On 19/12/2018 6:48 AM, Luigi Marongiu wrote:
Thank you,
that worked fine for me.
Best wishes of merry Christmas and happy new year,
Luigi
Actually it's wrong! Sorry about that.
If you look at my.data.new$column_2, you'll see that the levels have
changed:
> my.data
column_1 column_2 column_3
1 A B A
2 B B A
3 C C B
4 D E B
5 E E A
> my.data.new
column_1 column_2 column_3
1 A A A
2 B A A
3 C B B
4 D C B
5 E C A
What you want is this instead:
my.data.new <- as.data.frame(lapply(my.data, function(x) {factor(x,
levels = thelevels)}))
The last example in the ?levels help page does this too. I wonder if
that is intentional?
levels> ## we can add levels this way:
levels> f <- factor(c("a","b"))
levels> levels(f) <- c("c", "a", "b")
levels> f
[1] c a
Levels: c a b
levels> f <- factor(c("a","b"))
levels> levels(f) <- list(C = "C", A = "a", B = "b")
levels> f
[1] A B
Levels: C A B
Duncan Murdoch
On Wed, Dec 19, 2018 at 12:19 PM Duncan Murdoch
<murdoch.dun...@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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