如何安全地使用 coxph 和子集或权重? [英] How do I use safely with coxph and subset or weights?

问题描述

我正在尝试将 purrr::safely 与 coxph 一起使用，以便我可以捕获错误消息.我制作了一个安全版本的 coxph 如下

I'm trying to use purrr::safely with coxph so that I can capture error messages. I've made a safe version of coxph as follows

library(survival)
library(purrr)

coxph_safe <- safely(coxph)

当我的唯一输入是公式和数据时，这非常有效，但是，如果我添加另一个输入(例如子集或权重)，则会收到以下错误消息:

This works perfectly when my only inputs are the formula and data, however, if I add another input such as subset or weights, I get the following error message:

eval(substitute(subset), data, env) 中的 simpleError: ..3 在不正确的上下文中使用，没有......查看

simpleError in eval(substitute(subset), data, env): ..3 used in an incorrect context, no ... to look in

有谁知道在需要额外输入时如何安全地应用于 coxph?我也会得到同样的错误，而不是安全地使用，并且如果我制作了一个安全版本的 lm 并指定了一个子集.我正在使用 R 3.6.1 和 purrr 0.3.2.目前，我已经编写了一个变通方法，即在应用 coxph_safe 之前对数据进行子集化，但最好知道是否有更好的解决方案.

Does anyone know how to apply safely to coxph when additional inputs are required? I also get the same error using quietly instead of safely, and also if I make a safe version of lm and specify a subset. I'm using R 3.6.1 and purrr 0.3.2. For now, I've programmed a workaround, where I subset the data before applying coxph_safe, but it would be good to know if there was a better solution.

这是一个简单的例子:

test1 <- list(time=c(4,3,1,1,2,2,3), 
              status=c(1,1,1,0,1,1,0), 
              x=c(0,2,1,1,1,0,0), 
              sex=c(0,0,0,0,1,1,1))

# Without subset
coxph(Surv(time, status) ~ x, test1) # Works as expected
coxph_safe(Surv(time, status) ~ x, test1) # Works as expected

# With subset
coxph(Surv(time, status) ~ x, test1, subset = !sex) # Works as expected
coxph_safe(Surv(time, status) ~ x, test1, subset = !sex) # Error!

编辑

在相关说明中，将方差分析应用于通过 coxph_safe 生成的 coxph 对象时，我也遇到了类似的错误.

On a related note, I also get a similar error when applying anova to a coxph object generated via coxph_safe.

cox_1 <- coxph(Surv(time, status) ~ x, test1) # Works as expected
anova(cox_1) # Works as expected

cox_1s <- coxph_safe(Surv(time, status) ~ x, test1) # Works as expected
anova(cox_1s$result) # Error in is.data.frame(data) : ..2 used in an incorrect context, no ... to look in

据我所知，这与呼叫的存储方式有关.我可以通过覆盖调用来修复它.

As far as I can tell, this has something to do with how the call is stored. I can fix it by over-writing the call.

cox_1$call # coxph(formula = Surv(time, status) ~ x, data = test1)
cox_1s$result$call # .f(formula = ..1, data = ..2)
cox_1s$result$call <- cox_1$call
anova(cox_1s$result) # Now works as expected

有没有更好的方法来解决这个问题?

Is there a better way around this?

推荐答案

这实际上与 purrr::safely 无关.问题是函数嵌套.考虑:

This actually has nothing to do with purrr::safely. The issue is function nesting. Consider:

f <- function(...) {coxph(...)}

f(Surv(time, status) ~ x, test1)               # Works
f(Surv(time, status) ~ x, test1, subset=!sex)  # Error

失败的真正原因与有关substitute() 在嵌套函数中的行为.coxph() 使用 substitute()，并且 safely() 创建了一个嵌套函数，导致了我链接中描述的场景.

The real reason for why it fails has to do with the behavior of substitute() inside nested functions. coxph() uses substitute(), and safely() creates a nested function, leading to the scenario described in my link.

为了解决这个问题，我们需要将 coxph() 包装成一个正确处理非标准评估 (NSE) 的函数:

To address this issue, we need to wrap coxph() into a function that properly handles non-standard evaluation (NSE):

coxph_nse <- function(...) {eval(rlang::expr(coxph( !!!rlang::enexprs(...) )))}

新函数不再遇到相同的嵌套问题，可以安全地传递给safely():

The new function no longer suffers the same nesting issues and can be safely passed to safely():

coxph_safe <- safely(coxph_nse)

coxph_safe(Surv(time, status) ~ x, test1)                     # works
cx1 <- coxph_safe(Surv(time, status) ~ x, test1, subset=!sex) # now also works!

anova(cx1$result)                                             # works as well!

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