Convert column classes in data table

Question

I have a problem using data table  How do I convert column classes  Here is a simple example  With data frame I don t have a problem converting it  with data table I just don t know how   df  lt - data frame ID c rep  A   5   rep  B  5    Quarter c 1 5  1 5   value rnorm 10    One way  http   stackoverflow com questions 2851015 r-convert-data-frame-columns-from-factors-to-characters df  lt - data frame lapply df  as character   stringsAsFactors FALSE   Another way df    value    lt - as numeric df    value     library data table  dt  lt - data table ID c rep  A   5   rep  B  5    Quarter c 1 5  1 5   value rnorm 10   dt  lt - data table lapply dt  as character   stringsAsFactors FALSE    Error in rep     ncol xi     invalid  times  argument  Produces error  does data table not have the option stringsAsFactors  dt    ID   with FALSE   lt - as character dt    ID   with FALSE     Produces error  Error in    lt - data table    tmp       ID   with   FALSE  value    c 1  1  1  1  1  2  2  2  2  2        unused argument s   with   FALSE    Do I miss something obvious here    Update due to Matthew s post  I used an older version before  but even after updating to 1 6 6  the version I use now  I still get an error   Update 2  Let s say I want to convert every column of class  factor  to a  character  column  but don t know in advance which column is of which class  With a data frame  I can do the following   classes  lt - as character sapply df  class   colClasses  lt - which classes   factor   df   colClasses   lt - sapply df   colClasses   as character    Can I do something similar with data table   Update 3       sessionInfo         R version 2 13 1  2011-07-08        Platform  x86 64-pc-mingw32 x64  64-bit    locale   1  C  attached base packages   1  stats     graphics  grDevices utils     datasets  methods   base       other attached packages   1  data table 1 6 6  loaded via a namespace  and not attached    1  tools 2 13 1

User · Accepted Answer

For a single column   dtnew  lt - dt   Quarter  as character Quarter   str dtnew   Classes    data table    and  data frame    10 obs  of  3 variables     ID       Factor w  2 levels  A   B   1 1 1 1 1 2 2 2 2 2    Quarter  chr   1   2   3   4         value    num  -0 838 0 146 -1 059 -1 197 0 282         Using lapply and as character   dtnew  lt - dt   lapply  SD  as character   by ID  str dtnew   Classes    data table    and  data frame    10 obs  of  3 variables     ID       Factor w  2 levels  A   B   1 1 1 1 1 2 2 2 2 2    Quarter  chr   1   2   3   4         value    chr   1 487145280568   -0 827845218358881   0 028977182770002   1 35392750102305

User · Answer

Raising Matt Dowle s comment to Geneorama s answer  https   stackoverflow com a 20808945 4241780  to make it more obvious  as encouraged   you can use for     set         library data table   DT   data table a   LETTERS c 3L 1 3    b   4 7  c   letters 1 4   DT1  lt - copy DT  names factors  lt - c  a    c    for col in names factors    set DT  j   col  value   as factor DT  col      sapply DT  class    gt          a         b         c    gt    factor   integer    factor    Created on 2020-02-12 by the reprex package  v0 3 0   See another of Matt s comments at https   stackoverflow com a 33000778 4241780 for more info   Edit    As noted by Espen and in help set   j may be  Column name s   character  or number s   integer  to be assigned value when column s  already exist   So  names factors  lt - c 1L  3L  will also work

User · Answer

I tried several approaches     BY  dplyr  data table ID        c rep  A   5   rep  B  5                Quarter   c 1 5  1 5               value     rnorm 10   - gt  df1 df1   lt  gt   dplyr  mutate ID        as factor ID                          Quarter   as character Quarter     check classes dplyr  glimpse df1    Observations  10   Variables  3     ID       fctr  A  A  A  A  A  B  B  B  B  B     Quarter  chr   1    2    3    4    5    1    2    3    4    5      value    dbl  -0 07676732  0 25376110  2 47192852  0 84929175  -0 13567312   -0 94224435  0 80213218  -0 89652819        or otherwise    from list to data table using data table  setDT list ID        as factor c rep  A   5   rep  B  5           Quarter   as character c 1 5  1 5          value     rnorm 10     gt   setDT list df  - gt  df2 class df2     1   data table   data frame

User · Answer

Try this  DT  lt - data table X1   c  a    b    X2   c 1 2   X3   c  hello    you    changeCols  lt - colnames DT  which as vector DT  lapply  SD  class        character     DT   changeCols    lapply  SD  as factor    SDcols   changeCols

User · Answer

If you have a list of column names in data table  you want to change the class of do   convert to character  lt - c  Quarter    value    dt   convert to character   lt - dt   lapply  SD  as character    SDcols   convert to character

User · Answer

try   dt  lt - data table A   c 1 5                     B  c 11 15    x  lt - ncol dt   for i in 1 x          dt  i    lt - as character dt  i

User · Answer

Here is the same way as  Nera suggested to check the class first but instead of using  SD is to use the fast loop of data table with set as  Matt Dowle solution with added class check  for  j in seq len ncol DT       if class DT  j        factor       set DT  j   j  value   as character DT  j

User · Answer

I provide a more general and safer way to do this stuff         lt - function  x       stopifnot inherits x   character      stopifnot length x     1    get x  parent frame 4       set colclass  lt - function x  class     stopifnot all class  in  c  integer    numeric    double   factor   character       for i in intersect names class   names x         f  lt - get paste0  as    class i        x        i         f   get     i            invisible x      The function    makes sure we get a variable out of the scope of data table  set colclass will set the classes of your cols  You can use it like this   dt  lt - data table i 1 3 f 3 1  set colclass dt  c i  character    class dt i

User · Answer

This is a BAD way to do it  I m only leaving this answer in case it solves other weird problems  These better methods are the probably partly the result of newer data table versions    so it s worth while to document this hard way  Plus  this is a nice syntax example for eval substitute syntax   library data table  dt  lt - data table ID   c rep  A   5   rep  B  5                      fac1   c 1 5  1 5                     fac2   c 1 5  1 5    2                    val1   rnorm 10                    val2   rnorm 10    names factors   c  fac1    fac2   names values   c  val1    val2    for  col in names factors     e   substitute X    as factor X   list X   as symbol col      dt    eval e     for  col in names values     e   substitute X    as numeric X   list X   as symbol col      dt    eval e      str dt    which gives you  Classes    data table    and  data frame    10 obs  of  5 variables     ID    chr   A   A   A   A         fac1  Factor w  5 levels  1   2   3   4      1 2 3 4 5 1 2 3 4 5    fac2  Factor w  5 levels  2   4   6   8      1 2 3 4 5 1 2 3 4 5    val1  num  0 0459 2 0113 0 5186 -0 8348 -0 2185        val2  num  -0 0688 0 6544 0 267 -0 1322 -0 4893      - attr      internal selfref    lt externalptr gt

[r] Convert column classes in data.table

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