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





















          Your Answer






          StackExchange.ifUsing("editor", function () {
          StackExchange.using("externalEditor", function () {
          StackExchange.using("snippets", function () {
          StackExchange.snippets.init();
          });
          });
          }, "code-snippets");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "1"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53991996%2fset-all-vector-elements-to-na-in-a-list-of-vectors%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Stack Overflow!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53991996%2fset-all-vector-elements-to-na-in-a-list-of-vectors%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Eastern Orthodox Church

                                          Image
                                          Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
                                          Read more

                                          Zagreb

                                          Image
                                          This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
                                          Read more

                                          Understanding the information contained in the Deep Space Network XML data?

                                          Image
                                          up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
                                          Read more