R help - Dec 2018 - convert columns of dataframe to same factor levels

Dear all,
I have a data frame with character values where each character is a
level; however, not all columns of the data frame have the same
characters thus, when generating the data frame with stringsAsFactors
= TRUE, the levels are different for each column.
Is there a way to provide a single vector of levels and assign the
characters so that they match such vector?
Is there a way to do that not only when setting the data frame but
also when reading data from a file with read.table()?

For instance, I have:
column_1 = c("A", "B", "C", "D",
"E")
column_2 = c("B", "B", "C", "E",
"E")
column_3 = c("C", "C", "D", "D",
"C")
my.data <- data.frame(column_1, column_2, column_3, stringsAsFactors =
TRUE)
> str(my.data)
'data.frame': 5 obs. of  3 variables:
 $ column_1: Factor w/ 5 levels
"A","B","C","D",..: 1 2 3 4 5
 $ column_2: Factor w/ 3 levels "B","C","E": 1 1 2
3 3
 $ column_3: Factor w/ 2 levels "C","D": 1 1 2 2 1

Thank you
-- 
Best regards,
Luigi

On 19/12/2018 5:58 AM, Luigi Marongiu wrote:
> Dear all,
> I have a data frame with character values where each character is a
> level; however, not all columns of the data frame have the same
> characters thus, when generating the data frame with stringsAsFactors
> = TRUE, the levels are different for each column.
> Is there a way to provide a single vector of levels and assign the
> characters so that they match such vector?
> Is there a way to do that not only when setting the data frame but
> also when reading data from a file with read.table()?
> 
> For instance, I have:
> column_1 = c("A", "B", "C", "D",
"E")
> column_2 = c("B", "B", "C", "E",
"E")
> column_3 = c("C", "C", "D", "D",
"C")
> my.data <- data.frame(column_1, column_2, column_3, stringsAsFactors =
TRUE)
>> str(my.data)
> 'data.frame': 5 obs. of  3 variables:
>   $ column_1: Factor w/ 5 levels
"A","B","C","D",..: 1 2 3 4 5
>   $ column_2: Factor w/ 3 levels "B","C","E":
1 1 2 3 3
>   $ column_3: Factor w/ 2 levels "C","D": 1 1 2 2 1
> 
> Thank you
> 
I don't think read.table() can do it for you automatically.  To do it 
yourself, you need to get a vector of the levels.  If you know this, 
just assign it to a variable; if you don't know it, compute it as

   thelevels <- unique(unlist(lapply(my.data, levels)))

Then set the levels of each column to thelevels:

   my.data.new <- as.data.frame(lapply(my.data, function(x) {levels(x) 
<- thelevels; x}))

Duncan Murdoch

Thank you,
that worked fine for me.
Best wishes of merry Christmas and happy new year,
Luigi

On Wed, Dec 19, 2018 at 12:19 PM Duncan Murdoch
<murdoch.duncan at gmail.com> wrote:
>
> On 19/12/2018 5:58 AM, Luigi Marongiu wrote:
> > Dear all,
> > I have a data frame with character values where each character is a
> > level; however, not all columns of the data frame have the same
> > characters thus, when generating the data frame with stringsAsFactors
> > = TRUE, the levels are different for each column.
> > Is there a way to provide a single vector of levels and assign the
> > characters so that they match such vector?
> > Is there a way to do that not only when setting the data frame but
> > also when reading data from a file with read.table()?
> >
> > For instance, I have:
> > column_1 = c("A", "B", "C",
"D", "E")
> > column_2 = c("B", "B", "C",
"E", "E")
> > column_3 = c("C", "C", "D",
"D", "C")
> > my.data <- data.frame(column_1, column_2, column_3,
stringsAsFactors = TRUE)
> >> str(my.data)
> > 'data.frame': 5 obs. of  3 variables:
> >   $ column_1: Factor w/ 5 levels
"A","B","C","D",..: 1 2 3 4 5
> >   $ column_2: Factor w/ 3 levels
"B","C","E": 1 1 2 3 3
> >   $ column_3: Factor w/ 2 levels "C","D": 1 1 2 2
1
> >
> > Thank you
> >
>
> I don't think read.table() can do it for you automatically.  To do it
> yourself, you need to get a vector of the levels.  If you know this,
> just assign it to a variable; if you don't know it, compute it as
>
>    thelevels <- unique(unlist(lapply(my.data, levels)))
>
> Then set the levels of each column to thelevels:
>
>    my.data.new <- as.data.frame(lapply(my.data, function(x) {levels(x)
> <- thelevels; x}))
>
> Duncan Murdoch


-- 
Best regards,
Luigi

You can abuse the S4 class system to do this.

setClass("Size") # no representation, no prototype
setAs(from="character", to="Size", # nothing but a coercion
method
  function(from){
    ret <- factor(from,
levels=c("Small","Medium","Large"), ordered=TRUE)
    class(ret) <- c("Size", class(ret))
    ret
  })
z <- read.table(colClasses=c("integer", "Size"),
text="7 Medium\n5 Large\n3
Large")
dput(z)
#structure(list(V1 = c(7L, 5L, 3L), V2 = structure(c(2L, 3L, 3L
#), .Label = c("Small", "Medium", "Large"), class
= c("Size",
#"ordered", "factor"))), class = "data.frame",
row.names = c(NA,
#-3L))

I wonder if this behavior is intended or if there is a more sanctioned way
to get read.table(colClasses=...) to make a factor with a specified set of
levels.

Bill Dunlap
TIBCO Software
wdunlap tibco.com


On Wed, Dec 19, 2018 at 3:19 AM Duncan Murdoch <murdoch.duncan at
gmail.com>
wrote:
> On 19/12/2018 5:58 AM, Luigi Marongiu wrote:
> > Dear all,
> > I have a data frame with character values where each character is a
> > level; however, not all columns of the data frame have the same
> > characters thus, when generating the data frame with stringsAsFactors
> > = TRUE, the levels are different for each column.
> > Is there a way to provide a single vector of levels and assign the
> > characters so that they match such vector?
> > Is there a way to do that not only when setting the data frame but
> > also when reading data from a file with read.table()?
> >
> > For instance, I have:
> > column_1 = c("A", "B", "C",
"D", "E")
> > column_2 = c("B", "B", "C",
"E", "E")
> > column_3 = c("C", "C", "D",
"D", "C")
> > my.data <- data.frame(column_1, column_2, column_3,
stringsAsFactors > TRUE)
> >> str(my.data)
> > 'data.frame': 5 obs. of  3 variables:
> >   $ column_1: Factor w/ 5 levels
"A","B","C","D",..: 1 2 3 4 5
> >   $ column_2: Factor w/ 3 levels
"B","C","E": 1 1 2 3 3
> >   $ column_3: Factor w/ 2 levels "C","D": 1 1 2 2
1
> >
> > Thank you
> >
>
> I don't think read.table() can do it for you automatically.  To do it
> yourself, you need to get a vector of the levels.  If you know this,
> just assign it to a variable; if you don't know it, compute it as
>
>    thelevels <- unique(unlist(lapply(my.data, levels)))
>
> Then set the levels of each column to thelevels:
>
>    my.data.new <- as.data.frame(lapply(my.data, function(x) {levels(x)
> <- thelevels; x}))
>
> Duncan Murdoch
>
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