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Rename objects in environment r

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1

I'd like to rename objects in environment r. For example,

y1 <- vector('list', 3)

x1 <- matrix(0, 3, 3)
x2 <- matrix(1, 3, 3)
x3 <- matrix(2, 3, 3)

y1[[1]] <- x1
y1[[2]] <- x2
y1[[3]] <- x3

y2 <- vector('list', 3)

y2[[1]] <- x1
y2[[2]] <- x2
y2[[3]] <- x3

y <- new.env()
y$y1 <- y1
y$y2 <- y2

names(y)

names(y) <- c('a', 'b')

I expected that the name of lists inside y was a and b, that is, names(y) equals c('a', 'b'),

Obs.: I can't rename manually the variables y1 and y2, I need to change them inside the environment.

r environment-variables

share|improve this question

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I mean, why assign them as y1 and y2 in the first place, if you want to rename them? Because otherwise the easiest way is to simply assign them as y$a and y$b.
  
  – Konrad Rudolph
  Mar 22 at 19:06
  

  
  
  
  

add a comment | 

1

I'd like to rename objects in environment r. For example,

y1 <- vector('list', 3)

x1 <- matrix(0, 3, 3)
x2 <- matrix(1, 3, 3)
x3 <- matrix(2, 3, 3)

y1[[1]] <- x1
y1[[2]] <- x2
y1[[3]] <- x3

y2 <- vector('list', 3)

y2[[1]] <- x1
y2[[2]] <- x2
y2[[3]] <- x3

y <- new.env()
y$y1 <- y1
y$y2 <- y2

names(y)

names(y) <- c('a', 'b')

I expected that the name of lists inside y was a and b, that is, names(y) equals c('a', 'b'),

Obs.: I can't rename manually the variables y1 and y2, I need to change them inside the environment.

r environment-variables

share|improve this question

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I mean, why assign them as y1 and y2 in the first place, if you want to rename them? Because otherwise the easiest way is to simply assign them as y$a and y$b.
  
  – Konrad Rudolph
  Mar 22 at 19:06
  

  
  
  
  

add a comment | 

I'd like to rename objects in environment r. For example,

y1 <- vector('list', 3)

x1 <- matrix(0, 3, 3)
x2 <- matrix(1, 3, 3)
x3 <- matrix(2, 3, 3)

y1[[1]] <- x1
y1[[2]] <- x2
y1[[3]] <- x3

y2 <- vector('list', 3)

y2[[1]] <- x1
y2[[2]] <- x2
y2[[3]] <- x3

y <- new.env()
y$y1 <- y1
y$y2 <- y2

names(y)

names(y) <- c('a', 'b')

I expected that the name of lists inside y was a and b, that is, names(y) equals c('a', 'b'),

Obs.: I can't rename manually the variables y1 and y2, I need to change them inside the environment.

r environment-variables

share|improve this question

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

y1 <- vector('list', 3)

x1 <- matrix(0, 3, 3)
x2 <- matrix(1, 3, 3)
x3 <- matrix(2, 3, 3)

y1[[1]] <- x1
y1[[2]] <- x2
y1[[3]] <- x3

y2 <- vector('list', 3)

y2[[1]] <- x1
y2[[2]] <- x2
y2[[3]] <- x3

y <- new.env()
y$y1 <- y1
y$y2 <- y2

names(y)

names(y) <- c('a', 'b')

share|improve this question

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

share|improve this question

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

asked Mar 22 at 18:38

Wagner JorgeWagner Jorge

84

2 Answers
2

active

oldest

votes

1

If you can’t assign them directly with the correct name, then the easiest is to replace the environment by a new one. If you absolutely need to preserve the environment (because it’s referenced elsewhere), you can replace its contents using the same trick:

objs = mget(ls(env), env)
rm(list = ls(env), envir = env)
list2env(setNames(objs, new_names), env)

The relevant part here is the last parameter to list2env: if you leave it off, this just creates a new environment. If you specify an existing environment, the names are added to that instead.

This code will leave hidden names (i.e. names starting with .) untouched — to change this, provide the all.names argument to ls, or use names.

share|improve this answer

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

add a comment | 

0

R doesn't really have a built in operation to rename variables in any environment. YOu could write a simple helper function to do that.

env_rename <- function(e, new_names, old_names = names(e)) 
 stopifnot(length(new_names)==length(old_names)) 
 orig_val <- mget(old_names, envir=e)
 rm(list=old_names, envir=e)
 for(i in seq_along(old_names)) 
 assign(new_names[i], orig_val[[i]], envir=e)
 

and call that with

env_rename(y, c("a","b"))

share|improve this answer

edited Mar 22 at 19:10

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This won’t work if the old and new names overlap.
  
  – Konrad Rudolph
  Mar 22 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KonradRudolph Good point. I'll fix.
  
  – MrFlick
  Mar 22 at 19:09
  

  
  
  
  

add a comment | 

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1

If you can’t assign them directly with the correct name, then the easiest is to replace the environment by a new one. If you absolutely need to preserve the environment (because it’s referenced elsewhere), you can replace its contents using the same trick:

objs = mget(ls(env), env)
rm(list = ls(env), envir = env)
list2env(setNames(objs, new_names), env)

The relevant part here is the last parameter to list2env: if you leave it off, this just creates a new environment. If you specify an existing environment, the names are added to that instead.

This code will leave hidden names (i.e. names starting with .) untouched — to change this, provide the all.names argument to ls, or use names.

share|improve this answer

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

add a comment | 

1

If you can’t assign them directly with the correct name, then the easiest is to replace the environment by a new one. If you absolutely need to preserve the environment (because it’s referenced elsewhere), you can replace its contents using the same trick:

objs = mget(ls(env), env)
rm(list = ls(env), envir = env)
list2env(setNames(objs, new_names), env)

The relevant part here is the last parameter to list2env: if you leave it off, this just creates a new environment. If you specify an existing environment, the names are added to that instead.

This code will leave hidden names (i.e. names starting with .) untouched — to change this, provide the all.names argument to ls, or use names.

share|improve this answer

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

add a comment | 

If you can’t assign them directly with the correct name, then the easiest is to replace the environment by a new one. If you absolutely need to preserve the environment (because it’s referenced elsewhere), you can replace its contents using the same trick:

objs = mget(ls(env), env)
rm(list = ls(env), envir = env)
list2env(setNames(objs, new_names), env)

The relevant part here is the last parameter to list2env: if you leave it off, this just creates a new environment. If you specify an existing environment, the names are added to that instead.

This code will leave hidden names (i.e. names starting with .) untouched — to change this, provide the all.names argument to ls, or use names.

share|improve this answer

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

objs = mget(ls(env), env)
rm(list = ls(env), envir = env)
list2env(setNames(objs, new_names), env)

share|improve this answer

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

answered Mar 22 at 19:10

Konrad RudolphKonrad Rudolph

405k1017961042

0

R doesn't really have a built in operation to rename variables in any environment. YOu could write a simple helper function to do that.

env_rename <- function(e, new_names, old_names = names(e)) 
 stopifnot(length(new_names)==length(old_names)) 
 orig_val <- mget(old_names, envir=e)
 rm(list=old_names, envir=e)
 for(i in seq_along(old_names)) 
 assign(new_names[i], orig_val[[i]], envir=e)
 

and call that with

env_rename(y, c("a","b"))

share|improve this answer

edited Mar 22 at 19:10

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This won’t work if the old and new names overlap.
  
  – Konrad Rudolph
  Mar 22 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KonradRudolph Good point. I'll fix.
  
  – MrFlick
  Mar 22 at 19:09
  

  
  
  
  

add a comment | 

0

R doesn't really have a built in operation to rename variables in any environment. YOu could write a simple helper function to do that.

env_rename <- function(e, new_names, old_names = names(e)) 
 stopifnot(length(new_names)==length(old_names)) 
 orig_val <- mget(old_names, envir=e)
 rm(list=old_names, envir=e)
 for(i in seq_along(old_names)) 
 assign(new_names[i], orig_val[[i]], envir=e)
 

and call that with

env_rename(y, c("a","b"))

share|improve this answer

edited Mar 22 at 19:10

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This won’t work if the old and new names overlap.
  
  – Konrad Rudolph
  Mar 22 at 19:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KonradRudolph Good point. I'll fix.
  
  – MrFlick
  Mar 22 at 19:09
  

  
  
  
  

add a comment | 

R doesn't really have a built in operation to rename variables in any environment. YOu could write a simple helper function to do that.

env_rename <- function(e, new_names, old_names = names(e)) 
 stopifnot(length(new_names)==length(old_names)) 
 orig_val <- mget(old_names, envir=e)
 rm(list=old_names, envir=e)
 for(i in seq_along(old_names)) 
 assign(new_names[i], orig_val[[i]], envir=e)
 

and call that with

env_rename(y, c("a","b"))

share|improve this answer

edited Mar 22 at 19:10

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

env_rename <- function(e, new_names, old_names = names(e)) 
 stopifnot(length(new_names)==length(old_names)) 
 orig_val <- mget(old_names, envir=e)
 rm(list=old_names, envir=e)
 for(i in seq_along(old_names)) 
 assign(new_names[i], orig_val[[i]], envir=e)
 

env_rename(y, c("a","b"))

share|improve this answer

edited Mar 22 at 19:10

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176

answered Mar 22 at 19:02

MrFlickMrFlick

126k12143176