similar to: rsq.rpart not found ...

If there is a helpfile open (f.e ?glm) and it is the top window, then an exception error occurs (closing RGUI) when I hit the open file button. If the helpfile is not the top window (of the RGUI) I am able to open a new script without any error. The RGUI is not closing complete there is a blank screen left which I have to close with the X Button or Taskmanager Windows XP Home - German Version

I searched the help list and the manual, but I do not find my mistake. Switch is working with character , with integer, but not in the third example Regards Knut count1 <- 0 test

Is there any function to show the points like this example of SPSS? http://biostatistic.de/temp/reg.jpg The point size should represent the number of data at this point. with regards Knut Krueger

Is there any possibility to get high resolution plots in a windows xp system? I tried it with the device function png(filename = "c:/r/highresplot%d.png",pointsize=12, res=900) but when I try to set: width = 480, height = 480 or pointsize = 12, the text is not scaled in the same way as the plots. with regards Knut Krueger http://www.biostatistic.de

scaling<-4 xywidth<-480 resolution<-150 png(filename = "c:/r/anschluss/plots/4.png", width = xywidth*scaling, height = xywidth*scaling,pointsize = 12, bg = "white", res = resolution*scaling) ...... barplot(xrow,col = barcolors,cex.axis=scaling, ylab="mean time till attachment in sec",cex.lab=1.2*scaling) I tried to scale the barplot but there is one

I was wondering why it is possible to read long labels from the CVS files but not from the SPSS files. I did not have much time to search for the code but I found: in foreign_0.8-10 source file var.h.in > /* Definition of the max length of a short string value, generally > eight characters. */ > #define MAX_SHORT_STRING ((SIZEOF_DOUBLE)>=8 ? ((SIZEOF_DOUBLE)+1)/2*2 > : 8)

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 amazed that the traffic decreased the last days. I am afraid that there is any problem with the spamfilter of my provider. There are not messages from march 04rd until march 15th, Two messages from march 16 and nothing else since march 03rd Please answer also with PM to the mail address. I hope I will solve the problem. Thank you Knut

Dear All, I have a question regarding predict.rpart. I use rpart to build classification and regression trees and I deal with data with relatively large number of input variables (predictors). For example, I build an rpart model like this rpartModel <- rpart(Y ~ X, method="class", minsplit =1, minbucket=nMinBucket,cp=nCp); and get predictors used in building the model like

There is a difference in the p- value from 0.000 and 0.012 when I am using SPSS. 0.000 when I am using the independent variable as scaled 0.012 if I am using the variable as ordinal. The independent variable is ordinal but it seems that R is using the variable as an scaled, because the P- Value is computed with 4.66e-06 so I am not sure which description I am misunderstanding: SPSS;:

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

Hi to all, I am looking for a randomisation procedure for a single matrix, including a possibility to set the number of randomisations and the to set the number of row and columns . Knut

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 to all, test <- data.frame("X"=c(1:4),"Y"=c(5:8),"Z"=c(8:11)) test <- test[,-2] Is there a way to specify the col name "Y" to delete instead the number? Kind regards Knut

Hi to all, I am just looking for more efficient ways ;-) is there a better way instead a loop to decrease x if greater y test <- c(1,3,5,7,9) decrease if greater 1 to test2 <- c(1,2,4,6,8) Kind regards Knut

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

Hi to all, does anybody knows whether there is a package to plot those http://www.equine-science.de/temp/graph.jpg graphs. the thickness of the points and/or the lines should be represent the numbers of behaviours With kind regards Knut

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

Hi to all developer, sure that I will not be the first one with this idea but I did not found treads about this. I realized that there are some user interfaces like Brodgar and SciViews. There was used a lot of manpower to develop those programs. On the other side I realized that it could be very helpful to have such programs if somebody would like to change the statistical software from SPSS

. I am looking for a histogram or box plot with the adding normal distribution curve I think that must be possible, but I am not able to find out how to do. Regards Knut