Set all vector elements to NA in a list of vectors

How can I set all vector elements to NA in a list of vectors?

Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.

my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))

ret_list <- my_list



# Approach 1

for (element_name in names(my_list)) {

  ret_list[[element_name]] <- NA

}



ret_list

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA 



# Approach 2    

lapply(my_list, function(x) {x <- NA; return(x)})

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

1

both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?

– Ben Bolker
Dec 31 '18 at 23:27

1

Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.

– lowndrul
Dec 31 '18 at 23:40

add a comment |

How can I set all vector elements to NA in a list of vectors?

my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))

ret_list <- my_list



# Approach 1

for (element_name in names(my_list)) {

  ret_list[[element_name]] <- NA

}



ret_list

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA 



# Approach 2    

lapply(my_list, function(x) {x <- NA; return(x)})

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

1

both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?

– Ben Bolker
Dec 31 '18 at 23:27

1

Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.

– lowndrul
Dec 31 '18 at 23:40

add a comment |

How can I set all vector elements to NA in a list of vectors?

my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))

ret_list <- my_list



# Approach 1

for (element_name in names(my_list)) {

  ret_list[[element_name]] <- NA

}



ret_list

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA 



# Approach 2    

lapply(my_list, function(x) {x <- NA; return(x)})

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

How can I set all vector elements to NA in a list of vectors?

my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))

ret_list <- my_list



# Approach 1

for (element_name in names(my_list)) {

  ret_list[[element_name]] <- NA

}



ret_list

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA 



# Approach 2    

lapply(my_list, function(x) {x <- NA; return(x)})

# $A

# a  b  c 

# NA NA NA 

# 

# $B

# x  y 

# NA NA

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

edited Jan 1 at 0:02

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

asked Dec 31 '18 at 22:56

lowndrul

1,24032037

1

both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?

– Ben Bolker
Dec 31 '18 at 23:27

1

Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.

– lowndrul
Dec 31 '18 at 23:40

add a comment |

1

both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?

– Ben Bolker
Dec 31 '18 at 23:27

1

Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.

– lowndrul
Dec 31 '18 at 23:40

both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?

– Ben Bolker
Dec 31 '18 at 23:27

Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.

– lowndrul
Dec 31 '18 at 23:40

add a comment |

5 Answers
5

active

oldest

votes

Here's another one for numeric vectors:

lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.

# $A

#  a  b  c 

# NA NA NA 

#

# $B

#  x  y 

# NA NA

More generally,

lapply(my_list, replace, TRUE, NA)

and

lapply(ret_list, ifelse, NA, NA)

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

1

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

add a comment |

You can use function is.na<- in a lapply loop.

ret_list <- lapply(my_list, `is.na<-`)

ret_list

#$A

# a  b  c 

#NA NA NA 

#

#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

add a comment |

another alternative with dplyr:

lapply(my_list, function(x) dplyr::na_if(x,x))

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

add a comment |

If the list is not restricted to one level then use rapply.

# test data modified from question

my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))



rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")

which can also be written as:

rapply(my_list2, replace, list = TRUE, values = NA, how = "list")

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

add a comment |

Another way around

relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)



#$A

# a  b  c 

#NA NA NA 



#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:30

989

9,02751834

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

Here's another one for numeric vectors:

lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.

# $A

#  a  b  c 

# NA NA NA 

#

# $B

#  x  y 

# NA NA

More generally,

lapply(my_list, replace, TRUE, NA)

and

lapply(ret_list, ifelse, NA, NA)

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

1

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

add a comment |

Here's another one for numeric vectors:

lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.

# $A

#  a  b  c 

# NA NA NA 

#

# $B

#  x  y 

# NA NA

More generally,

lapply(my_list, replace, TRUE, NA)

and

lapply(ret_list, ifelse, NA, NA)

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

1

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

add a comment |

Here's another one for numeric vectors:

lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.

# $A

#  a  b  c 

# NA NA NA 

#

# $B

#  x  y 

# NA NA

More generally,

lapply(my_list, replace, TRUE, NA)

and

lapply(ret_list, ifelse, NA, NA)

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

Here's another one for numeric vectors:

lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.

# $A

#  a  b  c 

# NA NA NA 

#

# $B

#  x  y 

# NA NA

More generally,

lapply(my_list, replace, TRUE, NA)

and

lapply(ret_list, ifelse, NA, NA)

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

edited Jan 1 at 14:28

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

answered Dec 31 '18 at 22:58

Julius Vainora

38k76584

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

1

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

add a comment |

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

1

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

what is * called as in this context ?

– YOLO
Dec 31 '18 at 23:12

@YOLO, I'm not sure what you mean. The first approach is the same as lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.

– Julius Vainora
Dec 31 '18 at 23:15

add a comment |

You can use function is.na<- in a lapply loop.

ret_list <- lapply(my_list, `is.na<-`)

ret_list

#$A

# a  b  c 

#NA NA NA 

#

#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

add a comment |

You can use function is.na<- in a lapply loop.

ret_list <- lapply(my_list, `is.na<-`)

ret_list

#$A

# a  b  c 

#NA NA NA 

#

#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

add a comment |

You can use function is.na<- in a lapply loop.

ret_list <- lapply(my_list, `is.na<-`)

ret_list

#$A

# a  b  c 

#NA NA NA 

#

#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

You can use function is.na<- in a lapply loop.

ret_list <- lapply(my_list, `is.na<-`)

ret_list

#$A

# a  b  c 

#NA NA NA 

#

#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

answered Dec 31 '18 at 23:16

Rui Barradas

17.4k51731

add a comment |

another alternative with dplyr:

lapply(my_list, function(x) dplyr::na_if(x,x))

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

add a comment |

another alternative with dplyr:

lapply(my_list, function(x) dplyr::na_if(x,x))

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

add a comment |

another alternative with dplyr:

lapply(my_list, function(x) dplyr::na_if(x,x))

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

another alternative with dplyr:

lapply(my_list, function(x) dplyr::na_if(x,x))

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

answered Dec 31 '18 at 23:36

Mankind_008

1,5582312

add a comment |

If the list is not restricted to one level then use rapply.

# test data modified from question

my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))



rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")

which can also be written as:

rapply(my_list2, replace, list = TRUE, values = NA, how = "list")

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

add a comment |

If the list is not restricted to one level then use rapply.

# test data modified from question

my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))



rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")

which can also be written as:

rapply(my_list2, replace, list = TRUE, values = NA, how = "list")

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

add a comment |

If the list is not restricted to one level then use rapply.

# test data modified from question

my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))



rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")

which can also be written as:

rapply(my_list2, replace, list = TRUE, values = NA, how = "list")

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

If the list is not restricted to one level then use rapply.

# test data modified from question

my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))



rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")

which can also be written as:

rapply(my_list2, replace, list = TRUE, values = NA, how = "list")

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

answered Jan 1 at 0:10

G. Grothendieck

150k10134238

add a comment |

Another way around

relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)



#$A

# a  b  c 

#NA NA NA 



#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:30

989

9,02751834

add a comment |

Another way around

relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)



#$A

# a  b  c 

#NA NA NA 



#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:30

989

9,02751834

add a comment |

Another way around

relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)



#$A

# a  b  c 

#NA NA NA 



#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:30

989

9,02751834

Another way around

relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)



#$A

# a  b  c 

#NA NA NA 



#$B

# x  y 

#NA NA

answered Dec 31 '18 at 23:30

989

9,02751834

answered Dec 31 '18 at 23:30

989

9,02751834

answered Dec 31 '18 at 23:30

989

9,02751834

answered Dec 31 '18 at 23:30

989

9,02751834

add a comment |

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