Set all vector elements to NA in a list of vectors












6














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





r







share|improve this question




















  • 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
    10 hours ago








  • 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
    10 hours ago
















6














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





r







share|improve this question




















  • 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
    10 hours ago








  • 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
    10 hours ago














6












6








6







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





r







share|improve this question















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





r




r






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 9 hours ago

























asked 10 hours ago









lowndrul

1,16221735




1,16221735








  • 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
    10 hours ago








  • 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
    10 hours ago














  • 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
    10 hours ago








  • 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
    10 hours ago








1




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
10 hours ago






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
10 hours ago






1




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
10 hours ago




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
10 hours ago












5 Answers
5






active

oldest

votes


















8














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)





share|improve this answer























  • what is * called as in this context ?
    – YOLO
    10 hours ago






  • 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
    10 hours ago





















6














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





share|improve this answer





























    4














    another alternative with dplyr:



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





    share|improve this answer





























      4














      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")





      share|improve this answer





























        3














        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





        share|improve this answer





















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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8














          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)





          share|improve this answer























          • what is * called as in this context ?
            – YOLO
            10 hours ago






          • 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
            10 hours ago


















          8














          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)





          share|improve this answer























          • what is * called as in this context ?
            – YOLO
            10 hours ago






          • 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
            10 hours ago
















          8












          8








          8






          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)





          share|improve this answer














          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)






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 10 hours ago

























          answered 10 hours ago









          Julius Vainora

          32.3k75979




          32.3k75979












          • what is * called as in this context ?
            – YOLO
            10 hours ago






          • 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
            10 hours ago




















          • what is * called as in this context ?
            – YOLO
            10 hours ago






          • 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
            10 hours ago


















          what is * called as in this context ?
          – YOLO
          10 hours ago




          what is * called as in this context ?
          – YOLO
          10 hours ago




          1




          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
          10 hours ago






          @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
          10 hours ago















          6














          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





          share|improve this answer


























            6














            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





            share|improve this answer
























              6












              6








              6






              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





              share|improve this answer












              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






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 10 hours ago









              Rui Barradas

              16.1k41730




              16.1k41730























                  4














                  another alternative with dplyr:



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





                  share|improve this answer


























                    4














                    another alternative with dplyr:



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





                    share|improve this answer
























                      4












                      4








                      4






                      another alternative with dplyr:



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





                      share|improve this answer












                      another alternative with dplyr:



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






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 10 hours ago









                      Mankind_008

                      1,4682312




                      1,4682312























                          4














                          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")





                          share|improve this answer


























                            4














                            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")





                            share|improve this answer
























                              4












                              4








                              4






                              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")





                              share|improve this answer












                              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")






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 9 hours ago









                              G. Grothendieck

                              145k9126231




                              145k9126231























                                  3














                                  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





                                  share|improve this answer


























                                    3














                                    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





                                    share|improve this answer
























                                      3












                                      3








                                      3






                                      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





                                      share|improve this answer












                                      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






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 10 hours ago









                                      989

                                      8,91251834




                                      8,91251834






























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