89.84580670720979,45.35828361091658,1
83.48916274498238,48.38028579728175,1
42.2617008099817,87.10385094025457,1
99.31500880510394,68.77540947206617,1
55.34001756003703,64.9319380069486,1
74.77589300092767,89.52981289513276,1
python代码实现：
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.optimize as opt#读取数据集
path = "C:/Users/Administrator/Desktop/吴恩达机器学习数据集/week1/ex2data1.txt"
data = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
print(data.head())#找出结果分别为1和0的数据集并且赋值以名称，
positive = data[data['Admitted'].isin([1])] #取Admitted这一行，用isin函数查看是否存在1，若存在返回布尔值true1
negative = data[data['Admitted'].isin([0])]#定义代价函数
def sigmoid(z):return 1/(1 + np.exp(-z))def computeCost(theta, X, y):theta = np.matrix(theta)X = np.matrix(X)y = np.matrix(y)first = np.multiply(y, np.log(sigmoid(X * theta.T)))second = np.multiply((1-y), np.log(1-sigmoid(X * theta.T)))return -(np.sum(first + second) / (len(X)))#初始化数据
data.insert(0, 'Ones', 1)
cols = data.shape[1]
X = data.iloc[:, 0:cols-1]
y = data.iloc[:, cols-1:cols]
theta = np.zeros(3)
X = np.matrix(X.values)
y = np.matrix(y.values)
print("计算机初始代价：")
print(computeCost(theta, X, y))#高级优化算法 共轭梯度法求theta,先对梯度进行计算（此处使用函数名为gredientDecent，其实这个函数并没有下降theta的作用），循环次数为theta值的个数，计算出的就是每个theta值对应的梯度（代价函数的导数），计算后返回计算出的梯度向量。
def gredientDecent(theta, X, y):X = np.matrix(X)y = np.matrix(y)theta = np.matrix(theta)parameters =int(theta.ravel().shape[1])#多维降一维grad = np.zeros(parameters)error = sigmoid(X * theta.T) - yfor i in range(parameters):term = np.multiply(error, X[:, i])grad[i] = np.sum(term) / len(X)return grad
#Scipy高级优化算法求theta[0]：fmin_tnc()有约束的多元函数问题，提供梯度信息，使用截断牛顿法。'''func：优化的目标函数x0：初值fprime：提供优化函数func的梯度函数，不然优化函数func必须返回函数值和梯度，或者设置approx_grad=Trueapprox_grad :如果设置为True，会给出近似梯度args：元组，是传递给优化函数的参数'''
result = opt.fmin_tnc (func = computeCost, x0 = theta, fprime = gredientDecent, args = (X, y))
print("返回高级算法的结果：")
print(result)
print("计算合适theta值后的代价为：")
print(computeCost(result[0], X, y))#绘制散点图，并且绘制决策线
plotting_x1 = np.linspace(30, 100, 100) #构造等差数列，在30到100分之间，分隔100份
plotting_h1 = (-(result[0][0] + result[0][1] * plotting_x1)) / result[0][2]
#根据h1公式计算，-（theta[0]+theta[1]*x）/theta[2]，此公式可以得到散点图的决策边界，是提前给的fig,ax = plt.subplots(figsize = (8,6)) #8行6列
ax.plot(plotting_x1, plotting_h1, 'y', label = 'Prediction')
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=50, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
plt.show()#预测通过的概率
def hfunc(theta, X):return sigmoid(X * theta.T)
pre = [1,45,85]
pre = np.matrix(pre)
the = np.matrix(result[0])
print("该同学通过的概率为：", hfunc(the, pre))#计算准确率
def predict(theta, X):probability = sigmoid(X * theta.T) #theta为1*3，X为100*3,theta.T为3*1矩阵return [1 if x >= 0.5 else 0 for x in probability] #如果x>0.5,则输出1，否则x位于probability则为0theta_min = np.matrix(result[0])
predictions = predict(theta_min, X) #此函数theta_min为1*3的矩阵，X必须在之前设置为array或matrix，否则dataframe不可用，为100*3
correct = [1 if ((a == 1 and b == 1) or (a == 0 and b == 0)) else 0 for (a, b) in zip(predictions, y)]
accuracy = (sum(map(int, correct)) % len(correct))
print ('accuracy = {0}%'.format(accuracy))
题目二：
正则化逻辑回归在练习的这一部分，您将实施正则化逻辑回归来预测制造工厂的微芯片是否通过质量保证。在质量保证过程中，每个微芯片都要经过各种测试，以确保其功能正常。假设你是工厂的产品经理，你有两个不同测试的一些微芯片的测试结果。从这两个测试中，你可以决定微芯片应该被接受还是被拒绝。为了帮助您做出决定，您有一个过去微芯片的测试结果数据集，从中可以构建逻辑回归模型
数据集：
0.051267,0.69956,1
-0.092742,0.68494,1
-0.21371,0.69225,1
-0.375,0.50219,1
-0.51325,0.46564,1
-0.52477,0.2098,1
-0.39804,0.034357,1
-0.30588,-0.19225,1
0.016705,-0.40424,1
0.13191,-0.51389,1
0.38537,-0.56506,1
0.52938,-0.5212,1
0.63882,-0.24342,1
0.73675,-0.18494,1
0.54666,0.48757,1
0.322,0.5826,1
0.16647,0.53874,1
-0.046659,0.81652,1
-0.17339,0.69956,1
-0.47869,0.63377,1
-0.60541,0.59722,1
-0.62846,0.33406,1
-0.59389,0.005117,1
-0.42108,-0.27266,1
-0.11578,-0.39693,1
0.20104,-0.60161,1
0.46601,-0.53582,1
0.67339,-0.53582,1
-0.13882,0.54605,1
-0.29435,0.77997,1
-0.26555,0.96272,1
-0.16187,0.8019,1
-0.17339,0.64839,1
-0.28283,0.47295,1
-0.36348,0.31213,1
-0.30012,0.027047,1
-0.23675,-0.21418,1
-0.06394,-0.18494,1
0.062788,-0.16301,1
0.22984,-0.41155,1
0.2932,-0.2288,1
0.48329,-0.18494,1
0.64459,-0.14108,1
0.46025,0.012427,1
0.6273,0.15863,1
0.57546,0.26827,1
0.72523,0.44371,1
0.22408,0.52412,1
0.44297,0.67032,1
0.322,0.69225,1
0.13767,0.57529,1
-0.0063364,0.39985,1
-0.092742,0.55336,1
-0.20795,0.35599,1
-0.20795,0.17325,1
-0.43836,0.21711,1
-0.21947,-0.016813,1
-0.13882,-0.27266,1
0.18376,0.93348,0
0.22408,0.77997,0
0.29896,0.61915,0
0.50634,0.75804,0
0.61578,0.7288,0
0.60426,0.59722,0
0.76555,0.50219,0
0.92684,0.3633,0
0.82316,0.27558,0
0.96141,0.085526,0
0.93836,0.012427,0
0.86348,-0.082602,0
0.89804,-0.20687,0
0.85196,-0.36769,0
0.82892,-0.5212,0
0.79435,-0.55775,0
0.59274,-0.7405,0
0.51786,-0.5943,0
0.46601,-0.41886,0
0.35081,-0.57968,0
0.28744,-0.76974,0
0.085829,-0.75512,0
0.14919,-0.57968,0
-0.13306,-0.4481,0
-0.40956,-0.41155,0
-0.39228,-0.25804,0
-0.74366,-0.25804,0
-0.69758,0.041667,0
-0.75518,0.2902,0
-0.69758,0.68494,0
-0.4038,0.70687,0
-0.38076,0.91886,0
-0.50749,0.90424,0
-0.54781,0.70687,0
0.10311,0.77997,0
0.057028,0.91886,0
-0.10426,0.99196,0
-0.081221,1.1089,0
0.28744,1.087,0
0.39689,0.82383,0
0.63882,0.88962,0
0.82316,0.66301,0
0.67339,0.64108,0
1.0709,0.10015,0
-0.046659,-0.57968,0
-0.23675,-0.63816,0
-0.15035,-0.36769,0
-0.49021,-0.3019,0
-0.46717,-0.13377,0
-0.28859,-0.060673,0
-0.61118,-0.067982,0
-0.66302,-0.21418,0
-0.59965,-0.41886,0
-0.72638,-0.082602,0
-0.83007,0.31213,0
-0.72062,0.53874,0
-0.59389,0.49488,0
-0.48445,0.99927,0
-0.0063364,0.99927,0
0.63265,-0.030612,0
python代码实现
import numpy as np
import pandas as pd
import matplotlib
import matplotlib.pyplot as plt
import scipy.optimize as opt
matplotlib.rcParams['font.sans-serif']=['SimHei']#用黑体显示中文
matplotlib.rcParams['axes.unicode_minus']=False # 可以显示负号】、#第一步 引入数据
df=pd.read_csv('C:/Users/Administrator/Desktop/吴恩达机器学习数据集/week1/ex2data2.txt',names=['测试一','测试二','接受'])
print(df.head())#第一部分 数据可视化 绘制散点图
positive=df[df['接受'].isin([1])]
negetive=df[df['接受'].isin([0])]
fig,ax=plt.subplots(figsize=(8,5))
ax.scatter(positive['测试一'],positive['测试二'],s=50, c='b',label='接受')
ax.scatter(negetive['测试一'],negetive['测试二'],s=50, c='r',marker='x',label='未接受')
ax.legend()
ax.set_xlabel('测试一')
ax.set_ylabel('测试二')
plt.show()#第二部分 特征映射 为了能很好的拟合，所以我们要进行特征映射，从而得到非线性的决策边
#特征映射将低维特征向量（本例中为二维）转化为高维特征向量（本例中为28维），因为特征向量的个数变多，所以需要用正则化，否则容易出现过拟合。
#1.定义特征映射函数
def feature_mapping(x1,x2,power):data={}for i in np.arange(power+1):for p in np.arange(i+1):data["f{}{}".format(i-p,p)]=np.power(x1,i-p)*np.power(x2,p)return pd.DataFrame(data)
#2、将特征变量进行特征映射
x1=df['测试一']
x1=x1.values
x2=df['测试二']
x2=x2.values
df2=feature_mapping(x1,x2,6)
print(df2.head())
#3、重新选取变量
X=df2.values # 将特征映射后得到的变量定义为X
y=df['接受'].values
theta=np.zeros(X.shape[1])#第三部分 正则化代价函数
#1、定义sigmoid函数
def sigmoid(z):return 1/(1+np.exp(-z))
#2、定义代价函数
def cost(theta,X,y):first=y@np.log(sigmoid(X@theta)) # @表示矩阵乘法second=(1-y)@np.log(1-sigmoid(X@theta))return (first+second)/(-len(X))
#3、定义正则化代价函数
def costReg(theta,X,y,lam=1):_theta=theta[1:]reg=(lam/(2*len(X)))*(_theta@_theta)return cost(theta,X,y)+reg
#4、代价函数初始值
costReg(theta,X,y)#第四部分 高级优化算法求theta
#1、定义正则化后的梯度函数（偏导数）
def gradient(theta,X,y):return (X.T@(sigmoid(X@theta)-y))/(len(X))
def gradientReg(theta,X,y,lam=1):reg=(lam/len(X))*thetareg[0]=0 # 第一项没有惩罚因子return gradient(theta,X,y)+reg#2、使用高级优化算法求theta
result=opt.fmin_tnc(func=costReg,x0=theta,fprime=gradientReg,args=(X,y,2))#第五部分 求预测准确率（进行评估）
#1、定义预测函数，将求得的参数theta（result） 和X（特征映射后）带入预测函数，并计算预测精度
def predict(theta,X):probability=sigmoid(X@theta)return [1 if x>=0.5 else 0 for x in probability ]
finally_theta=result[0]
predictions=predict(finally_theta,X)
correct=[1 if a==b else 0 for (a,b) in zip(predictions,y)]
accuracy=sum(correct)/len(X)
print(accuracy)#或者用sklearn中的方法进行评估
'''
from sklearn.metrics import classification_report
print(classification_report(predictions,y))
'''#第六部分 绘制决策边界
x = np.linspace(-1, 1.5, 250)
xx, yy = np.meshgrid(x, x) #生成网格点坐标矩阵
# 得到的xx,yy是array形式，维度为(250,250)
z = feature_mapping(xx.ravel(), yy.ravel(), 6).values # xx.ravel()将xx一维化(62500,)
z = z @ finally_theta #z.shape (62500,)
z = z.reshape(xx.shape)fig,ax=plt.subplots(figsize=(8,6))
ax.scatter(positive['测试一'],positive['测试二'],label='接受')
ax.scatter(negetive['测试一'],negetive['测试二'],marker='x',label='未接受')
# 设置图例显示在图的上方
box = ax.get_position() #获取轴
ax.set_position([box.x0,box.y0,box.width,box.height*0.8]) #.set_position()功能用于设置轴位置
ax.legend(loc='center left',bbox_to_anchor=(0.2,1.12),ncol=3)
ax.set_xlabel('测试一')
ax.set_ylabel('测试二')
ax.set_title('决策边界lambda为2')
plt.contour(xx,yy,z,0) # 生成等高线，0是指图像一分为二
plt.show()
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