使用 rm() 删除多个对象

Question

My memory is getting clogged by a bunch of intermediate files (call them temp1 , temp2 , etc.). and I would like to know if it is possible to remove them from memory without doing repeated rm calls (ie rm(temp1) , rm(temp2) )?

I tried rm(list(temp1, temp2, etc.)) , but that doesn't seem to work.

Answer 1

Make the list a character vector (not a vector of names)

rm(list = c('temp1','temp2'))

or

rm(temp1, temp2)

Answer 2

另一个解决方案rm(list=ls(pattern="temp")) ，删除与模式匹配的所有对象。

Answer 3

Or using regular expressions

"rmlike" <- function(...) {
  names <- sapply(
    match.call(expand.dots = FALSE)$..., as.character)
  names = paste(names,collapse="|")
  Vars <- ls(1)
  r <- Vars[grep(paste("^(",names,").*",sep=""),Vars)]
  rm(list=r,pos=1)
}

rmlike(temp)

Answer 4

Another variation you can try is(expanding @mnel's answer) if you have many temp'x'.

here "n" could be the number of temp variables present

rm(list = c(paste("temp",c(1:n),sep="")))