similar to: "rpart": how to use each variable only once?

Hi list, Could someone explain to me why the following result is not a integer? > difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004", >format="%d%b%Y"), units="days") Time difference of 195.0417 days I'm using R2.12.0 on WinXP. Thanks! ...Tao

Are they the same with .RData being the newer format?? Thanks, ...Tao

Hi list, I want to draw a bar plot with color indicating one grouping and different shading on top of the color indicating another grouping.? How should I proceed? Thanks! ...Tao

Hi list, Is there a way to load one specific object from a .RData file which contains multiple data objects?? Thanks, ...Tao

Hi list, I've encounter this problem (see below).? I know it's particularly R-related and it's easy to get by but it still bothers me a lot.? It looks the last character of "N.C. " is a space to me, but it's clearly not.? Can someone tell me a way to figure out what character is in the last position. Thanks! Tao > levels(dat$flag)[3] [1] "N.C.?" >

HI, Dear R community, Last friday, I used the codes, it works, but today, it does not run? > fit.dimer <- rpart(outcome ~., method="class", data=p.df) Error in `[.data.frame`(frame, predictors) : undefined columns selected DOEs anyone have comments or suggestions? Thanks in advance! -- Sincerely, Changbin -- [[alternative HTML version deleted]]

Hi list, Can somebody explain why there are these warning messages?? I just don't get it.? I'm using R 2.15.1 on WinXP. Thanks! Tao > x [1] -2.143510 -1.157450 -1.315581? 1.033562 -1.225440 -1.179909 >? ifelse(x>0, log2(x), -log2(-x)) [1] -1.099975 -0.210950 -0.395700? 0.047625 -0.293300 -0.238675 Warning messages: 1: In ifelse(x > 0, log2(x), -log2(-x)) : NaNs produced

--- included message ---- Thus, my question is: *What common measures exists for ranking/measuring variable importance of participating variables in a CART model? And how can this be computed using R (for example, when using the rpart package)* ---end ---- Consider the following printout from rpart summary(rpart(time ~ age + ph.ecog + pat.karno, data=lung)) Node number 1: 228 observations,

Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]]

Hi, I am trying to make a multi-class classification tree by using rpart. I used MASS package'd data: fgl to test and it works well. However, when I used my small-sampled data as below, the program seems to take forever. I am not sure if it is due to slowness or there is something wrong with my codes or data manipulation. Please be advised ! The data is described as the output from str()

Hi list, Is there a R function I can use to extract the worksheet names from an Excel file?? If no, any other automatic ways (not using R) to do this? thanks! ...Tao

HI, Dear Greg, I AM A NEW to GBM package. Can boosting decision tree be implemented in 'gbm' package? Or 'gbm' can only be used for regression? IF can, DO I need to combine the rpart and gbm command? Thanks so much! -- Sincerely, Changbin -- [[alternative HTML version deleted]]

Hi list, I was trying to use "predict.coxph" to calculate martingale residuals on a test data, however, as pointed out before http://tolstoy.newcastle.edu.au/R/e4/help/08/06/13508.html predict(mycox1, newdata, type="expected") is not implemented yet. Dieter suggested to use 'cph' and 'predict.Design', but from my reading so far, I'm not sure they can

Hi, When I look at the summary of an rpart object run on my data, I get 7 nodes but when I plot the rpart object, I get only 3 nodes. Should the number of nodes not match in the results of the 2 functions (summary and plot) or it is not always the same? Look forward to your reply, Carol -------------------------------------------- ?summary(rpart.res) Call: rpart(formula = mydata$class ~ ., data

After fitting and pruning an rpart model, it is often the case that one or more of the original predictors is not used by any of the splits of the final tree. It seems logical, therefore, that values for these "unused" predictors would not be needed for prediction. But when predict() is called on such models, all predictors seem to be required. Why is that, and can it be easily

Dear all, I am having some trouble with getting the rpart function to work as expected. I am trying to use rpart to combine levels of a factor to reduce the number of levels of that factor. In exploring the code I have noticed that it is possible for chisq.test to return a statistically significant result whilst the rpart method returns only the root node (i.e. no split is made). The following

I am performing a binary classification using a classification tree. Ironically, the data themselves are 2483 tree (real biological ones) locations as described by a suite of environmental variables (slope, soil moisture, radiation load, etc). I want to separate them from an equal number of random points. Doing eda on the data shows that there is substantial difference between the tree and random

Hi, Suppose i have generated an object using the following : fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) And when i print fit, i get the following : n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 17 absent (0.7901235 0.2098765) 2) Start>=8.5 62 6 absent (0.9032258 0.0967742) 4) Start>=14.5 29 0 absent (1.0000000

Wondered about the best way to control for input variables that have a large number of levels in 'rpart' models. I understand the algorithm searches through all possible splits (2^(k-1) for k levels) and so variables with more levels are more prone to be good spliters... so I'm looking for ways to compensate and adjust for this complexity. For example, if two variables produce

I've been using the package rpart with R 1.3.0 for Windows to produce simple classification trees for some measurement data from paleontological specimens. Both the rpart documentation and the output confirm that the program produces splits on continuous data that leave "holes" in the data. It is probably of little practical importance, but is there a reason why the binary