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











up vote
6
down vote

favorite
1












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
    1 hour ago










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










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















up vote
6
down vote

favorite
1












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
    1 hour ago










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










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













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





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 1 hour ago









zx8754

28.9k76395




28.9k76395










asked 1 hour ago









Omry Atia

672411




672411








  • 3




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










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










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














  • 3




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










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










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








3




3




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




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












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




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












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




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












5 Answers
5






active

oldest

votes

















up vote
5
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
    1 hour ago










  • @OmryAtia : edited. it should work now ;)
    – digEmAll
    1 hour ago










  • Awesome... so simple :)
    – Omry Atia
    1 hour ago






  • 3




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










  • @Cath: sure, added in the answer, thanks !
    – digEmAll
    38 mins ago


















up vote
6
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
      2
      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
        52 mins ago










      • @digEmAll indeed and much shorter thanks!
        – Cath
        51 mins 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





















        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',
        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%2f53719387%2fa-function-to-fill-in-a-column-with-na-of-the-same-type%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








        up vote
        5
        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
          1 hour ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          1 hour ago










        • Awesome... so simple :)
          – Omry Atia
          1 hour ago






        • 3




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










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          38 mins ago















        up vote
        5
        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
          1 hour ago










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          1 hour ago










        • Awesome... so simple :)
          – Omry Atia
          1 hour ago






        • 3




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










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          38 mins ago













        up vote
        5
        down vote



        accepted







        up vote
        5
        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 38 mins ago

























        answered 1 hour ago









        digEmAll

        46.1k984120




        46.1k984120












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










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          1 hour ago










        • Awesome... so simple :)
          – Omry Atia
          1 hour ago






        • 3




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










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          38 mins ago


















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










        • @OmryAtia : edited. it should work now ;)
          – digEmAll
          1 hour ago










        • Awesome... so simple :)
          – Omry Atia
          1 hour ago






        • 3




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










        • @Cath: sure, added in the answer, thanks !
          – digEmAll
          38 mins ago
















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




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












        @OmryAtia : edited. it should work now ;)
        – digEmAll
        1 hour ago




        @OmryAtia : edited. it should work now ;)
        – digEmAll
        1 hour ago












        Awesome... so simple :)
        – Omry Atia
        1 hour ago




        Awesome... so simple :)
        – Omry Atia
        1 hour ago




        3




        3




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




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












        @Cath: sure, added in the answer, thanks !
        – digEmAll
        38 mins ago




        @Cath: sure, added in the answer, thanks !
        – digEmAll
        38 mins ago












        up vote
        6
        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
          6
          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
            6
            down vote










            up vote
            6
            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 1 hour ago









            Sotos

            27.2k51640




            27.2k51640






















                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 1 hour ago

























                    answered 1 hour ago









                    zx8754

                    28.9k76395




                    28.9k76395






















                        up vote
                        2
                        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
                          52 mins ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          51 mins ago















                        up vote
                        2
                        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
                          52 mins ago










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          51 mins ago













                        up vote
                        2
                        down vote










                        up vote
                        2
                        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 49 mins ago

























                        answered 54 mins ago









                        Cath

                        19.5k43464




                        19.5k43464








                        • 2




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










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          51 mins ago














                        • 2




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










                        • @digEmAll indeed and much shorter thanks!
                          – Cath
                          51 mins ago








                        2




                        2




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




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












                        @digEmAll indeed and much shorter thanks!
                        – Cath
                        51 mins ago




                        @digEmAll indeed and much shorter thanks!
                        – Cath
                        51 mins 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 1 hour ago









                            alex_555

                            666315




                            666315






























                                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%2f53719387%2fa-function-to-fill-in-a-column-with-na-of-the-same-type%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

                                Trompette piccolo

                                Slow SSRS Report in dynamic grouping and multiple parameters

                                Simon Yates (cyclisme)