对数据进行生存分析,考虑Cox比例为先模型的多重分数多项式模型,
library(mfp)
f=mfp(Surv(TIME,as.numeric(STATUS))~fp(AGE,df=4,select=0.05)+INST+SEX+TX+GRADE+COND+SITE+T_STAGE+N_STAGE,family=cox,data=u)
print(f)
(rsq=1-sum((f$residuals)^2)/sum((u$TIME-mean(u$TIME))^2))
结果如下:
Call:
mfp(formula = Surv(TIME, as.numeric(STATUS)) ~ fp(AGE, df = 4,
select = 0.05) + INST + SEX + TX + GRADE + COND + SITE +
T_STAGE + N_STAGE, data = u, family = cox)
Deviance table:
Resid. Dev
Null model 1303.317
Linear model 1248.388
Final model 1248.416
Fractional polynomials:
df.initial select alpha df.final power1 power2
COND2 1 1.00 0.05 1 1 .
SEX2 1 1.00 0.05 1 1 .
T_STAGE2 1 1.00 0.05 1 1 .
T_STAGE3 1 1.00 0.05 1 1 .
T_STAGE4 1 1.00 0.05 1 1 .
N_STAGE1 1 1.00 0.05 1 1 .
N_STAGE2 1 1.00 0.05 1 1 .
N_STAGE3 1 1.00 0.05 1 1 .
GRADE2 1 1.00 0.05 1 1 .
GRADE3 1 1.00 0.05 1 1 .
INST2 1 1.00 0.05 1 1 .
INST3 1 1.00 0.05 1 1 .
INST4 1 1.00 0.05 1 1 .
INST5 1 1.00 0.05 1 1 .
INST6 1 1.00 0.05 1 1 .
TX2 1 1.00 0.05 1 1 .
SITE2 1 1.00 0.05 1 1 .
SITE4 1 1.00 0.05 1 1 .
AGE 4 0.05 0.05 0 . .
Transformations of covariates:
formula
AGE T_STAGE
INST INST
SEX SEX
TX TX
GRADE GRADE
COND COND
SITE SITE
T_STAGE T_STAGE
N_STAGE N_STAGE
coef exp(coef) se(coef) z p
COND2.1 1.18540 3.2720 0.2139 5.54227 2.99e-08
SEX2.1 -0.27923 0.7564 0.2207 -1.26519 2.06e-01
T_STAGE2.1 -0.31705 0.7283 0.5000 -0.63416 5.26e-01
T_STAGE3.1 -0.23767 0.7885 0.4718 -0.50373 6.14e-01
T_STAGE4.1 0.16458 1.1789 0.4955 0.33214 7.40e-01
N_STAGE1.1 -0.28642 0.7509 0.3134 -0.91404 3.61e-01
N_STAGE2.1 -0.04994 0.9513 0.3410 -0.14644 8.84e-01
N_STAGE3.1 0.22688 1.2547 0.2567 0.88399 3.77e-01
GRADE2.1 0.09141 1.0957 0.2253 0.40576 6.85e-01
GRADE3.1 -0.30871 0.7344 0.3064 -1.00764 3.14e-01
INST2.1 0.31219 1.3664 0.3456 0.90327 3.66e-01
INST3.1 -0.12545 0.8821 0.3615 -0.34706 7.29e-01
INST4.1 -0.22755 0.7965 0.3872 -0.58775 5.57e-01
INST5.1 -0.22744 0.7966 0.3861 -0.58911 5.56e-01
INST6.1 0.03597 1.0366 0.3696 0.09731 9.22e-01
TX2.1 0.08314 1.0867 0.1829 0.45452 6.49e-01
SITE2.1 -0.13228 0.8761 0.2413 -0.54817 5.84e-01
SITE4.1 -0.02888 0.9715 0.2326 -0.12420 9.01e-01
Likelihood ratio test=54.9 on 18 df, p=1.327e-05 n= 193
> (rsq=1-sum((f$residuals)^2)/sum((u$TIME-mean(u$TIME))^2))
[1] 0.9999954
请教一下结果对于变量的诸如
INST2 1 1.00 0.05 1 1 .
INST3 1 1.00 0.05 1 1 .
INST4 1 1.00 0.05 1 1 .
INST5 1 1.00 0.05 1 1 .
INST6 1 1.00 0.05 1 1 .
是如何进行解释理解的?其中原数据是口咽癌数据中机构代码变量INST是定性变量,取值1,2,3,4,5,6;TIME是生存时间,STAUS表示0:数据缺失,1:死亡。求指导