Myujth

This is certainly a simple question but I can't find a solution.

I want to clean my environment by removing some variables I don't need anymore and keep some others.

I unterstand ls() can list them and ls()[[i]] returns the name of the variable, as a string.

So If I want to remove the 10th, let's say it's the variable age , ls()[[10]] will return "age", and  I would like to do something like rm(ls()[[10]), but it does not work. I can't figure out to force rm(ls([10])) to be be equivalent to rm(age).

I guess I need to force some evaluation of string "age" to return the variable age but can't find the proper function in R documentation.

Thanks if you can help.

How about the following:

1: Grab the list in the environment,

2: Define the items you want to remove,

3: Filter the list by the items you want to remove

4: Then remove them

 list <- ls()
 to_remove <- c("Item1", "Item2")
 list_to_remove <- list[ list %in% to_remove]
 list_to_remove

 rm(list=list_to_remove)

The list argument of rm will help you. It accepts a character vector. Consider:

age <- 1
rm(list = "age") # Same effect as rm(age)
age
#Error: object 'age' not found

So running e.g.

rm(list = ls())

will clear all visible objects in the specified environment.

In your case rm(list = ls()[10]) will do what you want. However, note that ls() always returns a sorted character vector, so the 10th entry can change rather easily. You probably want to do the following

objects_to_remove <- c("age", "another_object") # etc
rm(list = objects_to_remove)