similar to: tree

Hi, In Splus if I call the function crosstabs() the output is a contigency table; in each cell of the table is printed: N, N/RowTotal, N/ColTotal, N/Total. N is the number of observations in each cell. The same call to xtabs() in R will produce the contigency table but the only entry in each cell is N. How can I get the same relative frequencies that crosstabs() gives? Thanks, mike --

Hi, I have a data set with 1000 observations and 260 predictors. The predictor variables are all ordinal. There are 2 classes labeled as, F and T with class proportions of 0.44 and 0.56, respectively. In a call to the function randomForest() with mytry=1 and nodesize=1 and ntree=100 the resulting classifier puts all observations in class T. When I change nodesize to nodesize=5 I get the

Good Day, Included below is some code to generate data and to fit a mixed effects model to this fake data. The code works as expected when I call the function "lme" in Splus but not in R. The error message from calling lme in R is: "Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups" I installed the nlme package for R around 20 August

Good Day, I'm using the randomForest package to perform a classification. If I supply weights to the optional classwt argument are proximity values computed as a weighted average? I understand that the forest will possibly change as a function of the particular weights I supply. Thanks in advance. Mike Michael Fugate Los Alamos National Laboratory Mail Stop MS-F600, Los Alamos, NM

Date: Tue, 27 Jun 2000 16:09:43 -0600 (MDT) From: "W. Scott Wilburn" <wilburn at lanl.gov> To: ssh at clinet.fi Subject: F-secure -> Openssh Compatibility We have Macintoshes running Fsecure SSH client 1.0.1 which are unable to connect to a server running Openssh 2.1.1 on Red Hat 6.2. I believe that the problem is a bug with Fsecure, since a 30-day trial version of 1.0.2 works

I'm new to this list, so please forgive me if this has been discussed before. It appears that one of the (commendable) design goals of OpenSSH is to re-use existing open-source libraries wherever possible in order to simplify the OpenSSH code and hopefully improve security in the process. As exhibited by the current, non-open SSH, supporting all of the nuances of authentication and logins

Is there anyway to plot a matrix using a 3d bar plot. Something like bar3 in matlab? The example in demo hist3d does a 3d barplot for binned data, but has anyone tried something for a simple matrix with spaces betwen bars and axis labels using matrix dimnames or 1,2,3? stages<-letters[1:3] A<-matrix(c( 0.21, 0.21,0.03, 0.55, 0.58, 0.09, 1.30, 1.35, 0.22), nrow=3, byrow=TRUE,

Hi, I need to resample rows in a data frame by subsets L3 <- LETTERS[1:3] d <- data.frame(cbind(x=1, y=1:10), fac=sample(L3, 10, repl=TRUE)) x y fac 1 1 1 A 2 1 2 A 3 1 3 A 4 1 4 A 5 1 5 C 6 1 6 C 7 1 7 B 8 1 8 A 9 1 9 C 10 1 10 A I have seen this used to sample rows with replacement d[sample(nrow(d), replace=T), ] x y fac 7 1 7 B 2

I wanted to print the first and last rows of some dataframes in Sweave using dots in columns to separate the two parts. Head and tail almost work, but I have problems with factors and row names. z<-data.frame(id=letters[1:26], x=sample(1:26,26)) rbind(head(z,3), ".", tail(z,1)) id x 1 a 18 2 b 8 3 c 14 4 <NA> . 26 z 10 Warning message: invalid

O.K. I know it was too late to make changes for the next release, but thankfully there werent any changes needed. Kerberos over ssh protocol 1 worked out of the box from CVS today (2002/10/04-14:30) The krb5 patches we were using (I think from Olaf K) was all in the code. The only thing I had to patch to get things working was removing the scard-install from the Makefile.in The compiled code

Hi, I would like to construct a transition matrix from a data frame with annual transitions of marked plants. plant<-c(1:6) class<-c("seed","seed", "seed", "veg", "rep", "rep") fate<-c("dead", "veg","veg","rep", "rep", "veg") trans<-data.frame(plant, class, fate)

All, I apologize if this is an inappropriate question for this mailing list. If so, please recommend an appropriate place to post the question. I'm also somewhat new to LLVM, so I could have some pretty fundamental misunderstandings about what I am trying to do. I have searched for information on the llvm.org website (user's guide, programmer's manual and doxygen documentation),

Dear All, We came across a problem when using the "tree" package to analyze our data set. First, in the "tree" function, if we use the default value "mindev=0.01", the resulting regression tree has a single node. So, we set "mindev=0", and obtain a tree with 931 terminal nodes. However, when we further use the "cv.tree" function to run a 10-fold

I'm trying to read some web tables directly into R. These are both genome sequencing projects (eukaryotes and metagenomes) from NCBI and look very similar; however, only the first one works. http://www.ncbi.nlm.nih.gov/genomes/leuks.cgi http://www.ncbi.nlm.nih.gov/genomes/lenvs.cgi I added ?dump=selected to the end of the url string to get a tab- delimited file (which is what happens

Hi, everyone, The paper submission deadline for this year's LLVM in HPC workshop has been further extended to September 14th (AoE). We're looking for a few additional submissions, so if you have anything that could be submitted as a paper by the beginning of next week, please take advantage of this opportunity. If you have any questions, please let me know. SC20 is now a virtual

Hello, Thank you very much for your help. How can I draw a Lorenz curve with several replications ? Here is an example with 4 replications: hosts=c(23,31,19,10,7,7,3, 39,40,8,3,6,2,2, 47,17,8,10,6,11,1, 30,30,10,0,15,15,0) parasites=rep(seq(from=0,to=6,by=1),4) replications=c(rep(1,7),rep(2,7),rep(3,7),rep(4,7)) test <- cbind(parasites,hosts,replications) Should I

Hello dear R-help members, I would appreciate any help in understanding how the rpart function computes the "improve" (which is given in fit$split) when using the split='information' parameter. Thanks to Professor Atkinson help, I was able to find how this is done in the case that split='gini'. By following the explanation here:

Hi folks, I've got an idea for something I may be able to do as a Link-Time- Optimization pass, but I can't seem to find how LTO gets used. My crude approach has been to compile LLVM for debugging, and then run gdb on llvm-ld with -O5 and two .bc files, trying to catch LTO in action. I've also scanned the compiled tools, and none of them appear to use the symbol

Hi, I would like to change permissions of /dev/sda* and /dev/hda* files to 0644. I added the following rules to 99-udev.rules files: KERNEL=="[hs]d*", OWNER="root" GROUP="disk", MODE="0644 After restarting udev by strat_udev, I checked that permissions were changed: [root at gini rules.d]# ll /dev/sda* brw-r--r-- 1 root disk 8, 0 Sep 2 2009 /dev/sda

Hi all, I apologies in advance if I am missing something very simple here, but since I failed at resolving this myself, I'm sending this question to the list. I would appreciate any help in understanding how the rpart function is (exactly) computing the "improve" (which is given in fit$split), and how it differs when using the split='information' vs split='gini'