On 2020-08-21 16:22 +1200, Abby Spurdle wrote:
| On Fri, Aug 21, 2020 at 4:16 PM Abby Spurdle <spurdl...@gmail.com> wrote:
| | On Fri, Aug 21, 2020 at 1:06 PM Sparks, John <jspa...@uic.edu> wrote:
| | |
| | | Hi R Helpers,
| | |
| | | I wanted to try the rotationForest
| | | package.
| | |
| | | I pointed it at my data set and
| | | got the error message "Error in if
| | | (K >= ncol(x)) stop("K should not
| | | be greater than or equal to the
| | | number of columns in x") :
| | | argument is of length zero'.
| | |
| | | My dataset has 3688 obs. of 111 variables.
| | |
| | | Would a quick adjustment to the
| | | default value of K resolve this?
| | |
| | | If anybody with more experience
| | | with the package than me has a
| | | general suggestion I would
| | | appreciate it.
| |
| | Note that I'm not familiar with this
| | package or the method. Also note
| | that you haven't told anyone what
| | function you're using, or what your
| | call was.
| |
| | I'm assuming that you're using the
| | rotationForest() function.
| | According to its help page, the
| | default is:
| |
| | K = round(ncol(x)/3, 0)
| |
| | There's no reason why the default K
| | value should be higher than the
| | number of columns, unless:
| | (1) There's a bug with the package; or
| | (2) There's a problem with your input.
| |
| | I note that the package is only
| | version 0.1.3, so a bug is not out
| | of the question. Also, I'm a little
| | surprised the author didn't use
| | integer division:
| |
| | K = ncol(x) %/% 3
| |
| | You could just set K to the above
| | value, and see what happens...
|
| Just re-read your question and
| realized I misread the error message.
| The argument is of zero length.
|
| But the conclusion is the same, either
| a bug in the package, or a problem
| with your input.
Dear John,
check to see if your columns only has
numbers in them like the *de jure* iris
dataset used in the example in
?rotationForest::rotationForest.
idx <- 1:100
idx.new <- which(!(1:nrow(iris) %in% idx))
y <- as.factor(ifelse(iris$Species[idx]=="setosa", 0, 1))
x <- iris[idx, -5]
newdata <- iris[idx.new, -5]
K <- ncol(x) %/% 3
L <- 100
rF <-
rotationForest::rotationForest(
x=x,
y=y,
K=K,
L=L)
predict(object=rF, newdata=newdata)
Best,
Rasmus
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