similar to: Rpart description of tree groups

My apologies if this is obvious - I assume it is but still can't seem to make it work. I've got a data frame acps.df; I want to run an analysis on a subset of that data frame, defined as the subset for which the value of OGMCAT (a column in the data frame) is anything OTHER than 9, 11, or 12. So I'd like to define a new data frame that consists of the same variables as in acps.df,

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323

If I do not add "ress" into the data frame "res", there is no quote in the data frame. However, I add "ress", all column were found the quote. How to remove it? If you can delete the quote in the file "ress", that is better. Thanks. > ress[1:10] [1] "ABHO.ABNE" "ABHO.ACBA" "ABHO.ACGI" "ABHO.ACKO"

Greetings. I'm trying to replicate an analysis I did a few years ago, then in SPSS, using the SPSS GLM command: GLM n_diffpt WITH age_i inc_i join_i work_i educ_i give_i cs_i eff_i age_a inc_a join_a work_a educ_a give_a cs_a eff_a /METHOD = SSTYPE(3) /INTERCEPT = INCLUDE /PRINT = PARAMETER ETASQ RSSCP GEF /CRITERIA = ALPHA(.05) /DESIGN = age_i*age_a inc_i*inc_a

All - I think I'm being dense, but for the life of me, I can't figure out why I get error message with the code below. I have data that looks like param level perc.surv asin.tran DO 3 0.6864407 0.9764544 DO 3 0.1250000 0.3613671 DO 3 0.8738739 1.2077299 DO 4 0.4615385 0.7468986 DO 4 0.5691057 0.8547258 DO 4 0.8504673 1.1737517 DO 6 0.5494505 0.8349297 DO

Hello everyone. When I execute the following in R > (18-46)/(45-93) [1] 0.5833333 I get small precision for what I am trying to deal with . Is it possible to increase the precision for this and for other operations? For example openoffice calc for this operation returns 0.58333333333333300000 I I would like to thank you for your help [[alternative HTML version

Hi, I have run mars() with the default degree=1, and have constructed splines which are similar to what I would get from Salford System's MARS. I then use these splines as potential predictors, along with rpart( ) terminal nodes and other potential predictors, in logistic regression in SAS. The output of mars() and the splines I created are shown below (pardon the lengthy output): >