作者luckysnow ()
看板Statistics
标题[问题] sas违线性 polynomial&spline transformat指令
时间Mon Jan 21 10:47:05 2008
根据
http://sugiclub.blogspot.com/2007/06/solutions-to-violations-of-assumptions.html
当线性关系不存在时,最明显的影响是会导致参数估计值产生偏误。另一方面,
R-square 值也会被低估。有几个方法可以来解决这个问题:
二、对自变数做变数变换。常见的变数变换有 log, inverse 或 polynomial。另外,
spline transformation(使用 PROC TRANSREG)也是个不错的点子。
但我取log跟没取之前~
都跑PROC CORR~结果类似
PROC CORR DATA = a;
VAR LPq LPp LCp LBp LDp LFp G1;
应变数是LPq 自变数是LPp LCp LBp LDp LFp 虚拟变数是G1
要加入G1去跑吗?还是都可以?
如果取LOG的结果一样~请问inverse 或 polynomial,spline transformation
(使用 PROC TRANSREG)指令该怎麽下?
我有找SAS HELP~但事倍功半><请帮忙!
感谢
DATA a;
INFILE 'k:\sas.txt' FIRSTOBS=2;
INPUT YEAR YEARC Pq Pp Cp Bp Dp Fp Time;
IF YEAR>=1997.04 THEN G1 =1;
/*IF year>=1997.04 THEN will be divided to G1 =1*/
ELSE G1=0;
/*IF year<1997.04 THEN will be divided to G1 =0*/
LPq=LOG(Pq);
LPp=LOG(Pp);
LCp=LOG(Cp);
LBp=LOG(Bp);
LDp=LOG(Dp);
LFp=LOG(Fp);
/*ASSUMPTION 1: LINEARITY*/
PROC CORR DATA = a;
/*Pearson Correlation Coefficients <0.0001,so NO */
VAR LPq LPp LCp LBp LDp LFp G1;
PROC CORR DATA = a;
VAR Pq Pp Cp Bp Dp Fp G1;
RUN;
LPq LPp LCp LBp LDp LFp G1
LPq 1.00000 0.61851 0.55951 0.34521 0.74705 0.26216 0.74961
<.0001 <.0001 <.0001 <.0001 0.0003 <.0001
LPp 0.61851 1.00000 0.85927 0.41899 0.86538 0.50256 0.79639
<.0001 <.0001 <.0001 <.0001 <.0001 <.0001
LCp 0.55951 0.85927 1.00000 0.40777 0.81449 0.55848 0.80321
<.0001 <.001 <.0001 <.0001 <.0001 <.0001
LBp 0.34521 0.41899 0.40777 1.00000 0.62815 0.70867 0.35859
<.0001 <.0001 <.0001 <.0001 <.0001 <.0001
LDp 0.74705 0.86538 0.81449 0.62815 1.00000 0.60114 0.85977
<.0001 <.0001 <.0001 <.0001 <.0001 <.0001
LFp 0.26216 0.50256 0.55848 0.70867 0.60114 1.00000 0.38153
0.0003 <.0001 <.0001 <.0001 <.0001 <.0001
G1 0.74961 0.79639 0.80321 0.35859 0.85977 0.38153 1.00000
<.0001 <.0001 <.0001 <.0001 <.0001 <.0001
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