A function to fill in a column with NA of the same type











up vote
12
down vote

favorite
2












I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.



for example:



df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))

df[, 1:2] <- NA


yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:



df[,1] = as.numeric(NA)
df[,2] = as.character(NA)


But how do I do this collectively in a loop for all columns with all possible types of NA?










share|improve this question




















  • 3




    Good question +1, but why does this matter?
    – Tim Biegeleisen
    18 hours ago










  • It's a very weird problem, I later need to join the data frame with another frame of the original type...
    – Omry Atia
    18 hours ago






  • 1




    But why? Please give us more context, seems like pointless (but fun) step.
    – zx8754
    17 hours ago










  • I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
    – Omry Atia
    16 hours ago















up vote
12
down vote

favorite
2












I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.



for example:



df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))

df[, 1:2] <- NA


yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:



df[,1] = as.numeric(NA)
df[,2] = as.character(NA)


But how do I do this collectively in a loop for all columns with all possible types of NA?










share|improve this question




















  • 3




    Good question +1, but why does this matter?
    – Tim Biegeleisen
    18 hours ago










  • It's a very weird problem, I later need to join the data frame with another frame of the original type...
    – Omry Atia
    18 hours ago






  • 1




    But why? Please give us more context, seems like pointless (but fun) step.
    – zx8754
    17 hours ago










  • I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
    – Omry Atia
    16 hours ago













up vote
12
down vote

favorite
2









up vote
12
down vote

favorite
2






2





I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.



for example:



df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))

df[, 1:2] <- NA


yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:



df[,1] = as.numeric(NA)
df[,2] = as.character(NA)


But how do I do this collectively in a loop for all columns with all possible types of NA?










share|improve this question















I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.



for example:



df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))

df[, 1:2] <- NA


yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:



df[,1] = as.numeric(NA)
df[,2] = as.character(NA)


But how do I do this collectively in a loop for all columns with all possible types of NA?







r dplyr na






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 17 hours ago









zx8754

28.9k76395




28.9k76395










asked 18 hours ago









Omry Atia

702412




702412








  • 3




    Good question +1, but why does this matter?
    – Tim Biegeleisen
    18 hours ago










  • It's a very weird problem, I later need to join the data frame with another frame of the original type...
    – Omry Atia
    18 hours ago






  • 1




    But why? Please give us more context, seems like pointless (but fun) step.
    – zx8754
    17 hours ago










  • I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
    – Omry Atia
    16 hours ago














  • 3




    Good question +1, but why does this matter?
    – Tim Biegeleisen
    18 hours ago










  • It's a very weird problem, I later need to join the data frame with another frame of the original type...
    – Omry Atia
    18 hours ago






  • 1




    But why? Please give us more context, seems like pointless (but fun) step.
    – zx8754
    17 hours ago










  • I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
    – Omry Atia
    16 hours ago








3




3




Good question +1, but why does this matter?
– Tim Biegeleisen
18 hours ago




Good question +1, but why does this matter?
– Tim Biegeleisen
18 hours ago












It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
18 hours ago




It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
18 hours ago




1




1




But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
17 hours ago




But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
17 hours ago












I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
16 hours ago




I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
16 hours ago












5 Answers
5






active

oldest

votes

















up vote
8
down vote



accepted










You can use this "trick" :



df[1:nrow(df),1] <- NA
df[1:nrow(df),2] <- NA


the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.



Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :



rowIdxs <- 1:nrow(df)
df[rowIdxs ,1] <- NA
df[rowIdxs ,2] <- NA
df[rowIdxs ,3] <- NA
...




As cleverly suggested by @RonakShah, you can also use :



df[TRUE, 1] <- NA
df[TRUE, 2] <- NA
...




As pointed out by @Cath both the methods still work when you select more than one column e.g. :



df[TRUE, 1:3] <- NA
# or
df[1:nrow(df), 1:3] <- NA





share|improve this answer























  • This doesn't seem to work... df is still logical :(
    – Omry Atia
    18 hours ago










  • @OmryAtia : edited. it should work now ;)
    – digEmAll
    18 hours ago










  • Awesome... so simple :)
    – Omry Atia
    18 hours ago






  • 3




    why not directly df[TRUE, 1:2] <- NA?
    – Cath
    17 hours ago










  • @Cath: sure, added in the answer, thanks !
    – digEmAll
    17 hours ago


















up vote
7
down vote













Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.



df[!is.na(df)] <- NA


which gives,




# A tibble: 3 x 2
x y
<dbl> <chr>
1 NA <NA>
2 NA <NA>
3 NA <NA>






share|improve this answer




























    up vote
    3
    down vote













    Using dplyr::na_if:



    library(dplyr)

    df %>%
    mutate(x = na_if(x, x),
    y = na_if(y, y))

    # # A tibble: 3 x 2
    # x y
    # <dbl> <chr>
    # 1 NA NA
    # 2 NA NA
    # 3 NA NA


    If we want to mutate only subset of columns to NA, then:



    # dataframe with extra column that stay unchanged
    df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))

    df %>%
    mutate_at(vars(x, y), funs(na_if(.,.)))

    # # A tibble: 3 x 3
    # x y z
    # <dbl> <chr> <int>
    # 1 NA NA 4
    # 2 NA NA 5
    # 3 NA NA 6





    share|improve this answer






























      up vote
      3
      down vote













      Another way to change all columns at once while keeping the variables' classes:



      df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})

      df
      # A tibble: 3 x 2
      # x y
      # <dbl> <chr>
      #1 NA <NA>
      #2 NA <NA>
      #3 NA <NA>




      As @digEmAll notified in comments, there is another similar but shorter way:



      df <- lapply(df, function(x) as(NA,class(x)))





      share|improve this answer



















      • 2




        Also lapply(df, function(x)as(NA,class(x))) should work
        – digEmAll
        17 hours ago










      • @digEmAll indeed and much shorter thanks!
        – Cath
        17 hours ago






      • 2




        another base option df <- lapply(df, replace, TRUE, NA)
        – docendo discimus
        15 hours ago










      • This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
        – Emil Bode
        8 hours ago


















      up vote
      0
      down vote













      Using bind_cols() from dplyr you can also do:



      df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))
      classes <- sapply(df, class)
      df[,1:2] <- NA

      bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))

      V1 V2
      <dbl> <chr>
      1 NA NA
      2 NA NA
      3 NA NA


      Please note that this will change the colnames.






      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








        up vote
        8
        down vote



        accepted










        You can use this "trick" :



        df[1:nrow(df),1] <- NA
        df[1:nrow(df),2] <- NA


        the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.



        Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :



        rowIdxs <- 1:nrow(df)
        df[rowIdxs ,1] <- NA
        df[rowIdxs ,2] <- NA
        df[rowIdxs ,3] <- NA
        ...




        As cleverly suggested by @RonakShah, you can also use :



        df[TRUE, 1] <- NA
        df[TRUE, 2] <- NA
        ...




        As pointed out by @Cath both the methods still work when you select more than one column e.g. :



        df[TRUE, 1:3] <- NA
        # or
        df[1:nrow(df), 1:3] <- NA





        share|improve this answer























        • This doesn't seem to work... df is still logical :(
          – Omry Atia
          18 hours ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          18 hours ago










        • Awesome... so simple :)
          – Omry Atia
          18 hours ago






        • 3




          why not directly df[TRUE, 1:2] <- NA?
          – Cath
          17 hours ago










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          17 hours ago















        up vote
        8
        down vote



        accepted










        You can use this "trick" :



        df[1:nrow(df),1] <- NA
        df[1:nrow(df),2] <- NA


        the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.



        Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :



        rowIdxs <- 1:nrow(df)
        df[rowIdxs ,1] <- NA
        df[rowIdxs ,2] <- NA
        df[rowIdxs ,3] <- NA
        ...




        As cleverly suggested by @RonakShah, you can also use :



        df[TRUE, 1] <- NA
        df[TRUE, 2] <- NA
        ...




        As pointed out by @Cath both the methods still work when you select more than one column e.g. :



        df[TRUE, 1:3] <- NA
        # or
        df[1:nrow(df), 1:3] <- NA





        share|improve this answer























        • This doesn't seem to work... df is still logical :(
          – Omry Atia
          18 hours ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          18 hours ago










        • Awesome... so simple :)
          – Omry Atia
          18 hours ago






        • 3




          why not directly df[TRUE, 1:2] <- NA?
          – Cath
          17 hours ago










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          17 hours ago













        up vote
        8
        down vote



        accepted







        up vote
        8
        down vote



        accepted






        You can use this "trick" :



        df[1:nrow(df),1] <- NA
        df[1:nrow(df),2] <- NA


        the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.



        Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :



        rowIdxs <- 1:nrow(df)
        df[rowIdxs ,1] <- NA
        df[rowIdxs ,2] <- NA
        df[rowIdxs ,3] <- NA
        ...




        As cleverly suggested by @RonakShah, you can also use :



        df[TRUE, 1] <- NA
        df[TRUE, 2] <- NA
        ...




        As pointed out by @Cath both the methods still work when you select more than one column e.g. :



        df[TRUE, 1:3] <- NA
        # or
        df[1:nrow(df), 1:3] <- NA





        share|improve this answer














        You can use this "trick" :



        df[1:nrow(df),1] <- NA
        df[1:nrow(df),2] <- NA


        the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.



        Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :



        rowIdxs <- 1:nrow(df)
        df[rowIdxs ,1] <- NA
        df[rowIdxs ,2] <- NA
        df[rowIdxs ,3] <- NA
        ...




        As cleverly suggested by @RonakShah, you can also use :



        df[TRUE, 1] <- NA
        df[TRUE, 2] <- NA
        ...




        As pointed out by @Cath both the methods still work when you select more than one column e.g. :



        df[TRUE, 1:3] <- NA
        # or
        df[1:nrow(df), 1:3] <- NA






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 17 hours ago

























        answered 18 hours ago









        digEmAll

        46.2k984120




        46.2k984120












        • This doesn't seem to work... df is still logical :(
          – Omry Atia
          18 hours ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          18 hours ago










        • Awesome... so simple :)
          – Omry Atia
          18 hours ago






        • 3




          why not directly df[TRUE, 1:2] <- NA?
          – Cath
          17 hours ago










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          17 hours ago


















        • This doesn't seem to work... df is still logical :(
          – Omry Atia
          18 hours ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          18 hours ago










        • Awesome... so simple :)
          – Omry Atia
          18 hours ago






        • 3




          why not directly df[TRUE, 1:2] <- NA?
          – Cath
          17 hours ago










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          17 hours ago
















        This doesn't seem to work... df is still logical :(
        – Omry Atia
        18 hours ago




        This doesn't seem to work... df is still logical :(
        – Omry Atia
        18 hours ago












        @OmryAtia : edited. it should work now ;)
        – digEmAll
        18 hours ago




        @OmryAtia : edited. it should work now ;)
        – digEmAll
        18 hours ago












        Awesome... so simple :)
        – Omry Atia
        18 hours ago




        Awesome... so simple :)
        – Omry Atia
        18 hours ago




        3




        3




        why not directly df[TRUE, 1:2] <- NA?
        – Cath
        17 hours ago




        why not directly df[TRUE, 1:2] <- NA?
        – Cath
        17 hours ago












        @Cath: sure, added in the answer, thanks !
        – digEmAll
        17 hours ago




        @Cath: sure, added in the answer, thanks !
        – digEmAll
        17 hours ago












        up vote
        7
        down vote













        Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.



        df[!is.na(df)] <- NA


        which gives,




        # A tibble: 3 x 2
        x y
        <dbl> <chr>
        1 NA <NA>
        2 NA <NA>
        3 NA <NA>






        share|improve this answer

























          up vote
          7
          down vote













          Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.



          df[!is.na(df)] <- NA


          which gives,




          # A tibble: 3 x 2
          x y
          <dbl> <chr>
          1 NA <NA>
          2 NA <NA>
          3 NA <NA>






          share|improve this answer























            up vote
            7
            down vote










            up vote
            7
            down vote









            Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.



            df[!is.na(df)] <- NA


            which gives,




            # A tibble: 3 x 2
            x y
            <dbl> <chr>
            1 NA <NA>
            2 NA <NA>
            3 NA <NA>






            share|improve this answer












            Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.



            df[!is.na(df)] <- NA


            which gives,




            # A tibble: 3 x 2
            x y
            <dbl> <chr>
            1 NA <NA>
            2 NA <NA>
            3 NA <NA>







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 18 hours ago









            Sotos

            27.3k51640




            27.3k51640






















                up vote
                3
                down vote













                Using dplyr::na_if:



                library(dplyr)

                df %>%
                mutate(x = na_if(x, x),
                y = na_if(y, y))

                # # A tibble: 3 x 2
                # x y
                # <dbl> <chr>
                # 1 NA NA
                # 2 NA NA
                # 3 NA NA


                If we want to mutate only subset of columns to NA, then:



                # dataframe with extra column that stay unchanged
                df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))

                df %>%
                mutate_at(vars(x, y), funs(na_if(.,.)))

                # # A tibble: 3 x 3
                # x y z
                # <dbl> <chr> <int>
                # 1 NA NA 4
                # 2 NA NA 5
                # 3 NA NA 6





                share|improve this answer



























                  up vote
                  3
                  down vote













                  Using dplyr::na_if:



                  library(dplyr)

                  df %>%
                  mutate(x = na_if(x, x),
                  y = na_if(y, y))

                  # # A tibble: 3 x 2
                  # x y
                  # <dbl> <chr>
                  # 1 NA NA
                  # 2 NA NA
                  # 3 NA NA


                  If we want to mutate only subset of columns to NA, then:



                  # dataframe with extra column that stay unchanged
                  df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))

                  df %>%
                  mutate_at(vars(x, y), funs(na_if(.,.)))

                  # # A tibble: 3 x 3
                  # x y z
                  # <dbl> <chr> <int>
                  # 1 NA NA 4
                  # 2 NA NA 5
                  # 3 NA NA 6





                  share|improve this answer

























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Using dplyr::na_if:



                    library(dplyr)

                    df %>%
                    mutate(x = na_if(x, x),
                    y = na_if(y, y))

                    # # A tibble: 3 x 2
                    # x y
                    # <dbl> <chr>
                    # 1 NA NA
                    # 2 NA NA
                    # 3 NA NA


                    If we want to mutate only subset of columns to NA, then:



                    # dataframe with extra column that stay unchanged
                    df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))

                    df %>%
                    mutate_at(vars(x, y), funs(na_if(.,.)))

                    # # A tibble: 3 x 3
                    # x y z
                    # <dbl> <chr> <int>
                    # 1 NA NA 4
                    # 2 NA NA 5
                    # 3 NA NA 6





                    share|improve this answer














                    Using dplyr::na_if:



                    library(dplyr)

                    df %>%
                    mutate(x = na_if(x, x),
                    y = na_if(y, y))

                    # # A tibble: 3 x 2
                    # x y
                    # <dbl> <chr>
                    # 1 NA NA
                    # 2 NA NA
                    # 3 NA NA


                    If we want to mutate only subset of columns to NA, then:



                    # dataframe with extra column that stay unchanged
                    df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))

                    df %>%
                    mutate_at(vars(x, y), funs(na_if(.,.)))

                    # # A tibble: 3 x 3
                    # x y z
                    # <dbl> <chr> <int>
                    # 1 NA NA 4
                    # 2 NA NA 5
                    # 3 NA NA 6






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 17 hours ago

























                    answered 17 hours ago









                    zx8754

                    28.9k76395




                    28.9k76395






















                        up vote
                        3
                        down vote













                        Another way to change all columns at once while keeping the variables' classes:



                        df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})

                        df
                        # A tibble: 3 x 2
                        # x y
                        # <dbl> <chr>
                        #1 NA <NA>
                        #2 NA <NA>
                        #3 NA <NA>




                        As @digEmAll notified in comments, there is another similar but shorter way:



                        df <- lapply(df, function(x) as(NA,class(x)))





                        share|improve this answer



















                        • 2




                          Also lapply(df, function(x)as(NA,class(x))) should work
                          – digEmAll
                          17 hours ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          17 hours ago






                        • 2




                          another base option df <- lapply(df, replace, TRUE, NA)
                          – docendo discimus
                          15 hours ago










                        • This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                          – Emil Bode
                          8 hours ago















                        up vote
                        3
                        down vote













                        Another way to change all columns at once while keeping the variables' classes:



                        df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})

                        df
                        # A tibble: 3 x 2
                        # x y
                        # <dbl> <chr>
                        #1 NA <NA>
                        #2 NA <NA>
                        #3 NA <NA>




                        As @digEmAll notified in comments, there is another similar but shorter way:



                        df <- lapply(df, function(x) as(NA,class(x)))





                        share|improve this answer



















                        • 2




                          Also lapply(df, function(x)as(NA,class(x))) should work
                          – digEmAll
                          17 hours ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          17 hours ago






                        • 2




                          another base option df <- lapply(df, replace, TRUE, NA)
                          – docendo discimus
                          15 hours ago










                        • This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                          – Emil Bode
                          8 hours ago













                        up vote
                        3
                        down vote










                        up vote
                        3
                        down vote









                        Another way to change all columns at once while keeping the variables' classes:



                        df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})

                        df
                        # A tibble: 3 x 2
                        # x y
                        # <dbl> <chr>
                        #1 NA <NA>
                        #2 NA <NA>
                        #3 NA <NA>




                        As @digEmAll notified in comments, there is another similar but shorter way:



                        df <- lapply(df, function(x) as(NA,class(x)))





                        share|improve this answer














                        Another way to change all columns at once while keeping the variables' classes:



                        df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})

                        df
                        # A tibble: 3 x 2
                        # x y
                        # <dbl> <chr>
                        #1 NA <NA>
                        #2 NA <NA>
                        #3 NA <NA>




                        As @digEmAll notified in comments, there is another similar but shorter way:



                        df <- lapply(df, function(x) as(NA,class(x)))






                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited 17 hours ago

























                        answered 17 hours ago









                        Cath

                        19.6k43464




                        19.6k43464








                        • 2




                          Also lapply(df, function(x)as(NA,class(x))) should work
                          – digEmAll
                          17 hours ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          17 hours ago






                        • 2




                          another base option df <- lapply(df, replace, TRUE, NA)
                          – docendo discimus
                          15 hours ago










                        • This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                          – Emil Bode
                          8 hours ago














                        • 2




                          Also lapply(df, function(x)as(NA,class(x))) should work
                          – digEmAll
                          17 hours ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          17 hours ago






                        • 2




                          another base option df <- lapply(df, replace, TRUE, NA)
                          – docendo discimus
                          15 hours ago










                        • This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                          – Emil Bode
                          8 hours ago








                        2




                        2




                        Also lapply(df, function(x)as(NA,class(x))) should work
                        – digEmAll
                        17 hours ago




                        Also lapply(df, function(x)as(NA,class(x))) should work
                        – digEmAll
                        17 hours ago












                        @digEmAll indeed and much shorter thanks!
                        – Cath
                        17 hours ago




                        @digEmAll indeed and much shorter thanks!
                        – Cath
                        17 hours ago




                        2




                        2




                        another base option df <- lapply(df, replace, TRUE, NA)
                        – docendo discimus
                        15 hours ago




                        another base option df <- lapply(df, replace, TRUE, NA)
                        – docendo discimus
                        15 hours ago












                        This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                        – Emil Bode
                        8 hours ago




                        This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
                        – Emil Bode
                        8 hours ago










                        up vote
                        0
                        down vote













                        Using bind_cols() from dplyr you can also do:



                        df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))
                        classes <- sapply(df, class)
                        df[,1:2] <- NA

                        bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))

                        V1 V2
                        <dbl> <chr>
                        1 NA NA
                        2 NA NA
                        3 NA NA


                        Please note that this will change the colnames.






                        share|improve this answer

























                          up vote
                          0
                          down vote













                          Using bind_cols() from dplyr you can also do:



                          df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))
                          classes <- sapply(df, class)
                          df[,1:2] <- NA

                          bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))

                          V1 V2
                          <dbl> <chr>
                          1 NA NA
                          2 NA NA
                          3 NA NA


                          Please note that this will change the colnames.






                          share|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Using bind_cols() from dplyr you can also do:



                            df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))
                            classes <- sapply(df, class)
                            df[,1:2] <- NA

                            bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))

                            V1 V2
                            <dbl> <chr>
                            1 NA NA
                            2 NA NA
                            3 NA NA


                            Please note that this will change the colnames.






                            share|improve this answer












                            Using bind_cols() from dplyr you can also do:



                            df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))
                            classes <- sapply(df, class)
                            df[,1:2] <- NA

                            bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))

                            V1 V2
                            <dbl> <chr>
                            1 NA NA
                            2 NA NA
                            3 NA NA


                            Please note that this will change the colnames.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 18 hours ago









                            alex_555

                            666315




                            666315






























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