[問題] 如何處理sas違線性 polynomial&spline
根據
http://sugiclub.blogspot.com/2007/06/solutions-to-violations-of-assumptions.html
當線性關係不存在時,最明顯的影響是會導致參數估計值產生偏誤。另一方面,
R-square 值也會被低估。有幾個方法可以來解決這個問題:
二、對自變數做變數變換。常見的變數變換有 log, inverse 或 polynomial。另外,
spline transformation(使用 PROC TRANSREG)也是個不錯的點子。
但我取log跟沒取之前~
都跑PROC CORR~結果類似
PROC CORR DATA = a;
VAR LPq LPp LCp LBp LDp LFp G1;
應變數是LPq 自變數是LPp LCp LBp LDp LFp 虛擬變數是G1
要加入G1去跑嗎?還是都可以?
如果取LOG的結果一樣~請問inverse 或 polynomial,spline transformation
(使用 PROC TRANSREG)指令該怎麼下?
我有找SAS HELP~但事倍功半><請幫忙!
感謝
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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※ 編輯: luckysnow 來自: 123.193.99.101 (01/21 10:48)