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Creating a sublist for elements within a list of lists after creation of list object


List of lists changes reflected across sublists unexpectedlyHow do I remove an element from a list by index in Python?Getting the last element of a list in PythonShuffling a list of objectsHow do I get the number of elements in a list in Python?Is there a simple way to delete a list element by value?How to Sort a List<T> by a property in the objectPython - comparing sublists within two different listsAugment sublists in a list of sublists with missing elements as NAProlog - longest sublist within a list













2















Sample list:



sublist <- list(foo = "foo", bar = "bar", baz = "baz")

sample_list <- list(a = sublist,
 b = sublist,
 c = sublist)


Here, I would like to create a sublist for the elements in a b and c within each of the aforementioned lists. I.e I would like to nest foo, bar, baz, one list further down, after I have created the list in the manner above.



Desired output:



sample_list <- list(a = list(a_down = sublist),
 b = list(b_down = sublist),
 c = list(c_down = sublist))





r list recursion







share|improve this question



















  • 1





    Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

    – Rui Barradas
    15 hours ago















2















Sample list:



sublist <- list(foo = "foo", bar = "bar", baz = "baz")

sample_list <- list(a = sublist,
 b = sublist,
 c = sublist)


Here, I would like to create a sublist for the elements in a b and c within each of the aforementioned lists. I.e I would like to nest foo, bar, baz, one list further down, after I have created the list in the manner above.



Desired output:



sample_list <- list(a = list(a_down = sublist),
 b = list(b_down = sublist),
 c = list(c_down = sublist))





r list recursion







share|improve this question



















  • 1





    Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

    – Rui Barradas
    15 hours ago













2












2








2


1






Sample list:



sublist <- list(foo = "foo", bar = "bar", baz = "baz")

sample_list <- list(a = sublist,
 b = sublist,
 c = sublist)


Here, I would like to create a sublist for the elements in a b and c within each of the aforementioned lists. I.e I would like to nest foo, bar, baz, one list further down, after I have created the list in the manner above.



Desired output:



sample_list <- list(a = list(a_down = sublist),
 b = list(b_down = sublist),
 c = list(c_down = sublist))





r list recursion







share|improve this question
















Sample list:



sublist <- list(foo = "foo", bar = "bar", baz = "baz")

sample_list <- list(a = sublist,
 b = sublist,
 c = sublist)


Here, I would like to create a sublist for the elements in a b and c within each of the aforementioned lists. I.e I would like to nest foo, bar, baz, one list further down, after I have created the list in the manner above.



Desired output:



sample_list <- list(a = list(a_down = sublist),
 b = list(b_down = sublist),
 c = list(c_down = sublist))





r list recursion




r list recursion






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 15 hours ago







VictorNautica

















asked 15 hours ago









VictorNauticaVictorNautica

526213




526213







  • 1





    Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

    – Rui Barradas
    15 hours ago












  • 1





    Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

    – Rui Barradas
    15 hours ago







1




1





Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

– Rui Barradas
15 hours ago





Something like lapply(1:3, function(i) setNames(list(sublist), letters[i]))?

– Rui Barradas
15 hours ago












1 Answer
1






active

oldest

votes


















1














We can use imap



library(purrr)
out2 <- imap(sample_list, ~ set_names(list(.x), paste0(.y, "_down")))


or making use of lst



imap(sample_list, ~ lst(!! paste0(.y, "_down") := .x))


-checking with OP's output



identical(out, out2)
#[1] TRUE





share|improve this answer






















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    We can use imap



    library(purrr)
    out2 <- imap(sample_list, ~ set_names(list(.x), paste0(.y, "_down")))


    or making use of lst



    imap(sample_list, ~ lst(!! paste0(.y, "_down") := .x))


    -checking with OP's output



    identical(out, out2)
    #[1] TRUE





    share|improve this answer



























      1














      We can use imap



      library(purrr)
      out2 <- imap(sample_list, ~ set_names(list(.x), paste0(.y, "_down")))


      or making use of lst



      imap(sample_list, ~ lst(!! paste0(.y, "_down") := .x))


      -checking with OP's output



      identical(out, out2)
      #[1] TRUE





      share|improve this answer

























        1












        1








        1







        We can use imap



        library(purrr)
        out2 <- imap(sample_list, ~ set_names(list(.x), paste0(.y, "_down")))


        or making use of lst



        imap(sample_list, ~ lst(!! paste0(.y, "_down") := .x))


        -checking with OP's output



        identical(out, out2)
        #[1] TRUE





        share|improve this answer













        We can use imap



        library(purrr)
        out2 <- imap(sample_list, ~ set_names(list(.x), paste0(.y, "_down")))


        or making use of lst



        imap(sample_list, ~ lst(!! paste0(.y, "_down") := .x))


        -checking with OP's output



        identical(out, out2)
        #[1] TRUE






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 15 hours ago









        akrunakrun

        415k13204278




        415k13204278





























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