search for: xapply

...[,2] [1,] List,2 List,2 [2,] List,2 List,2 By the way: how can I get this not to abbeviate the entries but instead give me something like: [,1] [,2] [1,] list(1, 3) list(2, 3) [2,] list(1, 4) list(2, 4) The closest I can get is matrix(as.character(mm),2,2) By-the-way^2: is there some Xapply function that maps a function over all the elements of a structure (vector, matrix, list, ...) and preserves the original structure? For example, I'd want Xapply(matrix(1:4^2,2,2),sqrt) == sqrt(matrix(1:4^2,2,2)). In this case, I'd use Xapply(as.character,mm), because as.character returns...

Dear list, I'm stuck with looking for a function of the *apply family, which I suppose exists already ? just I can't find it: What I'm looking for is somewhere between sweep and mapply that does a calculation vectorized over a matrix and a vector: It should work complementary to sweep: for each row of the matrix, a different value of the vector should be handed over. Close to

..., NA, 5) V2b <- c (37, NA, 15, NA, NA, NA, 90, NA, 161, NA) Supongamos que tengo la expresión (que no puedo asignarlo a ninguna variable): (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) Supongamos que tengo que utilizar esta expresión dos o más veces y no puedo utilizar ni un xapply () ni un bucle. Por ejemplo: V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - V1)/V1)*100 V4 <- ifelse (! is.na ((((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a))-V1)/V1)*100), ifelse ((((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a))-V1)/V1)*100 > 0, "1&qu...

..., NA, 5) V2b <- c (37, NA, 15, NA, NA, NA, 90, NA, 161, NA) Supongamos que tengo la expresión (que no puedo asignarlo a ninguna variable): (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) Supongamos que tengo que utilizar esta expresión dos o más veces y no puedo utilizar ni un xapply () ni un bucle. Por ejemplo: V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - V1)/V1)*100 V4 <- ifelse (! is.na ((((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a))-V1)/V1)*100), ifelse ((((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a))-V1)/V1)*100 > 0, "1&qu...

...NA, NA, NA, 90, NA, 161, NA) > > Supongamos que tengo la expresión (que no puedo asignarlo a ninguna > variable): > > (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) > > Supongamos que tengo que utilizar esta expresión dos o más veces y no > puedo utilizar ni un xapply () ni un bucle. Por ejemplo: > > V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - V1)/V1)*100 > V4 <- ifelse (! is.na ((((ifelse (is.na (V2a) & ! is.na (V2b), > V2b, V2a))-V1)/V1)*100), ifelse ((((ifelse (is.na (V2a) & ! is.na > (V2b), V2b, V2a))-V1)/...

...> > > Supongamos que tengo la expresión (que no puedo asignarlo a ninguna > > variable): > > > > (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) > > > > Supongamos que tengo que utilizar esta expresión dos o más veces y no > > puedo utilizar ni un xapply () ni un bucle. Por ejemplo: > > > > V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - V1)/V1)*100 > > V4 <- ifelse (! is.na ((((ifelse (is.na (V2a) & ! is.na (V2b), > > V2b, V2a))-V1)/V1)*100), ifelse ((((ifelse (is.na (V2a) & ! is.na > &gt...

...tengo la expresión (que no puedo asignarlo a ninguna > > > variable): > > > > > > (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) > > > > > > Supongamos que tengo que utilizar esta expresión dos o más veces y no > > > puedo utilizar ni un xapply () ni un bucle. Por ejemplo: > > > > > > V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - > V1)/V1)*100 > > > V4 <- ifelse (! is.na ((((ifelse (is.na (V2a) & ! is.na (V2b), > > > V2b, V2a))-V1)/V1)*100), ifelse ((((ifelse (is.na (V2...

Hi R-Group, I am stuck with the following problem: I am constructing a portfolio of 2 variables x and y x <- rnorm(100, mean=100, sd=4) y <- rnorm(100, mean=120, sd=10) which I am combining as follows to a portfolio for sampling purposes: portfolio <- c(rep(x, 8), rep(y, 2)) In this case I have assigned the weights of 8 and 2 to calculate the bootstrapped mean: mean.boot <-

I would like to give a probability distribution function of a function of (x,y) on the half-plane y>0, and a constant 0<c<1 and I would like to know the c percentile of the marginal distribution of x. I have tried along the lines of the following but I keep getting errors: # SIMPLIFIED PROBLEM # The plan is to solve for the .975 percentile "xc" of the marginal x distribution

Dear all, I am trying to learn lapply. I would like, as a test case, to try the lapply alternative for the Shadowlist<-array(data=NA,dim=c(dimx,dimy,dimmaps)) for (i in c(1:dimx)){ Shadowlist[,,i]<-i } ---so I wrote the following--- returni <-function(i,ShadowMatrix) {ShadowMatrix<-i} lapply(seq(1:dimx),Shadowlist[,,seq(1:dimx)],returni) So far I do not get same results

> From: Thomas Lumley > > On Sun, 30 Oct 2005, Jonathan Rougier wrote: > > > I'm not sure about this. Perhaps I am a dinosaur, but my feeling is > > that if people are writing functions in R that might be subject to > > simple operations like outer products, then they ought to be writing > > vectorised functions! > > I would agree. How about an

* What were your biggest misconceptions or stumbling blocks to getting up and running with R? * What documents helped you the most in this initial phase? I especially want to hear from people who are lazy and impatient. Feel free to write to me off-list. Definitely write off-list if you are just confirming what has been said on-list. -- Patrick Burns pburns at pburns.seanet.com