similar to: problem with lapply(x, subset, ...) and variable select argument

hi, did'nt see anything in the archive: map('world',pro='rectangular',para=0) yields a strange artifact (horizontal bar) extending over the whole map at a certain latitude range (approx 65 deg. north), whereas map('world',pro='rectangular',para=180) (which should be the same) does not show the artifact. the artifact shows up in other projections as well,

I have encountered the following problem: I need to extract from a list of lists equally named compenents who happen to be 'one row' data frames. a trivial example would be: a <- list(list( df = data.frame(A = 1, B = 2, C = 3)), list(df = data.frame(A = 4,B = 5,C = 6))) I want the extracted compenents to fill up a matrix or data frame row by row. the obvious thing to do seems: b

Full_Name: joerg van den hoff Version: 1.9.0 and 1.7.1 OS: MacOS (1.9.0), SunOS (1.7.1) Submission from: (NULL) (149.220.4.88) something like (sure not the originally intended input, but something like this can happen...): approx(c(1,2),c(NA,NA),1.5) crashes R (bus error under MacOS, segmentation fault under SunOS). search of the bug archive did not work. I hope this bug was not reported

I need to make plots similar to those produced by the s-plus rayplot function but can't seem to find it in R. These 'vector maps' plot a ray or vector at each specified location. Is there something similar in R ? --Rich Richard Kittler AMD TDG 408-749-4099

I recently coded a recursion algorithm in R and ir ran a few days without returning any result. So I decided to try a simple case of computing binomial coefficient using recusrive relationship choose(n,k) = choose(n-1, k)+choose(n-1,k-1) I implemented in R and Fortran 90 the same algorithm (code follows). The R code finishes 31 minutes and the Fortran 90 program finishes in 6 seconds. So the

I raised this question quite some time ago but it quitly went down the river. I'll give it a second try (before keeping my modified version of matplot for ever...): matplot supports vectors (and/or character strings) for a number of arguments namely `type', `lty', `lwd', `pch', `col', `cex'. all of them act consistently in such a way that the first entries are used

hi everybody, I noticed the following: in one of my scripts 'layout' is used to generate a (approx. square) grid of variable dimensions (depending on no. of input files). if the no. of subplots (grid cells) becomes moderately large (say > 9) I use a construct like ###layout grid computation and set up occurs here### ... opar <- par(no.readonly = T);

Full_Name: joerg van den hoff Version: 1.9.1 OS: MacOS and SunOS Submission from: (NULL) (149.220.4.88) follow up to ID 7166. something like approx(c(1,2),c(NA,NA),1.5,rule=2) crashes 1.9.1 on both systems (MacOS 10.3.5.: bus error, SunOS 5.9: segmentation fault) even if xout is within given x range (as in example above) where rule=2 seems not be relevant anyway.

If object is user defined is: object$df.residual the same thing as df.residual(object) This is my first time to encounter the $ sign in R, I'm new. I'm reviewing "summary.glm" and in most cases it looks as though the $ is used to extract some characteristic/property of the object, but I'm not positive. Thanks ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

hi list, used versions: 2.12.1 and 2.14.0 under ubuntu and macosx. I recently stumbled over a problem with `nls', which occurs if the model is not specified explicitly but via an evaluation of a 'call' object. simple example: 8<-------------------------------------------------------------------------------------- nlsProblem <- function (len = 5) {

Dear R developers, Reviewing my code, I have realized that about 80% of the time in the lapply I need to access the names of the objects inside the loop. In such cases I iterate over indexes or names: lapply(names(x), ... [i]), lapply(seq_along(x), ... x[[i]] ... names(x)[i] ), or for(i in seq_along(x)) ... which is rather inconvenient. How about an argument to lapply which would specify the

Dear R helpers: I wonder how to pass more than one argument to the function called by lapply. For example, #R code below --------------------------- indf <- data.frame(id=I(c('a','b')),y=c(1,10)) #I want to add an addition argument cutoff into the function called by lapply. outside.fun <- function(indf, cutoff) { unlist(lapply(split(indf, indf[,'id']),

|Hello, Given a function with several arguments, I would like to perform an lapply (or equivalent) while holding one or more arguments fixed to some common value, and I would like to do it in as elegant a fashion as possible, without resorting to wrapping a separate wrapper for the function if possible. Moreover I would also like it to work in cases where one or more arguments to the original

u<-c(0.1,0.2,0.3) v<-c(0.2,0.3,0.5) outer1<-outer(u,u,">=") outer2<-outer(v,v,">=") m<-nrow(outer1) j<-nrow(outer2) zz<-lapply(1:m, function(m) as.numeric(outer1[m,])) tt<-lapply(1:m, function(m) as.numeric(outer2[m,])) zz[[1]]*tt[[3]], e.g., is possible, but I want every products between two lists. Is there a way to do that? -- View this

Hi, I am probably making a simple mistake but I can't see it > X Error: Object "X" not found > exists("X") [1] FALSE > lapply("X", exists) [[1]] [1] TRUE Why is lapply producing true? Is it something to do with the first argument of lapply also being called 'X'? > version _ platform i386-pc-mingw32 arch i386

Dear all, I'd like to subset a df within a function, and use select for choosing the variable. Something like (simplified example): mydf <- data.frame(a= 0:9, b= 10:19) ttt <- function(vv) { tmpdf <- subset(mydf, select= vv) mean(tmpdf$vv) } ttt(mydf$b) But this is not the correct way. Any help? Thanks in advance Juli

There seems to be a funny interaction between lapply and "$" -- also, "$" doesn't signal an error in some cases where "[[" does. The $ operator accepts a string second argument in functional form: > `$`(list(a=3,b=4),"b") [1] 4 lapply(list(list(a=3,b=4)),function(x) `$`(x,"b")) [[1]] [1] 4 ... but using an lapply "..."

I've defined my own version of summary.default, that gives a better summary for highly skewed vectors. If I call summary(x) the method is used. If I call summary(data.frame(x)) the method is not used. I've traced this to lapply; this uses the new method: lapply(list(x), function(x) summary(x)) and this does not: lapply(list(x), summary) If I make a copy of lapply, WITHOUT the

I thought this is a common question but rseek/google searches don't yield any relevant hit. I have a matrix of character strings, which are time stamps, > time.m[1:5,1:5] [,1] [,2] [,3] [,4] [,5] [1,] "08:00:20.799" "08:00:20.799" "08:00:20.799" "08:00:20.799" "08:00:20.799" [2,]

Hi Bert, On Tue, Nov 14, 2017 at 8:11 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Could someone please explain the following? I did check bug reports, but > did not recognize the issue there. I am reluctant to call it a bug, as it > is much more likely my misunderstanding. Ergo my request for clarification: > > ## As expected: > >> lapply(1:3, rnorm, n = 3)