r - lapply 与 "$"函数

Question

假设我有一个 data.frames 列表

dflist <- list(data.frame(a=1:3), data.frame(b=10:12, a=4:6))

如果我想从列表中的每个项目中提取第一列，我可以这样做

lapply(dflist, `[[`, 1)
# [[1]]
# [1] 1 2 3
# 
# [[2]]
# [1] 10 11 12

为什么我不能以同样的方式使用“$”函数

lapply(dflist, `$`, "a")
# [[1]]
# NULL
# 
# [[2]]
# NULL

但这些都有效:

lapply(dflist, function(x) x$a)
`$`(dflist[[1]], "a")

我意识到在这种情况下可以使用

lapply(dflist, `[[`, "a")

但我正在使用似乎不允许通过 [[ 进行索引的 S4 对象.例如

library(adegenet)
data(nancycats)
catpop <- genind2genpop(nancycats)
mylist <- list(catpop, catpop)

#works
catpop[[1]]$tab

#doesn't work
lapply(mylist, "$", "tab")
# Error in slot(x, name) : 
#   no slot of name "..." for this object of class "genpop"

#doesn't work
lapply(mylist, "[[", "tab")
# Error in FUN(X[[1L]], ...) : this S4 class is not subsettable

Answer 1

对于第一个示例，您可以这样做:

lapply(dflist, `$.data.frame`, "a")

对于第二个，使用 slot()访问函数

lapply(mylist, "slot", "tab")

我不确定为什么方法分派(dispatch)在第一种情况下不起作用，但是 Note ?lapply的部分确实解决了它对原始函数(如 $)的 borked 方法调度的问题。 :

 Note:

 [...]

 For historical reasons, the calls created by ‘lapply’ are
 unevaluated, and code has been written (e.g., ‘bquote’) that
 relies on this.  This means that the recorded call is always of
 the form ‘FUN(X[[i]], ...)’, with ‘i’ replaced by the current
 (integer or double) index.  This is not normally a problem, but it
 can be if ‘FUN’ uses ‘sys.call’ or ‘match.call’ or if it is a
 primitive function that makes use of the call.  This means that it
 is often safer to call primitive functions with a wrapper, so that
 e.g. ‘lapply(ll, function(x) is.numeric(x))’ is required to ensure
 that method dispatch for ‘is.numeric’ occurs correctly.