主成分分析R语言

1、第一题> data=read.table("q1.txt",head=T)> head(data)#前六行数据xlx2x3x4x6x7x8x9xlO1000.590.340,总30.400.230.200.110 0T2S91QQ口 .42a51D43o3a03Ea .21Q0930-350.421.00. J J0.190.360 _730.240.44-D . 084340.511,00Q2診Qd匂a3夕0.17a .is50-630 理22000.3402301303号04005203宮04需0.341.000320330.180.00> attac

2、h(data)> opti on s(digits=2)> data=as.matrix(data)# 转化为矩阵> v1<-eige n(data)#计算相关系数矩阵特征值和特征向量> v1Rvalues(1) 3.79 1.52 1,11 0.31 0.72 0.59 0-53 0,30 0.2 0.21Sveccors<1山f3【".5L.M,B【】,10J(lrl0.355-C.17-0H930.216-九 15 j0.19101刖0 *3&-0,0750疋绅4DM05-0.14-9-0.182-00旷005-0hQ960.131-

3、0+333-a.129-0.120160.3610,4340 0450.LB3O-25S0.03 40.109-0,0530 7500.0237040.346-0.10-90.131-0.4140.274-0.3310.5320.3S2-0.133-0.112095J0.313-Q.W-0.079-0.13Q0.074-0.IOS0*255o.ose-0,«25S3£Q.353'Q+034-0.153-0.3619-054-0.321-0.7630.095a.oss0.102阳叫0.3150.4230.1040.2450-4200.293-0.1530.033-0.

4、595-0*OS63S80.277-0.0710.3S9-0.460-0.3220.665-0.0200.0470.0-340.00099D.2230.35-6Q.3520.245一069-0.010.027-0,14S0.0717310,Q.075-0040.-6230.292O.2S5-0.105-0.101-Q.13a.0300261> zhuchengfen<-princomp(covmat=data)#用相关系数矩阵做主成分分析> summary(zhuche ngfen)Iirp匚二t巳:匚亡 Qi 二匚ir£events;Coir.p > 1Cc

5、ir,p .2CoK.p - 33oir.p * 4Coirp .5Ccrnp-Con * 7Standard devLatian1.951.231-060. 9560.8490.7710.72CF-cportlun wf0.3S0.130.110.0310*0720.053053Cumulative Frcporciano.se0.530.640.733O.Q050.50.317Coir.p, eCcir.p, 9Cwsp 10Staridtrd deviation0,61908S0-456of Zmw0*0320.0240-021Cuiculative Prcporrion0,9560.9

6、791.000> x<-zhuchengfen$loadings ； x#主成分载荷LaadirijB：Dorrs 1CcmpniCcrp.3Camp* 4匚 amp. 5CompCCcrrp coir,p.eCorcp + 9 3orcp 10xl-0.355-C.1760.433-0.216-0.1530.151-0.18-0.2390.629-0.H90.162-0.1B30.S33-0.129 -0,120-0.3510却-0*1330.25E-0.102C.75C-0.346-C-109-0.1310.4140.274-0.3S1-0.532-0.3S2-0.133 -0

7、.112-.318-0.447-0.438-0.1300.10 =-0.2E5-0.626x6-0.3530.1530,361-0,3210 7650.103x7-0.319C.23-0.10-0.2450.4200.2530.153-0.595-0.277-0.3S90.460-0.3223.6£5-0.523C.35£-0.352-0.245-0.669-0,401-0.HBxlD-0.633-0.2920.225-0.1C50.1010.13003Coup. *_Zzir.z . 2亡 cm口. 3Coup.Coilj: . 5 Ccicp -百 Camp. 7Cam

8、p. 2Coicp. 9Coii.lO55 lcadinijj1,01.01.01,01.0 1,01.01.01.0Proparticn Vx0.10.10,10,10*1Q.l0.LQ,10,1C*1s 10.1C.20.30.4D.5o.e0,7D.E0.91.0> y<-zhuche ngfen $scores;y # 主成分得分NULL因为只有相关系数矩阵，所以没有主成分的得分> screeplot(zhuche ngfen ,type="l in es")zhucli«ngtenS4OLII:二 e a.Cqitp.3 Comp.5

9、Cflmp.7CompJCame.9根据累计方差贡献率和Kaiser准则，应该选择三个主成分，即PC1、PC2、PC3。累计方差贡献率为 64%。PC1、PC2、PC3对方差贡献率分别是 38%、15%、11%。结果如下：运动项目第一主成分第二主成分第三主成分100米-0.355-0.1760.493跳远-0.405-0.1490.182铅球-0.3610.434跳咼-0.346-0.109-0.131400米-0.318-0.447110米跨栏-0.3530.153铁饼-0.3190.423-0.104撑杆跳高-0.277-0.389标枪-0.2230.356-0.3521500 米-0.4

10、84-0.623方差贝献率0.380.150.11累积方差贝献率0.380.530.64解释：前三个主成分解释了大部分信息；第一次主成分测量的是综合运动能力，第二个主成分代表了爆发力臂力，第三主成分测量的是爆发性腿力。附录：> librarypsych>> zhu<-principa 1 (data,nf actors=3 f rojtare=rtnone" > zhuFrincipal Camponentn AnalysisCall: principal (r =nfactors = 3T rotate = "none")Stand

11、ardized loadings (pattern matrix based pon correlation iratiixPC1PC2PC 3h2u210.630 .22-0.520.750.2120.75o.ie-o.ig0.63o.ai30.70-0.530.050.780.22些0.670,130.140.515CL 62Q +55-CLOBa. 690.3160.690.04-0.160.500*5070. £2-0.520.110* 670.3320.54 .090.410.47a. S330. e3-0 .440.370.520.4S100.150 -0.施o.ai0.

12、19PCIPC2M3S3 Loadings3了91.522 丄1PrarartLefLVat0.380.150.11CumuldEzveVar0.380.530.64Properti口皿Explained0.590.240,17CumulativeFzcpoxtion05#0.331.00lest of the hi'potheslfl th&r 3 componencs are sufficientThe degrees of freedcm for uhe null nadtl are 45 and 匸h亡 objective funcrion wasThe degrees

13、 of freedcm fcr the itcdel are IS and th& ob_ e ctive function was 0.97Fit based upon off diagonal values = 0 92> |这是psych软件包中计算主成分的结果，主成分载荷与princomp得到的结果有很大的差 异，并且这个结果似乎更容易解释第二题> library(RODBC)> data=odbcCo nn ectExcel("G:/R/q2.xls")> data仁sqlFetch(data,"Sheet1")

14、> close(data) #关闭通道> attach(data1)> head(data1)#查看前六行数据> data2=data1,-1# 去掉第一列数据> optio ns(digits=2) #数据均保留小数点后两位> q2<-princomp(data2,cor=T) #用相关系数矩阵做主成分分析> optio ns(digits=2) #数据均保留小数点后两位> summary(q2)In:pcrts7.7e of compD二亡ts:Coirp. 12.41Standird devi日匸ionCorrp * 20.808Co

15、ir.p. 40.354Coir.p 50.2320Ccmp 60.1576Coirp."0.149SFrcpcxcion of VarianceCiralative Proportion0.0930.9230.966o,oie0.9940.00770.99123.00560.996S0.00321.0000PC1> x<-q2$loadi ngs;x # 主成分载荷LoadjLngsCorrj2 * 1Corr.f . 2Ccrr.p 3Corr-p + 4Ccir.p , 5Ccir.p + 后Coirp , 7100 m (£>-CL 36Sa. 4

16、500卫£石0*319-CL 231( 20200 m -0-3650.5370.230-C.711-0.109400 rr. (s)-0.3920.247-0.E15-0.340C.131800 m (ntin)一口.385-0,155-0585一0620-0,3151500 rT. (rain>-0389-D300*430-C.2310 . 933000 m train、-0.3fi9-0.34S0.1530.-4 £3-0.53SMaratLhcn (irin'j-0-357-0.590.484-0.72-0.1310.142匚臥91匚OH區28富电5匚

17、a匹HCQIT芒*后Gcrr.p .7351.001.0Q1.C01。打P工口p口工匸j.口口 VdKQH吒01呛0,3.叮丄电Q 二唱匸 Xu Vax0 404Q.570口O.S<:L.M令x , x2x7分别代表100 m (s)、200 m (s)Marathon (min)的成绩，那么第一主成分的表达式:卩6 = -0.36卑-0.36玄-0.382-0.38爲一0.3遽-0.38*-0.367第一主成分中各个运动项所占的比例非常接近，因此它衡量的是一个人的综合运动能力, 且方差的贡献度达到 83%。> y<-q2$scores;y,1# 第一主成分得分1 23456

18、78 &1011-0.5322.1131.3931.524-0.3910.119-1.6982 + «32-Q.553-0.6470.1431213151617IE192022-6.133-2,«433.0621,127-2.317£.5021.103.53S2.5535.S05-0.02223252627282930313233-3.2571.491-1.024-2.1321.12fl：0.1442.1S90.0600.435-1.245-0.4673536373339牝花43侶-1-314-2.352-4.2730,0的1.S111.5251.497-,018-1.6552.974547笙495051525354552.0431,9090.3591.5451.3590.505-1的1-1,6233.3673,497*3.410> order(y,1)# 排序1J 1J15 9 rl _2 4-rl rL55 12 36 41 23 13 35 16 26 46 51 7 畀 52 鹑 32 25 22 10 5 1 33 5 1143 21 20 14