Convert data frame columns from factors to characters

Question

I have a data frame  Let s call him bob    gt  head bob                   phenotype                         exclusion GSM399350 3- 4- 8- 25- 44  11b- 11c- 19- NK1 1- Gr1- TER119- GSM399351 3- 4- 8- 25- 44  11b- 11c- 19- NK1 1- Gr1- TER119- GSM399352 3- 4- 8- 25- 44  11b- 11c- 19- NK1 1- Gr1- TER119- GSM399353 3- 4- 8- 25  44  11b- 11c- 19- NK1 1- Gr1- TER119- GSM399354 3- 4- 8- 25  44  11b- 11c- 19- NK1 1- Gr1- TER119- GSM399355 3- 4- 8- 25  44  11b- 11c- 19- NK1 1- Gr1- TER119-   I d like to concatenate the rows of this data frame  this will be another question   But look    gt  class bob phenotype   1   factor    Bob s columns are factors  So  for example    gt  as character head bob    1   c 3  3  3  6  6  6          c 3  3  3  3  3  3          3   c 29  29  29  30  30  30     I don t begin to understand this  but I guess these are indices into the levels of the factors of the columns  of the court of king caractacus  of bob  Not what I need   Strangely I can go through the columns of bob by hand  and do  bob phenotype  lt - as character bob phenotype    which works fine  And  after some typing  I can get a data frame whose columns are characters rather than factors  So my question is  how can I do this automatically  How do I convert a data frame with factor columns into a data frame with character columns without having to manually go through each column    Bonus question  why does the manual approach work

User · Answer

This works transforming all to character and then the numeric to numeric:

makenumcols<-function(df){
  df<-as.data.frame(df)
  df[] <- lapply(df, as.character)
  cond <- apply(df, 2, function(x) {
    x <- x[!is.na(x)]
    all(suppressWarnings(!is.na(as.numeric(x))))
  })
  numeric_cols <- names(df)[cond]
  df[,numeric_cols] <- sapply(df[,numeric_cols], as.numeric)
  return(df)
}

Adapted from: Get column types of excel sheet automatically

User · Answer

If you would use data table package for the operations on data frame then the problem is not present       library data table  dt   data table col1   c  a   b   c    col2   1 3  sapply dt  class          col1        col2    character     integer     If you have a factor columns in you dataset already and you want to convert them to character you can do the following     library data table  dt   data table col1   factor c  a   b   c     col2   1 3  sapply dt  class        col1      col2     factor   integer   upd cols   sapply dt  is factor  dt   names dt  upd cols     lapply  SD  as character    SDcols   upd cols  sapply dt  class          col1        col2    character     integer

User · Answer

If you want a new data frame bobc where every factor vector in bobf is converted to a character vector  try this   bobc  lt - rapply bobf  as character  classes  factor   how  replace     If you then want to convert it back  you can create a logical vector of which columns are factors  and use that to selectively apply factor  f  lt - sapply bobf  class      factor  bobc  f   lt - lapply bobc  f   factor

User · Answer

If you understand how factors are stored  you can avoid using apply-based functions to accomplish this  Which isn t at all to imply that the apply solutions don t work well   Factors are structured as numeric indices tied to a list of  levels   This can be seen if you convert a factor to numeric  So    gt  fact  lt - as factor c  a   b   a   d    gt  fact  1  a b a d Levels  a b d   gt  as numeric fact   1  1 2 1 3   The numbers returned in the last line correspond to the levels of the factor    gt  levels fact   1   a   b   d    Notice that levels   returns an array of characters  You can use this fact to easily and compactly convert factors to strings or numerics like this    gt  fact character  lt - levels fact  as numeric fact    gt  fact character  1   a   b   a   d    This also works for numeric values  provided you wrap your expression in as numeric      gt  num fact  lt - factor c 1 2 3 6 5 4    gt  num fact  1  1 2 3 6 5 4 Levels  1 2 3 4 5 6  gt  num num  lt - as numeric levels num fact  as numeric num fact     gt  num num  1  1 2 3 6 5 4

User · Answer

At the beginning of your data frame include stringsAsFactors   FALSE to ignore all misunderstandings

User · Answer

Another way is to convert it using apply  bob2  lt - apply bob 2 as character    And a better one  the previous is of class  matrix    bob2  lt - as data frame as matrix bob  stringsAsFactors F

User · Answer

I typically make this function apart of all my projects  Quick and easy   unfactorize  lt - function df     for i in which sapply df  class      factor    df  i     as character df  i      return df

User · Answer

New function  quot across quot  was introduced in dplyr version 1 0 0  The new function will supersede scoped variables   if   at   all   Here s the official documentation library dplyr  bob  lt - bob   gt           mutate across where is factor   as character

User · Answer

You should use convert in hablar which gives readable syntax compatible with tidyverse pipes   library dplyr  library hablar   df  lt - tibble a   factor c 1  2  3  4                 b   factor c 5  6  7  8     df   gt   convert chr a b     which gives you     a     b        lt chr gt   lt chr gt  1 1     5     2 2     6     3 3     7     4 4     8

User · Answer

Or you can try transform   newbob  lt - transform bob  phenotype   as character phenotype     Just be sure to put every factor you d like to convert to character   Or you can do something like this and kill all the pests with one blow   newbob char  lt - as data frame lapply bob sapply bob  is factor    as character   stringsAsFactors   FALSE  newbob rest  lt - bob   sapply bob  is factor    newbob  lt - cbind newbob char  newbob rest    It s not good idea to shove the data in code like this  I could do the sapply part separately  actually  it s much easier to do it like that   but you get the point    I haven t checked the code   cause I m not at home  so I hope it works      This approach  however  has a downside    you must reorganize columns afterwards  while with transform you can do whatever you like  but at cost of  pedestrian-style-code-writting      So there

User · Answer

This function does the trick  df  lt - stacomirtools  killfactor df

User · Answer

With the dplyr-package loaded use  bob bob  gt  mutate at  phenotype   as character    if you only want to change the phenotype-column specifically

User · Answer

This works for me - I finally figured a one liner  df  lt - as data frame lapply df function  y  if class y    factor    as character y  else y  stringsAsFactors F

User · Answer

The global option     stringsAsFactors        The default setting for arguments of data frame and read table    may be something you want to set to FALSE in your startup files  e g     Rprofile   Please see help options

User · Answer

Just following on Matt and Dirk   If you want to recreate your existing data frame without changing the global option  you can recreate it with an apply statement   bob  lt - data frame lapply bob  as character   stringsAsFactors FALSE    This will convert all variables to class  character   if you want to only convert factors  see Marek s solution below   As  hadley points out  the following is more concise    bob    lt - lapply bob  as character    In both cases  lapply outputs a list  however  owing to the magical properties of R  the use of    in the second case keeps the data frame class of the bob object  thereby eliminating the need to convert back to a data frame using as data frame with the argument stringsAsFactors   FALSE

User · Answer

Update  Here s an example of something that doesn t work   I thought it would  but I think that the stringsAsFactors option only works on character strings - it leaves the factors alone   Try this   bob2  lt - data frame bob  stringsAsFactors   FALSE    Generally speaking  whenever you re having problems with factors that should be characters  there s a stringsAsFactors setting somewhere to help you  including a global setting

User · Answer

Maybe a newer option   library  tidyverse    bob  lt - bob   gt   group by if is factor  as character

User · Answer

To replace only factors  i  lt - sapply bob  is factor  bob i   lt - lapply bob i   as character   In package dplyr in version 0 5 0 new function mutate if was introduced  library dplyr  bob   gt   mutate if is factor  as character  - gt  bob     and in version 1 0 0 was replaced by across  library dplyr  bob   gt   mutate across where is factor   as character   - gt  bob  Package purrr from RStudio gives another alternative  library purrr  bob   gt   modify if is factor  as character  - gt  bob

[r] Convert data.frame columns from factors to characters

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