英文:
Using package::function inside formula in R gives different results (specifically survival::strata inside coxph)
问题
I believed the following two formula calls should be equivalent.
install.packages("survival")library(survival)
Model 1
coxph(Surv(futime, fustat) ~ age + strata(rx), data=ovarian)Call:coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian)coef exp(coef) se(coef) z page 0.13735 1.14723 0.04741 2.897 0.00376Likelihood ratio test=12.69 on 1 df, p=0.0003678n= 26, number of events= 12
Model 2
coxph(Surv(futime, fustat) ~ age + survival::strata(rx), data=ovarian)Call:coxph(formula = Surv(futime, fustat) ~ age + survival::strata(rx),data = ovarian)coef exp(coef) se(coef) z page 0.14733 1.15873 0.04615 3.193 0.00141survival::strata(rx)rx=2 -0.80397 0.44755 0.63205 -1.272 0.20337Likelihood ratio test=15.89 on 2 df, p=0.0003551n= 26, number of events= 12
However, they are not. The second model, which references 'strata' specifically using the "::" operator, treats it like a fixed effect variable in the model, rather than fitting the model separately within levels of 'strata'. I do not know if this is:
A) An issue with using 'package::function' type calls within a formula in R, or
B) An issue with the survival package and namespaces/dependencies
I want to use the call to Model 2, as I am building a package and don't want to have to attach the entire survival package (which is large) when loading mine.
Any help on this would be much appreciated.
英文:
I believed the following two formula calls should be equivalent.
install.packages("survival")library(survival)
Model 1
coxph(Surv(futime, fustat) ~ age + strata(rx), data=ovarian)Call:coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian)coef exp(coef) se(coef) z page 0.13735 1.14723 0.04741 2.897 0.00376Likelihood ratio test=12.69 on 1 df, p=0.0003678n= 26, number of events= 12
Model 2
coxph(Surv(futime, fustat) ~ age + survival::strata(rx), data=ovarian)Call:coxph(formula = Surv(futime, fustat) ~ age + survival::strata(rx),data = ovarian)coef exp(coef) se(coef) z page 0.14733 1.15873 0.04615 3.193 0.00141survival::strata(rx)rx=2 -0.80397 0.44755 0.63205 -1.272 0.20337Likelihood ratio test=15.89 on 2 df, p=0.0003551n= 26, number of events= 12
However they are not. The second model, which references 'strata' specifically using the "::" operator, treats it like a fixed effect variable in the model, rather than fitting the model separately within levels of 'strata'. I do not know if this is:
A) An issue with using 'package::function' type calls within a formula in R, or
B) An issue with the survival package and namespaces/dependencies
I want to use the call to Model 2, as I am building a package and don't want to have to attach the entire survival package (which is large) when loading mine.
Any help on this would be much appreciated.
答案1
得分: 3
In coxph()，这里有这些行：
special <- c("strata", "tt", "frailty", "ridge", "pspline")tform$formula <- if (missing(data))terms(formula, special)else terms(formula, special, data = data)
因此，在第一种情况下，coxph 正确检测到了 strata，但在第二种情况下没有。我们可以使用您的示例来验证这一点：
library(survival)formula1 <- Surv(futime, fustat) ~ age + strata(rx)formula2 <- Surv(futime, fustat) ~ age + survival::strata(rx)special <- c("strata", "tt", "frailty", "ridge", "pspline")data <- ovariant1 <- terms(formula1, special, data = data)t2 <- terms(formula2, special, data = data)attr(t1, "specials")$strata#> [1] 3attr(t2, "specials")$strata#> NULL
虽然我不了解 coxph() 的具体细节，但我会说这种差异后来导致了结果的不同。
因此，您应该避免在公式中使用前缀 survival::。如果由于某种原因，您必须使用这个前缀，可以使用以下技巧：
library(survival)# 在这里strata <- survival::stratacoxph(Surv(futime, fustat) ~ age + strata(rx), data=ovarian)#> Call:#> coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian)#>#> coef exp(coef) se(coef) z p#> age 0.13735 1.14723 0.04741 2.897 0.00376#>#> Likelihood ratio test=12.69 on 1 df, p=0.0003678#> n= 26, number of events= 12
英文:
In coxph(), there are these lines:
special <- c("strata", "tt", "frailty", "ridge", "pspline")tform$formula <- if (missing(data))terms(formula, special)else terms(formula, special, data = data)
Therefore, in the first case, coxph correctly detects the strata but not in the second case. We can see this using your example:
library(survival)formula1 <- Surv(futime, fustat) ~ age + strata(rx)formula2 <- Surv(futime, fustat) ~ age + survival::strata(rx)special <- c("strata", "tt", "frailty", "ridge", "pspline")data <- ovariant1 <- terms(formula1, special, data = data)t2 <- terms(formula2, special, data = data)attr(t1, "specials")$strata#> [1] 3attr(t2, "specials")$strata#> NULL
While I don't know the specifics of coxph(), I would say that this difference later leads to the differences in results.
Therefore, you should avoid using the prefix survival:: in the formula. If for some reason, you have to use this prefix, you can use the following trick:
library(survival)# herestrata <- survival::stratacoxph(Surv(futime, fustat) ~ age + strata(rx), data=ovarian)#> Call:#> coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian)#>#> coef exp(coef) se(coef) z p#> age 0.13735 1.14723 0.04741 2.897 0.00376#>#> Likelihood ratio test=12.69 on 1 df, p=0.0003678#> n= 26, number of events= 12
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论