Question

我想在模型列表中应用函数。我想通过将模型的名称放在其输出之前来组织输出。以下是基于this r-help thread使用match.call()的（非常）简化示例：

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))
}

如果我将它喂给一个模型，它可以正常工作：

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

但是，如果我将它应用于名单或名称，它不起作用：

> 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

您是否有理由不想在列表名称上使用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"

Answer 2

试试这个，

mLst <- list(modA=mod, mobB=mod)
p <- lapply(mLst, function(m) summary(m)$coefficients[2,4])
as.list(mapply(function(n,p) paste(n,p, sep=" ; "), names(mLst), p))

从列表中获取参数传递给函数内的函数

2 个答案: