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1、PCA主成分分析应用举例例1a=c(177,179,95,96,53,32,-7,-4,-3,179,419,245,131,181,127,-2,1,4,95,245,302,60,109,142,4,4,11,96,131,60,153,102,42,4,3,2,53,181,109,102,137,96,4,5,6,32,127,142,42,96,128,2,2,8,-7,-2,4,4,4,2,34,31,33,-4,1,4,3,5,2,31,39,39,-3,4,11,2,6,8,33,39,48)s=matrix(a,ncol=9)S为样本方差阵求方差阵S的特征值和
2、特征向量c=eigen(s)c样本前3个主成分的系数是：rho=diag(1/(sqrt(diag(s))))%*%s%*%diag(1/(sqrt(diag(s))))rho例2学生身体各指标的主成分分析.随机抽取30名某年级中学生,测量其身高(X1)、体重(X2)、胸围(X3)和坐高(X4)。试对中学生身体指标数据做主成分分析.30名中学生的四项身体指标####用数据框形式输入数据student<-data.frame(X1=c(148,139,160,149,159,142,153,150,151,139,140,161,158,140,137,152,149,145,1
3、60,156,151,147,157,147,157,151,144,141,139,148),X2=c(41,34,49,36,45,31,43,43,42,31,29,47,49,33,31,35,47,35,47,44,42,38,39,30,48,36,36,30,32,38),X3=c(72,71,77,67,80,66,76,77,77,68,64,78,78,67,66,73,82,70,74,78,73,73,68,65,80,74,68,67,68,70),X4=c(78,76,86,79,86,76,83,79,80,74,74,84,83,77,73,79
4、,79,77,87,85,82,78,80,75,88,80,76,76,73,78))>cor(student)X1X2X3X4X11.00000000.86316210.73211190.9204624X20.86316211.00000000.89650580.8827313X30.73211190.89650581.00000000.7828827X40.92046240.88273130.78288271.0000000>eigen(cor(student))$values[1]3.541098000.313383160.079408950.06610989$vect
5、ors[,1][,2][,3][,4][1,]-0.49696610.5432128-0.44962710.5057471[2,]-0.5145705-0.2102455-0.4623300-0.6908436[3,]-0.4809007-0.72462140.17517650.4614884[4,]-0.50692850.36829410.7439083-0.2323433>####作主成分分析>student.pr<-princomp(student,cor=TRUE)>>####并显示分析结果summary(student.pr,loadings=TRUE)Importa
6、nceofcomponents:Comp.1Comp.2Comp.3Comp.4Standarddeviation1.88178050.559806360.281795940.25711844ProportionofVariance0.88527450.078345790.019852240.01652747CumulativeProportion0.88527450.963620290.983472531.00000000Loadings:Comp.1Comp.2Comp.3Comp.4X1-0.4970.543-0.4500.506X2-0.515-0.210-0.462-
7、0.691X3-0.481-0.7250.1750.461X4-0.5070.3680.744-0.232PRINCOMP过程由相关阵出发进行主成分分析.由相关阵的特征值可以看出，第一主成分的贡献率已高达88.53%；且前二个主成分的累计贡献率已达96.36%.因此只须用两个主成分就能很好地概括这组数据.另由第三和四个特征值近似为0，可以得出这4个标准化后的身体指标变量(Xi*,i=1,2,3,4)有近似的线性关系(即所谓共线性),如0.505747X1*-0.690844X2*+0.46
