我有一个有很多对象的工作空间，我想删除所有的，但只有一个。理想情况下，我希望避免键入rm(obj。1、obj.2……obj.n)。是否有可能指示删除除这些对象之外的所有对象?
当前回答
从函数内部，rm .GlobalEnv中除函数外的所有对象
initialize <- function(country.name) {
if (length(setdiff(ls(pos = .GlobalEnv), "initialize")) > 0) {
rm(list=setdiff(ls(pos = .GlobalEnv), "initialize"), pos = .GlobalEnv)
}
}
其他回答
将v替换为要保留的对象的名称
rm(list=(ls()[ls()!="v"]))
hat-tip: http://r.789695.n4.nabble.com/Removing-objects-and-clearing-memory-tp3445763p3445865.html
使用gdata包中的keep函数非常方便。
> ls()
[1] "a" "b" "c"
library(gdata)
> keep(a) #shows you which variables will be removed
[1] "b" "c"
> keep(a, sure = TRUE) # setting sure to TRUE removes variables b and c
> ls()
[1] "a"
这利用了ls()的模式选项，在这种情况下，你有很多具有相同模式的对象，而你不想保留:
> foo1 <- "junk"; foo2 <- "rubbish"; foo3 <- "trash"; x <- "gold"
> ls()
[1] "foo1" "foo2" "foo3" "x"
> # Let's check first what we want to remove
> ls(pattern = "foo")
[1] "foo1" "foo2" "foo3"
> rm(list = ls(pattern = "foo"))
> ls()
[1] "x"
要保留所有名称与模式匹配的对象，可以使用grep，如下所示:
to.remove <- ls()
to.remove <- c(to.remove[!grepl("^obj", to.remove)], "to.remove")
rm(list=to.remove)
从函数内部，rm .GlobalEnv中除函数外的所有对象
initialize <- function(country.name) {
if (length(setdiff(ls(pos = .GlobalEnv), "initialize")) > 0) {
rm(list=setdiff(ls(pos = .GlobalEnv), "initialize"), pos = .GlobalEnv)
}
}
