r语言 - 设置此函数的参数以确定内存中对象的大小时出现错误

Question

从这个不错的回答开始，反过来基于这个问题的代码，我编写了以下函数来获取内存中对象的大小：

# improved list of objects
.ls.objects <- function (pos = 1, pattern, order.by,
decreasing=FALSE, head=FALSE, n=5) {
napply <- function(names, fn) sapply(names, function(x)
fn(get(x, pos = pos)))
names <- ls(pos = pos, pattern = pattern)
obj.class <- napply(names, function(x) class(x)[1])
obj.mode <- napply(names, mode)
obj.type <- ifelse(is.na(obj.class), obj.mode, obj.class)
obj.prettysize <- napply(names, function(x) {
format(utils::object.size(x), units = "auto") })
obj.size <- napply(names, object.size)
obj.dim <- t(napply(names, function(x)
as.numeric(dim(x))[1:2]))
vec <- is.na(obj.dim)[, 1] & (obj.type != "function")
obj.dim[vec, 1] <- napply(names, length)[vec]
out <- data.frame(obj.type, obj.size, obj.prettysize, obj.dim)
names(out) <- c("Type", "Size", "PrettySize", "Length/Rows", "Columns")
if (!missing(order.by))
out <- out[order(out[[order.by]], decreasing=decreasing), ]
if (head)
out <- head(out, n)
out
}
# shorthand
lsos <- function(..., n=10) {
.ls.objects(..., order.by="Size", decreasing=TRUE, head=TRUE, n=n)
}

但是，当我尝试使用参数调用lsos时，为了显示内存中的所有对象，而不仅仅是第一个n，我收到以下错误：

> lsos(head = FALSE)
Error in .ls.objects(..., order.by = "Size", decreasing = TRUE, head = TRUE,  : 
formal argument "head" matched by multiple actual arguments

发生了什么事情？

Answer 1

lsos函数调用中的...基本上是调用

.ls.objects(head = FALSE, order.by="Size", decreasing=TRUE, head=TRUE, n=n)

请注意，您尝试设置head=两次(一次在...中，一次在函数调用中硬编码(。省略号(...(不会试图计算出哪些命名参数已经存在或决定哪个值"获胜"。如果希望能够覆盖这些值，则需要将它们设置为显式参数。

lsos <- function(..., order.by="Size", decreasing=TRUE, head=TRUE, n=10) {
.ls.objects(..., order.by=order.by, decreasing=decreasing, head=head, n=n)
}