similar to: Running code sequentially from separate scripts (but not functions)

I think I am overlooking a call or concept in R to help me easily and quickly restructure my data.frame:   Sometimes the data I receive looks like: VariableName, Run1, Run2, Run3, Location temp,               15.0,  16.0,  17.0, There   And other times it looks like: VariableName, Run, Location temp,               17.0, There   I would like to use the header information in order to be able to

Hello I am new in R. I am trying to implement Biomod 2 package. However, I have a doubt. I want to calculate the mean and sd of "Testing.data" (ROC and TSS) > # let's print the ROC scores of all selected models > myBiomodModelEval_55["ROC","Testing.data",,,] RUN1 RUN2 RUN3 RUN4 RUN5 RUN6 RUN7 RUN8 RUN9 RUN10 0.938 0.938 0.926 0.931 0.939

I am writing an interprocedural compiler pass. Because the passneeds information from a FunctionPass, e.g., the post-dominance frontier (PDF), and because a ModulePass is not permitted to require a FunctionPass, I am forced to make my pass a FunctionPass and do majority of its work in the doFinalization() method. When I run "opt -mypass -verify -o code2.bc code1.bc" I get no

I followed what you said and called verifyModule() with the AbortProcessAction option. verifyModule() returns false, but does not abort and does not print out any information about what caused the verification to fail. Chris Lattner wrote: > On Wed, 21 Feb 2007, Ryan M. Lefever wrote: >> I am writing an interprocedural compiler pass. Because the passneeds >> information from a

Hi, I have a lattice lot conditioned on two variables. Example code is: library(lattice) x <- data.frame(d=runif(100), f1=sample(c('yes', 'no'),100,replace=TRUE), f2=c(rep('Run1',30),rep('Run2',30),rep('Run3',40))) histogram(~d | f1 + f2, x) In the plot, for a given value of f2, there are two panels, one for 'n'

I also tried iterating through the functions of the module and calling verifyFunction(), which also returns false, but does not cause an abort or report anything to stderr about what caused the verification to fail. From the doxygen for verifyFunction() and verifyModule(), it seems like they both should print information to stderr if the verification fails and should abort opt if

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Dear All Sorry for what appears a trivial matter - I'm new to R and am stumbling ahead. I have a table of numerical data (36 rows by 12 columns) such as below: GM1 GM2 GM3 GM4 GM5 ...etc GM12 Run1 1 2 1 2 3 ... Run2 2 1 3 2 1 ... ... Run36 2 1 1 1 1 I would like to plot simple line graphs of some of the runs or

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Dear R-ers, I have a large data frame (several thousands of rows and about 2.5 thousand columns). One variable ("group") is a grouping variable with over 30 levels. And I have a lot of NAs. For each variable, I need to divide each value by variable mean - by subgroup. I have the code but it's way too slow - takes me about 1.5 hours. Below is a data example and my code that is too

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a

Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g., x<-c(1,2,3,4). How can I get a TRUE or FALSE for whether x contains a 2? -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello! If I have in my data frame MyFrame a variable saved as a Date and want to translate it into years, I currently do it like this using "zoo": library(zoo) as.year <- function(x) as.numeric(floor(as.yearmon(x))) myFrame$year<-as.year(myFrame$date) Is there a function that would do it directly - like "as.yearmon" - but for years? Thank you! -- Dimitri

Hello. Until today I've been using R2.9 and since today R2.10 (on a PC). In both of them it takes about 20 sec for the prompt to appear IN R console after I start R. And every time it says: "Previous saved work space restored" - even if I have not saved any workspace or, in case of R2.10 - even though I have not used it once. In the older versions - R would start within 2-3 sec. Is

Dear R-ers, I have a data frame data with predictors x1 through x5 and the response variable y. I am running a simple regression: reg<-lm(y~x1, data=data) I would like to loop through all predictors. Something like: predictors<-c("x1","x2",... "x10) for(i in predictors){ reg<-lm(y~i) etc. } But it's not working. I am getting an error: Error in

On Wed, 21 Feb 2007, Ryan M. Lefever wrote: > I am writing an interprocedural compiler pass. Because the passneeds > information from a FunctionPass, e.g., the post-dominance frontier > (PDF), and because a ModulePass is not permitted to require a > FunctionPass, I am forced to make my pass a FunctionPass and do majority > of its work in the doFinalization() method. ok > When

Hello, dear R-ers! I have a data frame: x<-data.frame(a=c(4,2,4,1,3,4),b=c(1,3,4,1,5,0),c=c(NA,2,5,3,4,NA),d=rep(NA,6),e=rep(NA,6)) x When x$a==1, I would like to replace NAs in columns d and e with 8 and 9, respectively When x$a != 1, I would like to replace NAs in columns d and e 101 and 1022, respectively. However, I only want to do it for rows 2:5 - while ignoring what's happening in