What software are you using, exactly? I'm the maintainer of the
randomForest package, yet I do not know which "manual" you are quoting.
If you are using the randomForest package, the model object can be saved
to a file by save(Rfobject, file="myRFobject.rda"). If you need that to
be in ascii, us
Hello! As a new R user, I'm sure this will be a silly question for the
rest of you. I've been able to successfully run a forest but yet to
figure out proper command lines for the following:
1. saving the forest. The guide just says isavef=1. I'm unsure how
expand on this to create the com
I've been fixing some problems in the combine() function, but that's
only for regression data. Looks like you are doing classification, and
I don't see the problem:
R> library(randomForest)
randomForest 4.5-19
Type rfNews() to see new features/changes/bug fixes.
R> set.seed(1)
R> rflist <- repli
My apologies, subject corrected.
I'm building a RF 50 trees at a time due to memory limitations (I have
roughly .5 million observations and around 20 variables). I thought I
could combine some or all of my forests later and look at global
importance.
If I have say 2 forests : tree1 and tree2
I'm building a RF 50 trees at a time due to memory limitations (I have roughly
.5 million observations and around 20 variables). I thought I could combine
some or all of my forests later and look at global importance.
If I have say 2 forests : tree1 and tree2, they have similar Gini and Raw
im
In the words of Simpson (2007), "D'OH!"
I knew it had to be something simple!
On 4/29/07, Gavin Simpson <[EMAIL PROTECTED]> wrote:
>
> On Sat, 2007-04-28 at 21:13 -0400, David L. Van Brunt, Ph.D. wrote:
> > Just out of curiosity, I took the default "iris" example in the RF
> > helpfile...
> > but
On Sat, 2007-04-28 at 21:13 -0400, David L. Van Brunt, Ph.D. wrote:
> Just out of curiosity, I took the default "iris" example in the RF
> helpfile...
> but seeing the admonition against using the formula interface for large data
> sets, I wanted to play around a bit to see how the various options
Just out of curiosity, I took the default "iris" example in the RF
helpfile...
but seeing the admonition against using the formula interface for large data
sets, I wanted to play around a bit to see how the various options affected
the output. Found something interesting I couldn't find documentati
On Tue, 9 Jan 2007, Bálint Czúcz wrote:
There is an improved version of the original random forest algorithm
available in the "party" package (you can find some additional
information on the details here:
http://www.stat.uni-muenchen.de/sfb386/papers/dsp/paper490.pdf ).
I do not know whether i
There is an improved version of the original random forest algorithm
available in the "party" package (you can find some additional
information on the details here:
http://www.stat.uni-muenchen.de/sfb386/papers/dsp/paper490.pdf ).
I do not know whether it yields a solution to your problem about
mi
You can try randomForest in Fortran codes, which has that function
doing missing replacement automatically. There are two ways of
imputations (one is fast and the other is time-consuming) to do that.
I did it long time ago.
the link is below. If you have any question, just let me know.
http://www.
> If there is a way around that with randomForest, I'd be interested to
> know too.
>
> Hugues Sicotte
>
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Darin A. England
> Sent: Thursday, January 04, 2007 3:
Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Darin A. England
Sent: Thursday, January 04, 2007 3:13 PM
To: r-help@stat.math.ethz.ch
Subject: [R] randomForest and missing data
Does anyone know a reason why, in principle, a call to randomForest
cannot acce
Does anyone know a reason why, in principle, a call to randomForest
cannot accept a data frame with missing predictor values? If each
individual tree is built using CART, then it seems like this
should be possible. (I understand that one may impute missing values
using rfImpute or some other metho
With remiss, I haven't tried these R tools.
However, I tried a dozen Naive Bayes-like programs, often used to filter
email, where the serious problem with spam has resulted in many
innovations.
The most touted of the worldwide Naive Bayes programs seems to be
CRM114 (not in R, I expect, since its p
When mtry is equal to total number of features, you just get regular bagging
(in the R package -- Breiman & Cutler's Fortran code samples variable with
replacement, so you can't do bagging with that). There are cases when
bagging will do better than random feature selection (i.e., RF), even in
sim
Hello,
I've a question regarding randomForest (from the package with same name). I've
16 featurs (nominative), 159 positive and 318 negative cases that I'd like to
classify (binary classification).
Using the tuning from the e1071 package it turns out that the best performance
if reached when u
I can't add much to your question, being a complete novice at
classification, but I have tried both randomForest and SVM and I get better
results from randomForest than SVM (even after tuning). randomForest is
also much, much faster. I just thought randomForest was a much better
algorithm, althou
Hi
This is a question regarding classification performance using different methods.
So far I've tried NaiveBayes (klaR package), svm (e1071) package and
randomForest (randomForest). What has puzzled me is that randomForest seems to
perform far better (32% classification error) than svm and NaiveBa
From: Stephen Choularton
>
> Hi
>
> I am trying to use randomForest for classification. I am using this
> code:
>
> > set.seed(71)
> > rf.model <- randomForest(similarity ~ ., data=set1[1:100,],
> importance=TRUE, proximity=TRUE)
> Warning message:
> The response has five or fewer unique valu
Hi
I am trying to use randomForest for classification. I am using this
code:
> set.seed(71)
> rf.model <- randomForest(similarity ~ ., data=set1[1:100,],
importance=TRUE, proximity=TRUE)
Warning message:
The response has five or fewer unique values. Are you sure you want to
do regression? in:
If all you need the formula interface for is auto deletion of NAs, I'd
suggest doing something like:
varlist <- c("fruit", "apples", "oranges", "blueberries")
fruits.nona <- na.omit(fruits.data[varlist])
model.rf <- randomForest(fruits.data[-1], fruits.data[[1]], ...)
If you want to know the cal
mmv wrote:
> I'm attempting to pass a string argument into the function
> randomForest but I get an error:
>
> state <- paste(list("fruit ~", "apples+oranges+blueberries",
> "data=fruits.data, mtry=2, do.trace=100, na.action=na.omit,
> keep.forest=TRUE"), sep= " ", collapse="")
I really don't u
I'm attempting to pass a string argument into the function
randomForest but I get an error:
state <- paste(list("fruit ~", "apples+oranges+blueberries",
"data=fruits.data, mtry=2, do.trace=100, na.action=na.omit,
keep.forest=TRUE"), sep= " ", collapse="")
model.rf <- randomForest(state)
Error in
> From: Uwe Ligges
>
> [EMAIL PROTECTED] wrote:
>
> > Hello,
> >
> > I'm trying to find out the optimal number of splits (mtry parameter)
> > for a randomForest classification. The classification is binary and
> > there are 32 explanatory variables (mostly factors with each up to 4
> > levels bu
> From: [EMAIL PROTECTED]
>
> Hello,
>
> I'm trying to find out the optimal number of splits (mtry
> parameter) for a randomForest classification. The
> classification is binary and there are 32 explanatory
> variables (mostly factors with each up to 4 levels but also
> some numeric variables
See the tuneRF() function in the package for an implementation of
the strategy recommended by Breiman & Cutler.
BTW, "randomForest" is only for the R package. See Breiman's
web page for notice on trademarks.
Andy
> From: Weiwei Shi
>
> Hi,
> I found the following lines from Leo's randomFore
Hi,
I found the following lines from Leo's randomForest, and I am not sure
if it can be applied here but just tried to help:
mtry0 = the number of variables to split on at each node. Default is
the square root of mdim. ATTENTION! DO NOT USE THE DEFAULT VALUES OF
MTRY0 IF YOU WANT TO OPTIMIZE THE P
[EMAIL PROTECTED] wrote:
> Hello,
>
> I'm trying to find out the optimal number of splits (mtry parameter)
> for a randomForest classification. The classification is binary and
> there are 32 explanatory variables (mostly factors with each up to 4
> levels but also some numeric variables) and 575
Hello,
I'm trying to find out the optimal number of splits (mtry parameter) for a
randomForest classification. The classification is binary and there are 32
explanatory variables (mostly factors with each up to 4 levels but also some
numeric variables) and 575 cases.
I've seen that although th
Thanks.
Many people pointed that out. (It was due to that I only knew lappy by
that time :).
On 7/11/05, Martin Maechler <[EMAIL PROTECTED]> wrote:
> > "Duncan" == Duncan Murdoch <[EMAIL PROTECTED]>
> > on Thu, 07 Jul 2005 15:44:38 -0400 writes:
>
> Duncan> On 7/7/2005 3:38 PM, W
> "Duncan" == Duncan Murdoch <[EMAIL PROTECTED]>
> on Thu, 07 Jul 2005 15:44:38 -0400 writes:
Duncan> On 7/7/2005 3:38 PM, Weiwei Shi wrote:
>> Hi there:
>> I have a question on random foresst:
>>
>> recently i helped a friend with her random forest and i came with
With small sample sizes the variability for estimate of test set error will
be large. Instead of splitting the data once, you should consider
cross-validation or bootstrap for estimating performance.
AFAIK gbm as is won't handle more than two classes. You will need to do
quite a bit of work to g
thanks. but can you suggest some ways for the classification problems
since for some specific class, there are too few observations.
the following is from adding sample.size :
> najie.rf.2 <- randomForest(Diag~., data=one.df[ind==1,4:ncol(one.df)],
> importance=T, sampsize=unlist(sample.size))
>
On 7/7/2005 3:47 PM, Weiwei Shi wrote:
> it works.
> thanks,
>
> but: (just curious)
> why i tried previously and i got
>
>> is.vector(sample.size)
> [1] TRUE
>
> i also tried as.vector(sample.size) and assigned it to sampsz,it still
> does not work.
Sorry, I used "vector" incorrectly. Lists a
> From: Weiwei Shi
>
> it works.
> thanks,
>
> but: (just curious)
> why i tried previously and i got
>
> > is.vector(sample.size)
> [1] TRUE
Because a list is also a vector:
> a <- c(list(1), list(2))
> a
[[1]]
[1] 1
[[2]]
[1] 2
> is.vector(a)
[1] TRUE
> is.numeric(a)
[1] FALSE
Actually, t
it works.
thanks,
but: (just curious)
why i tried previously and i got
> is.vector(sample.size)
[1] TRUE
i also tried as.vector(sample.size) and assigned it to sampsz,it still
does not work.
On 7/7/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 7/7/2005 3:38 PM, Weiwei Shi wrote:
> > Hi the
On 7/7/2005 3:38 PM, Weiwei Shi wrote:
> Hi there:
> I have a question on random foresst:
>
> recently i helped a friend with her random forest and i came with this
> problem:
> her dataset has 6 classes and since the sample size is pretty small:
> 264 and the class distr is like this (Diag is th
Hi there:
I have a question on random foresst:
recently i helped a friend with her random forest and i came with this problem:
her dataset has 6 classes and since the sample size is pretty small:
264 and the class distr is like this (Diag is the response variable)
sample.size <- lapply(1:6, functi
The limitation comes from the way categorical splits are represented in the
code: For a categorical variable with k categories, the split is
represented by k binary digits: 0=right, 1=left. So it takes k bits to
store each split on k categories. To save storage, this is `packed' into a
4-byte in
Hello,
I'm using the random forest package. One of my factors in the data set contains
41 levels (I can't code this as a numeric value - in terms of linear models
this would be a random factor). The randomForest call comes back with an error
telling me that the limit is 32 categories.
Is there
All,
I'm trying to set up a function which calls the partialPlot function but
am getting an error that I can't seem to solve. Here's a simplified
version of the function and error...
> pplot <-
function(rf,pred.var){partialPlot(x=rf,pred.data=acoust,x.var=pred.var)}
>
> attach(acoust)
> acoust
> From: luk
>
> When I run randonForest with a 169453x5 matrix, I got the
> following message.
>
> Error in matrix(0, n, n) : matrix: too many elements specified
>
> Can you please advise me how to solve this problem?
>
> Thanks,
>
> Lu
1. When asking new questions, please don't reply to
When I run randonForest with a 169453x5 matrix, I got the following message.
Error in matrix(0, n, n) : matrix: too many elements specified
Can you please advise me how to solve this problem?
Thanks,
Lu
Uwe Ligges <[EMAIL PROTECTED]> wrote:
Vera Hofer wrote:
> Dear colleagues,
>
> I have
On Tue, 13 Apr 2004, Hui Han wrote:
> Hi,
>
> I am doing feature selection for my dataset. The following is
> the extreme case where only one feature is left. But I got
> the error below. So my question is that do I have to use
> more than one features?
>
> sample.subset
> udomain.edu hpclass
>
nal Message-
> > From: [EMAIL PROTECTED]
> > [mailto:[EMAIL PROTECTED] Behalf Of Hui Han
> > Sent: Tuesday, 13 April, 2004 17:16
> > To: [EMAIL PROTECTED]
> > Subject: [R] randomForest: more than one variable needed?
> >
> >
> > Hi,
> >
&g
gt;
> Philippe Grosjean
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Hui Han
> Sent: Tuesday, 13 April, 2004 17:16
> To: [EMAIL PROTECTED]
> Subject: [R] randomForest: more than one variable needed?
>
>
> Hi,
>
iable? there is no
point here!
Philippe Grosjean
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Hui Han
Sent: Tuesday, 13 April, 2004 17:16
To: [EMAIL PROTECTED]
Subject: [R] randomForest: more than one variable needed?
Hi,
I am doing feature selection
Hi,
I am doing feature selection for my dataset. The following is
the extreme case where only one feature is left. But I got
the error below. So my question is that do I have to use
more than one features?
sample.subset
udomain.edu hpclass
1-1.0 not
2-1.0 not
3-0
> From: Christian Schulz
>
> Hi,
>
> is it correct that i need ~ 2GB RAM that it's
> possible to work with the default setting
> ntree=500 and a data.frame with 100.000 rows
> and max. 10 columns for training and testing?
If you have the test set, and don't need the forest for predicting othe
> Hi,
>
> is it correct that i need ~ 2GB RAM that it's
> possible to work with the default setting
> ntree=500 and a data.frame with 100.000 rows
> and max. 10 columns for training and testing?
>
no. You may parallelize the computations: perform 5 runs of RF with `ntree
= 100' (or less) and sav
Hi,
is it correct that i need ~ 2GB RAM that it's
possible to work with the default setting
ntree=500 and a data.frame with 100.000 rows
and max. 10 columns for training and testing?
P.S.
It's possible calculate approximate the
memory demand for different settings with RF?
Many thanks & regar
simplify the splitting.
You might find the heuristics in the CART book, but I'm not sure.
HTH,
Andy
> -Original Message-
> From: Vladimir N. Kutinsky [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, August 20, 2003 10:26 AM
> To: Liaw, Andy; [EMAIL PROTECTED]
> Subject: R
Andy,
Does it mean that the error rate does increase as long as the aggregating
number of out-of-bag cases reaches the number of all cases? or, in other
words, because the number of points being predicted (right or wrong) gets
larger at the first steps of the process?
If it so then it's all clea
Andy,
First of all, thank you for you reply.
I'm using R1.6.1 for Windows. A few days ago I updated the randomForest
package from CRAN. It gives warning messages now that the package was built
under R1.6.2 but seems to work fine.
To make sure we're talking about the same thing, let's take iris
cla
> -Original Message-
> From: Vladimir N. Kutinsky [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, August 20, 2003 4:43 AM
> To: [EMAIL PROTECTED]
> Subject: [R] RandomForest
>
>
> Hello,
>
> When I plot or look at the error rate vector for a random forest
> (
Hello,
When I plot or look at the error rate vector for a random forest
(rf$err.rate) it looks like a descending function except for a few first
points of the vector with error rates values lower(sometimes much lower)
than the general level of error rates for a forest with such number of trees
whe
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