Hi,
Perhaps this is not the proper place to ask this
question but I am out of options, therefore I
apologize in advance.
I want to know how the (upper bound?) generalization
error of the random forest is determined using the
out-of-bag estimate. I read in Breiman's paper that s
and p determine
Hi,
Perhaps you should try the transpose function t(). It
converts a row into a column and vice versa.
HTH,
Martin Lam
--- Michael [EMAIL PROTECTED] wrote:
Hi all,
I tended to use rbind, or cbind to force a vector be
be deemed as a column
or row vector. This is very important if I want
Dear Claudia,
This is how I save the plots as *.eps.
postscript(file=testplot.eps,
paper=special,
width=10,
height=10,
horizontal=FALSE)
yvalues = runif(100)
plot(yvalues)
dev.off()
HTH,
Martin Lam
--- [EMAIL PROTECTED] wrote:
Hello!
I
(names %in%chars)] = 1
}
mymatrix
HTH,
Martin Lam
--- Dale Steele [EMAIL PROTECTED] wrote:
I have a data management problem which exceeds my
meager R programming
skills and would greatly appreciate suggestions on
how to proceed? The
data consists of a series of observation periods
Dear Gueorgui,
Is it true that R generally cannot handle medium
sized data sets(a
couple of hundreds of thousands observations) and
threrefore large
date set(couple of millions of observations)?
It depends on what you want to do with the data sets.
Loading the data sets shouldn't be any
Hi,
I have a problem when I want to add new points (or a
new line) to the graph. Some points (or parts of the
line) are not shown on the graph because they lie
beyond the scale of the axis. Is there a way to
overcome this so all points (or the entire line) are
shown on the graph? Here's an
Hi,
I was wondering, if it is possible to print out the
values of variables while you are in a for/while loop?
Like this for example:
for (i in 1:5) {
i
}
So what I want is this as output in the console:
1
2
3
4
5
Thanks in advance,
Martin
__
Hi,
I have written a piece of code, which is a variant of
the random forest (rf) package algorithm, entirely in
R. I know that some of the code in the rf package is
written in c or c++. The problem is that the execution
of my code in R takes a lot of time. To give you an
example, the building and
# create a 4x4 matrix with random values
smallmatrix = matrix(runif(16)*10, ncol = 4, nrow = 4)
# get the row and column number of the item in the
matrix with the highest value
which(smallmatrix == max(smallmatrix), arr.ind = 1)
HTH,
Martin
--- Jonathan Williams
[EMAIL PROTECTED] wrote:
Dear
Type ?mean and hit enter in the console to see the
description of the function.
# randomly draw 5 numbers between 1 and 10 without
replacement
randomlist = sample(1:10, 5)
# calculate the mean of the list
meanoflist = mean(randomlist)
# show the result
meanoflist
HTH,
Martin
--- Lisa Wang
Hi,
I was wondering if it is possible to get the
rowindices without using the function which because
I don't have a restriction criteria. Here's an example
of what I mean:
# take 10 randomly selected instances
iris[sample(1:nrow(iris), 10),]
# output
Sepal.Length Sepal.Width Petal.Length
Hi,
I was wondering if it's possible to get the row
numbers from a filtering. Here's an example:
# give me the rows with sepal.length == 6.2
iris[(iris[,1]==6.2),]
# output
Sepal.Length Sepal.Width Petal.Length Petal.Width
Species
69 6.2 2.2 4.5 1.5
Hi,
I was wondering if someone know how to sort a data set
by column.
I've tried sort() but without luck. I would think
there should be a function for it somewhere. An
example with the iris data set would be appreciated.
Thanks,
Martin
__
# to see the objects that are currently in memory
objects()
# to remove everything
rm(list = ls())
HTH,
Martin
--- Luis Ridao Cruz [EMAIL PROTECTED] wrote:
?rm
[EMAIL PROTECTED] 13/09/2005 12:08:55
Hi, I admit that I rather carelessly built lots of
large objects and therefore ran
-Original Message-
From: Martin Lam [mailto:[EMAIL PROTECTED]
Sent: Friday, September 09, 2005 9:04 AM
To: Liaw, Andy; r-help@stat.math.ethz.ch
Subject: RE: [R] Re-evaluating the tree in the
random forest
Hi,
Let me give a simple example, assume a dataset
containing 5 instances
was if there was a way to
re-evaluate the instances again into:
Root node: question: x1 1?
Left terminal node:
[0.5, A]
Leaf classification: A
Right terminal node:
[3.2, B]
[4.5, B]
[1.4, C]
[1.6, C]
[1.9, C]
Leaf classification: C
Cheers,
Martin
--- Liaw, Andy [EMAIL PROTECTED] wrote:
From: Martin
Hi,
This may not be the best solution, but at least it's
easy to see what i'm doing, assume that your data set
is called data:
# remove the 4th column
data1 = data[,-4]
# remove the 3rd column
data2 = data[,-3]
# use cbind to add an extra column with only X1
#elements
data1 = cbind(data1,
Dear mailinglist members,
I was wondering if there was a way to re-evaluate the
instances of a tree (in the forest) again after I have
manually changed a splitpoint (or split variable) of a
decision node. Here's an illustration:
library(randomForest)
forest.rf - randomForest(formula = Species ~
Hi,
I will explain my problem with this example:
library(randomForest)
# load the iris plant data set
dataset - iris
numberarray - array(1:nrow(dataset), nrow(dataset),
1)
# include only instances with Species = setosa or
virginica
indices - t(numberarray[(dataset$Species == setosa
|
Hi,
Since I've had no replies on my previous post about my
problem I am posting it again in the hope someone
notice it. The problem is that the randomForest
function doesn't take datasets which has instances
only containing a subset of all the classes. So the
dataset with instances that either
Thank you for this and earlier help Mr. Ripley.
Martin
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
Look at ?[.factor:
finaldataset$Species -
finaldataset$Species[,drop=TRUE]
solves this.
On Fri, 26 Aug 2005, Martin Lam wrote:
Hi,
Since I've had no replies on my
x - array(1:20, dim = c(4, 5))
ind - array(c(1:3, 3:1), dim = c(3, 2))
# instead of using ind (pairs of coordinates) for
getting the items in the matrix, you can convert it to
a list of single coordinates to point to the item in
the matrix:
# t = transpose
# nrow = get the number of rows
indices
Hmm, it seems that the only difference is the use of
as.data.frame(a) instead of a. Hence, the same
result can be done with:
a - rbind(c(T, T, F), c(F, F, T))
a[a==F]=
as.data.frame(a)
A possible drawback is that the mode changes from
logical to character when using a[a==F]=
Instead you could
Hi,
I am trying to use a vector of indices to select some
rows from a matrix. But before I can do that I somehow
need to convert 'combinations' into a list, since
'mode(combinations)' says it's 'numerical'. Any idea
how I can do that?
library(combinat)
combinations - t(combn(8,2))
indices -
# row bind
a - matrix(1:5)
a
a - rbind(a, 6)
a
# column bind
b - matrix(1:5)
b
b - cbind(b, 6:12)
b
b - cbind(b, 13)
b
Hope this helps,
Martin
--- Simone Gabbriellini [EMAIL PROTECTED] wrote:
dear list,
if I have a matrix
s-matrix(1:5, ncol=5)
how can I add another row with other
Hi,
I am somewhat new to R so this question may be
foolish, but is it possible to add decision trees into
a list, array or vector in R?
I am trying to build a collection (ensemble) of
decision trees. Every time a new instance arrive I
need to get the prediction of each decision tree. I
have
26 matches
Mail list logo
