Convert all data frame character columns to factors

Question

Given a  pre-existing  data frame that has columns of various types  what is the simplest way to convert all its character columns to factors  without affecting any columns of other types   Here s an example data frame   df  lt - data frame A   factor LETTERS 1 5                     B   1 5  C   as logical c 1  1  0  0  1                     D   letters 1 5                    E   paste LETTERS 1 5   letters 1 5                     stringsAsFactors   FALSE  df     A B     C D   E   1 A 1  TRUE a A a   2 B 2  TRUE b B b   3 C 3 FALSE c C c   4 D 4 FALSE d D d   5 E 5  TRUE e E e str df     data frame    5 obs  of  5 variables       A  Factor w  5 levels  A   B   C   D      1 2 3 4 5      B  int  1 2 3 4 5      C  logi  TRUE TRUE FALSE FALSE TRUE      D  chr   a   b   c   d           E  chr   A a   B b   C c   D d        I know I can do   df D  lt - as factor df D  df E  lt - as factor df E    Is there a way to automate this process a bit more

User · Accepted Answer

DF  lt - data frame x letters 1 5   y 1 5  stringsAsFactors FALSE   str DF    data frame    5 obs  of  2 variables      x  chr   a   b   c   d          y  int  1 2 3 4 5   The  annoying  default of as data frame is to turn all character columns into factor columns  You can use that here   DF  lt - as data frame unclass DF   str DF    data frame    5 obs  of  2 variables      x  Factor w  5 levels  a   b   c   d      1 2 3 4 5     y  int  1 2 3 4 5

User · Answer

I used to do a simple for loop  As  A5C1D2H2I1M1N2O1R2T1 answer  lapply is a nice solution  But if you convert all the columns  you will need a data frame before  otherwise you will end up with a list  Little execution time differences      mm2N mm2New  10 18   str mm2N   data frame     35487 obs  of  9 variables     bb      int  4 6 2 3 3 2 5 2 1 2        vabb    int  -3 -3 -2 -2 -3 -1 0 0 3 3        bb55    int  7 6 3 4 4 4 9 2 5 4        vabb55  int  -3 -1 0 -1 -2 -2 -3 0 -1 3        zr      num  0 -2 -1 1 -1 -1 -1 1 1 0        z55r    num  -2 -2 0 1 -2 -2 -2 1 -1 1        fechar  num  0 -1 1 0 1 1 0 0 1 0        varr    num  3 3 1 1 1 1 4 1 1 3        minmax  int  3 0 4 6 6 6 0 6 6 1         For solution  t1 Sys time    for i in 1 ncol mm2N   mm2N  i  as factor mm2N  i    Sys time  -t1 Time difference of 0 2020121 secs  str mm2N   data frame     35487 obs  of  9 variables     bb      Factor w  6 levels  1   2   3   4      4 6 2 3 3 2 5 2 1 2        vabb    Factor w  7 levels  -3   -2   -1      1 1 2 2 1 3 4 4 7 7        bb55    Factor w  8 levels  2   3   4   5      6 5 2 3 3 3 8 1 4 3        vabb55  Factor w  7 levels  -3   -2   -1      1 3 4 3 2 2 1 4 3 7        zr      Factor w  5 levels  -2   -1   0      3 1 2 4 2 2 2 4 4 3        z55r    Factor w  5 levels  -2   -1   0      1 1 3 4 1 1 1 4 2 4        fechar  Factor w  3 levels  -1   0   1   2 1 3 2 3 3 2 2 3 2        varr    Factor w  5 levels  1   2   3   4      3 3 1 1 1 1 4 1 1 3        minmax  Factor w  7 levels  0   1   2   3      4 1 5 7 7 7 1 7 7 2        lapply solution  mm2N mm2New  10 18   t1 Sys time    mm2N  lt - lapply mm2N  as factor   Sys time  -t1 Time difference of 0 209012 secs  str mm2N  List of 9    bb      Factor w  6 levels  1   2   3   4      4 6 2 3 3 2 5 2 1 2        vabb    Factor w  7 levels  -3   -2   -1      1 1 2 2 1 3 4 4 7 7        bb55    Factor w  8 levels  2   3   4   5      6 5 2 3 3 3 8 1 4 3        vabb55  Factor w  7 levels  -3   -2   -1      1 3 4 3 2 2 1 4 3 7        zr      Factor w  5 levels  -2   -1   0      3 1 2 4 2 2 2 4 4 3        z55r    Factor w  5 levels  -2   -1   0      1 1 3 4 1 1 1 4 2 4        fechar  Factor w  3 levels  -1   0   1   2 1 3 2 3 3 2 2 3 2        varr    Factor w  5 levels  1   2   3   4      3 3 1 1 1 1 4 1 1 3        minmax  Factor w  7 levels  0   1   2   3      4 1 5 7 7 7 1 7 7 2        data frame lapply solution  mm2N mm2New  10 18   t1 Sys time    mm2N  lt - data frame lapply mm2N  as factor    Sys time  -t1 Time difference of 0 2010119 secs  str mm2N   data frame     35487 obs  of  9 variables     bb      Factor w  6 levels  1   2   3   4      4 6 2 3 3 2 5 2 1 2        vabb    Factor w  7 levels  -3   -2   -1      1 1 2 2 1 3 4 4 7 7        bb55    Factor w  8 levels  2   3   4   5      6 5 2 3 3 3 8 1 4 3        vabb55  Factor w  7 levels  -3   -2   -1      1 3 4 3 2 2 1 4 3 7        zr      Factor w  5 levels  -2   -1   0      3 1 2 4 2 2 2 4 4 3        z55r    Factor w  5 levels  -2   -1   0      1 1 3 4 1 1 1 4 2 4        fechar  Factor w  3 levels  -1   0   1   2 1 3 2 3 3 2 2 3 2        varr    Factor w  5 levels  1   2   3   4      3 3 1 1 1 1 4 1 1 3        minmax  Factor w  7 levels  0   1   2   3      4 1 5 7 7 7 1 7 7 2

User · Answer

I noticed     indexing columns fails to create levels when iterating      for   a feature in convert to factors       feature df a feature    lt - factor feature df a feature        It creates  e g  for the  Status  column      Status   Factor w  1 level  c   Success      Fail       NA NA NA       Which is remedied by using      indexing      for   a feature in convert to factors       feature df  a feature     lt - factor feature df  a feature         Giving instead  as desired     Status   Factor w  2 levels  Success    Fail    1 1 2 1

User · Answer

The easiest way would be to use the code given below  It would automate the whole process of converting all the variables as factors in a dataframe in R  it worked perfectly fine for me  food cat here is the dataset which I am using  Change it to the one which you are working on       for i in 1 ncol food cat     food cat  i   lt - as factor food cat  i

User · Answer

As  Raf Z commented on this question  dplyr now has mutate if   Super useful  simple and readable    gt  str df   data frame     5 obs  of  5 variables     A  Factor w  5 levels  A   B   C   D      1 2 3 4 5    B  int  1 2 3 4 5    C  logi  TRUE TRUE FALSE FALSE TRUE    D  chr   a   b   c   d         E  chr   A a   B b   C c   D d        gt  df  lt - df   gt   mutate if is character as factor    gt  str df   data frame     5 obs  of  5 variables     A  Factor w  5 levels  A   B   C   D      1 2 3 4 5    B  int  1 2 3 4 5    C  logi  TRUE TRUE FALSE FALSE TRUE    D  Factor w  5 levels  a   b   c   d      1 2 3 4 5    E  Factor w  5 levels  A a   B b   C c      1 2 3 4 5

User · Answer

Working with dplyr library dplyr   df  lt - data frame A   factor LETTERS 1 5                     B   1 5  C   as logical c 1  1  0  0  1                     D   letters 1 5                    E   paste LETTERS 1 5   letters 1 5                     stringsAsFactors   FALSE   str df   we get   data frame     5 obs  of  5 variables     A  Factor w  5 levels  quot A quot   quot B quot   quot C quot   quot D quot      1 2 3 4 5    B  int  1 2 3 4 5    C  logi  TRUE TRUE FALSE FALSE TRUE    D  chr   quot a quot   quot b quot   quot c quot   quot d quot         E  chr   quot A a quot   quot B b quot   quot C c quot   quot D d quot       Now  we can convert all chr to factors  df  lt - df  gt  mutate if is character  as factor  str df   And we get   data frame     5 obs  of  5 variables     A  Factor w  5 levels  quot A quot   quot B quot   quot C quot   quot D quot      1 2 3 4 5    B  int  1 2 3 4 5    C  logi  TRUE TRUE FALSE FALSE TRUE    D  chr   quot a quot   quot b quot   quot c quot   quot d quot         E  chr   quot A a quot   quot B b quot   quot C c quot   quot D d quot       Let s provide also other solutions  With base package  df sapply df  is character    lt - lapply df sapply df  is character                                                                as factor   With dplyr 1 0 0 df  lt - df  gt  mutate across where is factor   as character    With purrr package  library purrr   df  lt - df  gt   modify if is factor  as character

User · Answer

Roland s answer is great for this specific problem  but I thought I would share a more generalized approach   DF  lt - data frame x   letters 1 5   y   1 5  z   LETTERS 1 5                     stringsAsFactors FALSE  str DF     data frame    5 obs  of  3 variables       x  chr   a   b   c   d           y  int  1 2 3 4 5      z  chr   A   B   C   D          The conversion DF sapply DF  is character    lt - lapply DF sapply DF  is character                                            as factor  str DF     data frame    5 obs  of  3 variables       x  Factor w  5 levels  a   b   c   d      1 2 3 4 5      y  int  1 2 3 4 5      z  Factor w  5 levels  A   B   C   D      1 2 3 4 5   For the conversion  the left hand side of the assign  DF sapply DF  is character    subsets the columns that are character  In the right hand side  for that subset  you use lapply to perform whatever conversion you need to do  R is smart enough to replace the original columns with the results   The handy thing about this is if you wanted to go the other way or do other conversions  it s as simple as changing what you re looking for on the left and specifying what you want to change it to on the right

[r] Convert all data frame character columns to factors

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