r - 从传递给函数内部函数的列表中获取参数

Question

我想在模型列表上应用一个函数。我想通过将模型名称放在输出之前来组织输出。这是一个(非常)简化的示例，使用基于 this r-help thread 的 match.call() :

x   <- rnorm(10)
y   <- rnorm(10)
mod <- lm(y~x)
fun <- function(model){
 name <- as.character(match.call()[[2]])
 p    <- summary(model)$coefficients[2,4]
 return(paste(name, "; p =", p))
}

如果我给它一个单一的模型，它工作正常:

> fun(model=mod)
[1] "mod ; p = 0.901618595026321"

但是如果我将它应用于一个列表，无论是否命名，它都不起作用:

> lapply(list(mod, mod), fun)
[[1]]
[1] "[[ ; p = 0.901618595026321" "X ; p = 0.901618595026321" 
[3] "i ; p = 0.901618595026321" 

[[2]]
[1] "[[ ; p = 0.901618595026321" "X ; p = 0.901618595026321" 
[3] "i ; p = 0.901618595026321" 

> lapply(list(modA=mod, modB=mod), fun)
$modA
[1] "[[ ; p = 0.901618595026321" "X ; p = 0.901618595026321" 
[3] "i ; p = 0.901618595026321" 

$modB
[1] "[[ ; p = 0.901618595026321" "X ; p = 0.901618595026321" 
[3] "i ; p = 0.901618595026321" 

为清楚起见，我想要的是:

[[1]]
[1] "modA ; p = 0.901618595026321"
[[2]]
[1] "modB ; p = 0.901618595026321"

Answer 1

您是否有理由不想对列表名称使用 lapply？

fun2 = function(models, name){
  model <- models[[name]]
  p <- summary(model)$coefficients[2,4]
  return(paste(name, "; p =", p))
}
mods = list(modA = mod, modB = mod)
lapply(names(mods), fun2, models = mods)
# [[1]]
# [1] "modA ; p = 0.828746440943558"

# [[2]]
# [1] "modB ; p = 0.828746440943558"