2013-10-14 48 views
5

我在使用coxph()時遇到了一些麻煩。 我有兩個分類變量:性別和可能原因,我想用作預測變量。性別只是典型的男性/女性,但可能的原因有5個選項。 我不知道警告消息有什麼問題。爲什麼cofidence間隔從0到Inf並且p值如此之高?R coxph()警告:Loglik在變量前收斂

下面的代碼和輸出:

> my_coxph <- coxph(Surv(tempo,status) ~ factor(Sexo)+ factor(Causa.provavel) ,   data=ceabn) 
Warning message: 
In fitter(X, Y, strats, offset, init, control, weights = weights, : 
Loglik converged before variable 2,3,5,6 ; beta may be infinite. 

> summary(my_coxph) 
Call: 
coxph(formula = Surv(tempo, status) ~ factor(Sexo) + factor(Causa.provavel), 
data = ceabn) 

n= 43, number of events= 31 

              coef exp(coef) se(coef)  z Pr(>|z|) 
factor(Sexo)macho      7.254e-01 2.066e+00 4.873e-01 1.488 0.137 
factor(Causa.provavel)caca    2.186e+01 3.107e+09 9.698e+03 0.002 0.998 
factor(Causa.provavel)colisao linha MT 1.973e+01 3.703e+08 9.698e+03 0.002 0.998 
factor(Causa.provavel)indeterminado 9.407e-01 2.562e+00 1.683e+04 0.000 1.000 
factor(Causa.provavel)predacao   2.170e+01 2.655e+09 9.698e+03 0.002 0.998 
factor(Causa.provavel)predado   2.276e+01 7.659e+09 9.698e+03 0.002 0.998 

             exp(coef) exp(-coef) lower .95 upper .95 
factor(Sexo)macho      2.065e+00 4.841e-01 0.7947  5.368 
factor(Causa.provavel)caca    3.107e+09 3.219e-10 0.0000  Inf 
factor(Causa.provavel)colisao linha MT 3.703e+08 2.701e-09 0.0000  Inf 
factor(Causa.provavel)indeterminado 2.562e+00 3.904e-01 0.0000  Inf 
factor(Causa.provavel)predacao   2.655e+09 3.766e-10 0.0000  Inf 
factor(Causa.provavel)predado   7.659e+09 1.306e-10 0.0000  Inf 

Concordance= 0.752 (se = 0.059) 
Rsquare= 0.608 (max possible= 0.987) 
Likelihood ratio test= 40.23 on 6 df, p=4.105e-07 
Wald test   = 7.46 on 6 df, p=0.2807 
Score (logrank) test = 30.48 on 6 df, p=3.183e-05 

謝謝

回答

8

當我問特里Therneau(PKG作者:生存)有關,幾年前,他說,測試被觸發產生該警告過於敏感。通常警告不正確。你通常可以看看你的係數,看看他們是不是無限的

然而,在你的情況下,它似乎正確地警告你,你的數據可能有問題,因爲你有難以置信的大系數。指數模型中的Beta係數爲2.276e + 01(= 22.7)的值很高。估計的相對風險超過100萬!您應該查看數據的表格分類以瞭解完全分離的問題。你的控制組是否有人死亡,呃有事件發生?

+0

我在43個31個事件。但我認爲你是對的,我搞砸了數據。 – JMarcelino