数据描述性分析

1、数据分析实验报告实验名称：数据描述性分析使用软件名称：Matlab11.熟练掌握Matlab软件进行距离判别与Bayes判别的方法与步骤验2.掌握判别分析的回代误判率与交叉误判率的编程目"3.掌握Bayes判别的平均误判率的计算以及灵敏度分析根据下面的数据，解决以下问题：判别两类总体的协方差矩阵是否相等，然后用马氏距离判别未知地区的类别，并计算回代误判率与交叉误判率；若两类总体服从正态分布，第一类与第二类的先验概率分别为13/27,14/27,利用Bayes判别未知地区的类别 先验概率=多少时，距离判别与Bayes判别相同，调整先验概率对判别结果的影响是什么？ 对第一类与第二类的先验

2、概率分别为13/27,14/27,计算误判概率表6.4.1各地区农、林、牧、渔各业数据类别农林牧渔类别农林牧渔289.709.50105.209.601405.9011.30236.405.80286.701.5060.8020.601450.6015.70224.6020.10295.503.5088.4040.101529.5073.70195.90308.802191.3012.3096.301.701688.0066.20371.60132.302307.6026.10216.206.001433.082.30215.50330.50实2141.3043.3058.2082.30140

3、5.9054.00226.10104.302250.4011.20154.4015.201658.3027.10352.60134.80验2337.4023.60114.103.801665.7051.90480.3085.202254.008.6080.901.101817.9056.80423.20390.10内228.901.8032.500.101439.9039.40292.30101.20249.403.5030.302.101769.9050.90605.0041.00容2348.8010.10134.003.90x431.3047.20210.6014.402899.4034.

4、00685.9061.20x1401.3047.20654.70350.7021142.7030.80448.50334.20x1331.6057.00693.8020.401503.1021.80332.30188.50x279.9015.10118.505.10首先判别两类总体的协方差距阵是否相等:clearclcA=89.709.50105.209.6086.701.5060.8020.6095.503.5088.4040.10191.3012.3096.301.70307.6026.10216.206.00141.3043.3058.2082.30250.4011.20154.4015

5、.20337.4023.60114.103.80254.008.6080.901.1028.901.8032.500.1049.403.5030.302.10348.8010.10134.003.90899.4034.00685.9061.201142.7030.80448.50334.20503.1021.80332.30188.50405.9011.30236.405.80450.6015.70224.6020.10529.5073.70195.90308.80688.0066.20371.60132.30433.2082.30215.50330.50405.9054.00226.1010

6、4.30658.3027.10352.60134.80665.7051.90480.3085.20817.9056.80423.20390.10439.9039.40292.30101.20769.9050.90605.0041.00;%已知样本X=431,3047.20210.6014.401401.3047.20654.70350.701331.6057.00693.8020.40279.9015.10118.505.10；%未知样本A2=A(1:14,:);嘛二类数据A1=A(15:26,:);愉一类数据n1=12;n2=14;p=4;s1=cov(A1);s2=cov(A2);s=(n

7、1-1)*s1+(n2-1)*s2)/(n1+n2-2);Q01=(n1-1)*(log(det(s)-10g(det(s1)-p+trace(inv(s)*s1);Q02=(n2-1)*(log(det(s)-10g(det(s2)-p+trace(inv(s)*s2);P=1-chi2cdf(Q01,Q02,p*(p+1)/2)%以上计算可得两总体的协方差距阵相等m1=mean(A1);m2=mean(A2);fori=1:4以未知样本到两类总d(i)=(X(i,:)-m1)*inv(s)*(X(i,:)-m1)'-(X(i,:)-m2)*inv(s)*(X(i,:)-m2)

8、9;体的马氏距离之差函数end以十算第一类误判为第二类的个数a1a1=0;fori=1:n1d1=(A1(i,:)-m1)*inv(s)*(A1(i,:)-m1)'-(A1(i,:)-m2)*inv(s)*(A1(i,:)-m2)'ifd1>0a1=a1+1;elsea1=a1;endend%十算第二类误判为第一类的个数a2a2=0;forj=1:n2d2=(A2(j,:)-m1)*inv(s)*(A2(j,:)-m1)'-(A2(j,:)-m2)*inv(s)*(A2(j,:)-m2)'ifd2<0a2=a2+1;elsea2=a2;endendw

9、ph=(a1+a2)/(n1+n2)%U代误判率%十算交叉误判率a11=0;fori=1:n2B1=A2(1:i-1,:);A2(i+1:n2,:);mb1=mean(B1);b1=mean(A1);sb1=cov(B1);l1=n2-1;l2=n1;p=4;T1=(l1-1)*sb1+(l2-1)*s1)/(l1+l2-2);Q1=(l1-1)*(log(det(T1)-10g(det(sb1)-p+trace(inv(T1)*sb1);Q2=(l2-1)*(log(det(T1)-10g(det(s1)-p+trace(inv(T1)*s1);P1=1-chi2cdf(Q1,p*(p+1)

10、/2);P2=1-chi2cdf(Q2,p*(p+1)/2);ifP1>0.05&&P2>0.05d11(i)=(A2(i,:)-mb1)*inv(T1)*(A2(i,:)-mb1)'-(A2(i,:)-b1)*inv(T1)*(A2(i,:)-b1)'elsed11(i)=mahal(A2(i,:),B1)-mahal(A2(i,:),A1);endenda12=0;forj=1:n1B2=A1(1:j-1,:);A1(j+1:n1,:);mb2=mean(B2);b2=mean(A2);sb2=cov(B2);l11=n1-1;l12=n2;p=

11、4;T2=(l11-1)*sb2+(l12-1)*s2)/(l11+l12-2);Q11=(l11-1)*(log(det(T2)-log(det(sb2)-p+trace(inv(T2)*sb2);Q12=(l12-1)*(log(det(T2)-log(det(s2)-p+trace(inv(T2)*s2);P11=1-chi2cdf(Q11,p*(p+1)/2);P12=1-chi2cdf(Q12,p*(p+1)/2);ifP11>0.05&&P12>0.05d12(j)=(A1(j,:)-mb2)*inv(T2)*(A1(j,:)-mb2)'-(A1

12、(j,:)-b2)*inv(T2)*(A1(j,:)-b2)'elsed12(j)=mahal(A1(j,:),B2)-mahal(A1(j,:),A2);endenda11=length(find(d11>=0);a12=length(find(d12>=0);wpj=(a11+a12)/(n1+n2)班叶斯判别bd=m1*inv(s)*X'-0.5*m1*inv(s)*m1'+log(13/27)-(m2*inv(s)*X'-0.5*m2*inv(s)*m2'+10g(14/27)mp=(m1-m2)*inv(s)*X'-0.5*

13、m1*inv(s)*m1'+0.5*m2*inv(s)*m2'p12=exp(-mp)%两个先验概率的比值%喻算平均误判率P1=13/27;P2=14/27;d0=log(P2/P1);r=(m1-m2)*inv(s)*(m1-m2)'pjwu=P1*normcdf(d0-r/2)/sqrt(r)+P2*(1-normcdf(d0+r/2)/sqrt(r)(1)运行程序得到：P=0.43350.3034对于a=0.05,0.4335a0.05,0.3034>0.05,所以，两类总体的协方差距阵相等。未知样本(行向量)到两类总体G1,G2马氏距离之差分别为：d=-1

14、.1647-5.5044-5.85383.9163所以前三个未知样本属于第一类，第四个未知样本属于第二类。运行程序得到，回代误判率为wph=0.1923交叉误判率为wpj=0.2692若两类总体服从正态分布，第一类与第二类的先验概率分别为13/27,14/27,利用贝叶斯判别可知距离只差分别为：实bd=0.54152.30152.3655-2.1129验前三个均大于0,第四个小于0,所以前三个未知样本均属于第一类，第四个未知样本属于第二类。结(2)如果先验概率有了变化，就会影响到最后判别的结果，由果，、，、c，-mpWi(x)W2(x)=0=p"p2=e分析其中，mp=(m1-m2)inv(s)X'_0.5mlinv(s)m1+0.5m2inv(s)m2'对于题中给出的四个未知样本，我们计算出临界值：表1先验概率的临界值未知地区X1X2X3X4P1/P20.54030.09300.08727.6816若p1/p2<0.5403,则未知地区X1从第一类=第二类若p1/p2<0.0930,则未知地区X2从第一类二第二类若p1/p2<0.0872,则未知地区X3从第一类二第二类若p1/p2<7,6816,则未知地区X4从第二类=第一类对于划分R=(R1,R2),将来自G1的样品x误判为来自G2的概率为P(2|1)=P