clear everything but one variable rm(list = ls()) [duplicate]











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  • How can I remove all objects but one from the workspace in R?

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I'm aware that rm(list = ls()) will clear my workspace.



I have a data frame x that I would like to preserve while removing everything else from memory. How can I do that?



rm(list = ls(!x)) #???





r







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marked as duplicate by Ronak Shah r
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 Nov 9 at 18:18


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    This question already has an answer here:




    • How can I remove all objects but one from the workspace in R?

      13 answers




    I'm aware that rm(list = ls()) will clear my workspace.



    I have a data frame x that I would like to preserve while removing everything else from memory. How can I do that?



    rm(list = ls(!x)) #???





    r







    share|improve this question













    marked as duplicate by Ronak Shah r
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     Nov 9 at 18:18


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      up vote
      0
      down vote

      favorite












      This question already has an answer here:




      • How can I remove all objects but one from the workspace in R?

        13 answers




      I'm aware that rm(list = ls()) will clear my workspace.



      I have a data frame x that I would like to preserve while removing everything else from memory. How can I do that?



      rm(list = ls(!x)) #???





      r







      share|improve this question














      This question already has an answer here:




      • How can I remove all objects but one from the workspace in R?

        13 answers




      I'm aware that rm(list = ls()) will clear my workspace.



      I have a data frame x that I would like to preserve while removing everything else from memory. How can I do that?



      rm(list = ls(!x)) #???




      This question already has an answer here:




      • How can I remove all objects but one from the workspace in R?

        13 answers







      r




      r






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 9 at 18:04









      Doug Fir

      5,2722780148




      5,2722780148




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          1 Answer
          1






          active

          oldest

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          up vote
          6
          down vote



          accepted










          Try: rm(list = setdiff(ls(), x))



          Edit based on mickey's comment:



          Three objects in environment: 



          ls()
          
[1] "data_df" "list_ls" "vector_v"


          Remove data_df:



          rm(list = setdiff(ls(), "data_df"))
          
ls()
          
[1] "data_df"


          Vector of things to keep:



          toKeep_v <- c("list_ls", "vector_v")
          
rm(list = setdiff(ls(), toKeep_v)
          
ls()
          
[1] "list_ls" "vector_v"





          share|improve this answer



















          • 5




            If x is the name of the data frame, it should be in quotes.
            – mickey
            Nov 9 at 18:08






          • 1




            While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
            – James
            Nov 9 at 18:10




















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          6
          down vote



          accepted










          Try: rm(list = setdiff(ls(), x))



          Edit based on mickey's comment:



          Three objects in environment: 



          ls()
          
[1] "data_df" "list_ls" "vector_v"


          Remove data_df:



          rm(list = setdiff(ls(), "data_df"))
          
ls()
          
[1] "data_df"


          Vector of things to keep:



          toKeep_v <- c("list_ls", "vector_v")
          
rm(list = setdiff(ls(), toKeep_v)
          
ls()
          
[1] "list_ls" "vector_v"





          share|improve this answer



















          • 5




            If x is the name of the data frame, it should be in quotes.
            – mickey
            Nov 9 at 18:08






          • 1




            While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
            – James
            Nov 9 at 18:10

















          up vote
          6
          down vote



          accepted










          Try: rm(list = setdiff(ls(), x))



          Edit based on mickey's comment:



          Three objects in environment: 



          ls()
          
[1] "data_df" "list_ls" "vector_v"


          Remove data_df:



          rm(list = setdiff(ls(), "data_df"))
          
ls()
          
[1] "data_df"


          Vector of things to keep:



          toKeep_v <- c("list_ls", "vector_v")
          
rm(list = setdiff(ls(), toKeep_v)
          
ls()
          
[1] "list_ls" "vector_v"





          share|improve this answer



















          • 5




            If x is the name of the data frame, it should be in quotes.
            – mickey
            Nov 9 at 18:08






          • 1




            While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
            – James
            Nov 9 at 18:10















          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          Try: rm(list = setdiff(ls(), x))



          Edit based on mickey's comment:



          Three objects in environment: 



          ls()
          
[1] "data_df" "list_ls" "vector_v"


          Remove data_df:



          rm(list = setdiff(ls(), "data_df"))
          
ls()
          
[1] "data_df"


          Vector of things to keep:



          toKeep_v <- c("list_ls", "vector_v")
          
rm(list = setdiff(ls(), toKeep_v)
          
ls()
          
[1] "list_ls" "vector_v"





          share|improve this answer














          Try: rm(list = setdiff(ls(), x))



          Edit based on mickey's comment:



          Three objects in environment: 



          ls()
          
[1] "data_df" "list_ls" "vector_v"


          Remove data_df:



          rm(list = setdiff(ls(), "data_df"))
          
ls()
          
[1] "data_df"


          Vector of things to keep:



          toKeep_v <- c("list_ls", "vector_v")
          
rm(list = setdiff(ls(), toKeep_v)
          
ls()
          
[1] "list_ls" "vector_v"






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 9 at 18:14

























          answered Nov 9 at 18:06









          Qwfqwf

          1488




          1488








          • 5




            If x is the name of the data frame, it should be in quotes.
            – mickey
            Nov 9 at 18:08






          • 1




            While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
            – James
            Nov 9 at 18:10
















          • 5




            If x is the name of the data frame, it should be in quotes.
            – mickey
            Nov 9 at 18:08






          • 1




            While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
            – James
            Nov 9 at 18:10










          5




          5




          If x is the name of the data frame, it should be in quotes.
          – mickey
          Nov 9 at 18:08




          If x is the name of the data frame, it should be in quotes.
          – mickey
          Nov 9 at 18:08




          1




          1




          While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
          – James
          Nov 9 at 18:10






          While rm has NSE, setdiff doesn't so this will have unexpected results unless x is "x"
          – James
          Nov 9 at 18:10





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