similar to: Dynamic reference, right-hand side of function

Hi! Thanks for the replies! I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The syntax from Peter Dalgaard was really clever, and I learned a lot from it, even though it didn't solve my problem (I guess it wasn't very well explained). My problem was basically that I have a data matrix

The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g. datanames <- paste0("aa_", 2000:2007) datalist <- lapply(datanames, get) names(datalist) <- datanames col1 <- lapply(datalist, "[[", 1) colnum <- lapply(col1, as.numeric) (The 2nd

Um, if you insist on doing it that way, at least use assign(varname, as.vector(get(varname))) -pd > On 4 Dec 2017, at 22:46 , Love Bohman <love.bohman at sociology.su.se> wrote: > > Hi! > Thanks for the replies! > I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The

:-) I don't insist on anything, I'm just struggling to learn a new language and partly a new way of thinking, and I really appreciate the corrections. I hope I someday will be able to handle lists in R as easy as I handle loops in Stata... Thanks again! Love -----Ursprungligt meddelande----- Fr?n: peter dalgaard [mailto:pdalgd at gmail.com] Skickat: den 4 december 2017 23:09 Till:

Hi again! I know you don't find loops evil (well, at least not diabolic :-) ). (After many hours googling I have realized that thinking about loops rather than lists is a newbie thing we Stata-users do, I just jokingly pointed it out). Anyway, I'm really happy that you try to teach me some R-manners. Since I still get questions about what the h**k I mean by my strange question, I sort it

Loops are not evil, and no-one in this thread said they are. But I believe your failure to provide a reproducible example is creating confusion, since you may be using words that mean one thing to you and something else to the readers here. ################################ # A reproducible example includes a tiny set of sample data # Since we cannot reproducibly refer to filenames (your

By the way, R 'vectors' are not the equivalents of mathematical 'vectors'. In R, a vector is something that can have arbitrary length and which has no 'attributes', other than perhaps element names. Vectors can be numeric, character, complex, lists, etc. Functions, names, and NULL are not vectors. In my opinion, the typical data scientist will rarely find the R vector

https://bugzilla.mindrot.org/show_bug.cgi?id=3629 Bug ID: 3629 Summary: Building with Clang-17 fails due to -fzero-call-used-regs Product: Portable OpenSSH Version: 9.5p1 Hardware: amd64 OS: Mac OS X Status: NEW Severity: critical Priority: P5 Component: Build system

I am writing a function where the arguments are names of objects or variable names in a data frame. To convert the strings to the objects I am using eval(parse(text=name)): f.graph.two.vbs<-function(dataname,v1){ val<-paste(dataname,v1,sep="$") val<-eval(parse(text=val)) val } However running this returns an error:

Hi, I have a circular shaped set of point on the plane (X,Y) centered in zero. The distribution is more dense close to zero and less dense far from zero. I need to find the radius of a circle centered in zero that contains 65% of the points in the sample. Is there any R directive that can do this? I wanna start with 2D set of points, but the real case scenario is with a 5D set of points.

I have an R script with the following applicable lines: xshort <- window(s, start=st, end=ed) . . . xshort <- ts(xshort, frequency=1, start=1) . . . m1 <- m2 <- m3 <- m4 <- m5 <- m6 <- NULL m1 <- tslm(xshort ~ trend) I get an error: Error in get(dataname) : object 'xshort' not found When I do traceback() I get: 3: get(dataname) 2: tslm(xshort ~

Hey everybody, i am using the rugarch-package and its great! I have a pretty easy problem, but i just dont get it, so thanks if you can help me. Normally i use: / data(DATANAME) spec = ugarchspec() fit = ugarchfit(data = x[,1], spec = spec) fit slotNames(fit) names(fit at fit) coef(fit) infocriteria(fit) likelihood(fit) nyblom(fit) signbias(fit) head(as.data.frame(fit)) head(sigma(fit))

Folks, I want to automate some graphing using ggplot. Here is my code graphChargeOffs2<-function(coffs) { ggplot(coffs, aes(levels)) dataNames<-names(coffs)[!names(coffs) == "levels"] for(i in dataNames) { thisData<-coffs[[i]] last_plot() + geom_line(aes(y = thisData, colour = i)) } last_plot() + ylab("Total Chargeoffs") } coffs is a data.frame.

Dear all, I have a function that for a variable number of inputs plots them to the same plot I am doing this quite simply by plot(seq(from=start, to=stop, length.out=np), datalist[[1]]$dataset                                                                  xlim=c(start, stop), ylim=c(0, 1), type="l")                                                                                      

I want to generate a list (called "dataList" below) where each of its levels is named. These names are assigned to nameList, which contains all possible permutations of size two taking letters from a larger alphabet, e.g., "aa",...,"Fd",..,"Z1",... One of these permutations is the character string "NA". It seems that when I try to name one

At 10:37 PM 5/1/00 -0400, E. S. Venkatraman wrote: >I have the following problem. I have survival data (time, status) along >with several covariates (X1, X2,..., Xn). I want to fit a Cox model for >each of the covariate (univariately) and obtain the fitted probability of >survival at a fixed time point t0 and covariate value Xi0. I tried to do >this in a for loop where the index

I have been using R the past couple of years to run models on data we are collecting. I recently got a new computer and updated to a new version of R (2.60 -> 2.90). Since the update, I cannot get my syntax to run. I have tried copying the file it is looking for into many different directories to try and run it. In the last version, I found that it was easiest if the file was copied into the R

Hello I have created a graph using the following commands: <<< startBReP3O1T <- diffs$BReP3O1T - diffs$diff_BReP3O1T endBReP3O1T <- diffs$BReP3O1T x <- seq(47,89, length = 10) ymin <- min(min(startBReP3O1T), min(endBReP3O1T)) ymax <- max(max(startBReP3O1T), max(endBReP3O1T)) y <- seq(ymin, ymax, length = 10) plot(x,y, type = 'n', xlab = 'Age', ylab =

Hello, i only got a small problem. i try to create automatic new dataframes, or png?s. the main problem i got is: how can i create automatic a new name for a file (read out by simply "for") - i tried to use "(paste...) but theres an errormessage, about a wrong declination. R told it is as.character, but need as.Real. Should i use another method than "paste"? i tried as

Dear all, I have a code that looks like the following (I am sorry that this is not a reproducible example)     indexSkipped<-NULL  ....code Skipped that might alter indexSkipped     if (length(indexSkipped)==0)         spatial_structure<-spatial_structures_from_measurements(DataList[[i]]$Lon,DataList[[i]]$Lat,meanVector)     else