search for: ans

...hink that the "dimname names" of tables and arrays could make aperm() and apply() (and probably some other functions) easier to use. (dimname names are, for example, created by table() ) The use would be something like: -- x <-table( from=sample(3,100,rep=T), to=sample(5,100,rep=T)) trans <- x / apply(x,"from",sum) y <- aperm( trans, c("from","to") ) z <- aperm(y, c("to","from") ) res <-apply( y, "to", sum) -- This makes the array much easier to handle than having to keep track which dimension currently means...

<<insert bug report here>> > apply(array(1:20,c(2,2,5)),2:3,function(x) x) Error: length of dimnames must match that of dims > Changing: dimnames = if (is.null(dn.ans)) list(ans.names, NULL) else c(list(ans.names), dn.ans) To: dimnames = if (length(dn)==0) NULL else if (is.null(dn.ans)) list(ans.names, NULL) else c(list(ans.names), dn.ans) seems to fix this. Chuck Berry --please do not edit the information below-- Version: platform = sparc-sun-solari...

...ed code below which reproduces the problem and below that the output from a run of that code on my computer. Any help would be greatly appreciated. Thanks in advance. -Ben ########################### begin code ################################### version dendro.leaf <- function(label) { ans <- list() attr(ans, 'members') <- 1 attr(ans, 'height') <- 0 attr(ans, 'leaf') <- T attr(ans, 'midpoint') <- 0 attr(ans, 'label') <- label attr(ans, 'class') <- 'dendrogram' ans } dendro.merge <- function...

I have a list 'ans' from the following code: tt <- rnorm(50) rr <- rnorm(50) ans <- lm(rr~tt) ans[1] is "$coefficients", ans[2] is "$residuals", ans[3] is "$effects", ... and so on up to ans[12]. Is there an easy way to display just these names and not the data they contai...

Hi, it looks like: > stats:::`[.formula` function (x, i) { ans <- NextMethod("[") if (length(ans) == 0L || as.character(ans[[1L]])[1L] == "~") { class(ans) <- "formula" environment(ans) <- environment(x) } ans } <bytecode: 0x556688d87cc0> <environment: namespace:stats> doesn't...

...#39;ve seen various posts on this question, but still can't get the code right. If I run the following code one line at a time, it works fine. If I run it together as a block, however, it doesn't wait for the input and gives an error. There must be a way to have are pause/wait for an answer, but I can't seem to find it. Thanks! J Code: choosefunction <- function(){readline("1. linear, 2. linear with lag, 3. nonlinear ")} ans <- as.integer(choosefunction()) if (ans==1){K2=x1} if (ans==2){K2=x2 } if (ans==3){K2=x3 } ans Error text: > ans <- as.intege...

...== 0) { r <- z$residuals n <- length(r) w <- z$weights if (is.null(w)) { rss <- sum(r^2) } else { rss <- sum(w * r^2) r <- sqrt(w) * r } resvar <- rss/(n - p) ans <- z[c("call", "terms")] class(ans) <- "summary.lm" ans$aliased <- is.na(coef(object)) ans$residuals <- r ans$df <- c(0, n, length(ans$aliased)) ans$coefficients <- matrix(NA, 0, 4) dimnames(ans$co...

...are the first few lines of the S-Plus file:   sshc _ function(rc, nc, d, method, alpha=0.05, power=0.8,              tol=0.01, tol1=.0001, tol2=.005, cc=c(.1,2), l.span=.5) { ### for method 1 if (method==1) { ne1 _ ss.rand(rc,nc,d,alpha=.05,power=.8,tol=.01) return(ne=ne1)                }   My translation looks like this:   sshc<-function(rc, nc=500, d=.5, method=3, alpha=0.05, power=0.8,               tol=0.01, tol1=.0001, tol2=.005, cc=c(.1,2), l.span=.5) { ### for method 1 if (method==1) {  ne1<-ss.rand(rc,nc,d,alpha=.05,power=.8,tol=.01)  return(ne=ne1)                 } The program...

...a plot where horizontal axes limits are c(0,10), top axis has ticks at odd integers, and bottom axis has ticks at even integers. library(lattice) df <- data.frame(x=1:10,y=1:10) xscale.components.A <- function(...,user.value=NULL) { # get default axes definition list; print user.value ans <- xscale.components.default(...) print(user.value) # start with the same definition of bottom and top axes ans$top <- ans$bottom # - bottom labels ans$bottom$labels$at <- seq(0,10,by=2) ans$bottom$labels$labels <- paste("B",seq(0,10,by=2),sep="-")...

Hi! I want to return a matrix. The code does the R interfacing. This version does it fine. SEXP ans,dim; PROTECT(ans = NEW_NUMERIC(count*2)); memcpy(NUMERIC_POINTER(ans),result,count*sizeof(double)); memcpy(&(NUMERIC_POINTER(ans)[count]),occur,count*sizeof(double)); /** PROTECT(dim=NEW_INTEGER(2)); INTEGER_POINTER(dim)[0]=2; INTEGER_POINTER(dim)[1]=count; setAttrib(ans,R_Dim...

...t; > Maybe this is something that should have been changed in R 2.0.0; we > squandered that change from 1.x.x to 2.x.x. > > Duncan Murdoch Hello, I'm happy with the modified ifelse: ifelse.o <- function (test, yes, no) { storage.mode(test) <- "logical" ans <- test nas <- is.na(test) if (any(test[!nas])) ans[test & !nas] <- rep(yes, length.out = length(ans))[test & !nas] if (any(!test[!nas])) ### Changed ans[!test & !nas] <- rep(no, length.out = length(ans[!test & !nas])) ans...

...nikkoam.com > <mailto:A4678959B3D65D449266DE8D3825E0821F7F47@nycmsg501.nikkoam.com>> > > Content-Type: text/plain; charset="us-ascii" > > > > This is deliberate behavior. If you check the code, the third-to-last > > line is as follows: > > ans = ans[!(as.POSIXlt(ans@Data)$wday == 0 | > > as.POSIXlt(ans@Data)$wday== 6)] > > > > > > You could make an alternate version of holidayNYSE that omits this line, > > if you like. > > > > -CN > > > > Charles Naylor > > Assistant Vice Presid...

Dear R-help, I've encounter what seems to me a strange problem with "names<-". Suppose I define the function: fun <- function(x, f) { m <- tapply(x, f, mean) ans <- x - m[match(f, unique(f))] names(ans) <- names(x) ans } which subtract out the means of `x' grouped by `f' (which is the same as, e.g., resid(lm(x~f)) if `f' is a factor). If `x' does not have names, then I'd expect the output of the function not to have names...

...ed vector of length >=2. I propose the source of the bug and the fix below, but let me stick to the facts first. Here is an example: >a<-array(1:24, dim=2:4) > func1<-function(x)c(a=mean(x),b=max(x)) > func2<-function(x)c(mean(x),max(x)) > apply(a,1:2,func1) Error in array(ans, c(len.a%/%d2, d.ans), if (is.null(dn.ans)) { : length of dimnames [2] must match that of dims [3] > apply(a,1:2,func2) , , 1 [,1] [,2] [1,] 10 11 [2,] 19 20 , , 2 [,1] [,2] [1,] 12 13 [2,] 21 22 , , 3 [,1] [,2] [1,] 14 15 [2,] 23 24 *********...

summary: The function 'by' inconsistently strips class from the data to which it is applied. quick reason: tapply strips class when simplify is set to TRUE (the default) due to the class stripping behaviour of unlist. quick answer: This can be fixed by invoking tapply with simplify=FALSE, or changing tapply to use do.call(c instead of unlist executable example: mytimes=data.frame(date = 1:3 + Sys.time(), set = c(1,1,2)) by(mytimes$date, mytimes$set, function(x)x) INDICES: 1 [1] "2008-04-15 11:41:38 BST" &...

...; >> If I run the following code one line at a time, it works fine. If I run >> >> it together as a block, however, it doesn't wait for the input and gives >> >> an error. >> >> >> >> There must be a way to have are pause/wait for an answer, but I can't >> >> seem to find it. Thanks! J >> >> >> >> Code: >> >> >> >> choosefunction <- function(){readline("1. linear, 2. linear with lag, 3. >> >> nonlinear ")} >> >> ans <-...

Hi, There is an object, ".Last.value" to which the result of the most recent evaluation is assigned. This is similar to "ans" in Matlab. In Matlab "ans" can be very useful and time-saving, but typing the larger R version is somewhat clunky and takes away from the usefulness. Is it possible to reassign '.Last.value' to something simpler, like 'ans' ? I'm aware of this thread: http:...

there is my code,  expect return value  is a data frame but R say it is list: SEXP Julia_R_MD_NA_DataFrame(jl_value_t* Var) {  SEXP ans,names,rownames;  char evalcmd[4096];  int i;  const char* dfname="DataFrameName0tmp";  jl_set_global(jl_main_module, jl_symbol(dfname), (jl_value_t*)Var);  //Get Frame cols   sprintf(evalcmd,"size(%s,2)",dfname);  jl_value_t* cols=jl_eval_string(evalcmd);  int collen=jl_unbox_lo...

Is there some fast (built-in?) way to get the OR of all the elements in a logical vector? In other words, is there some fast (built-in) version of the function vor below? Thanks. -Ben vor <- function(v) { ans <- v[1] if (length(v) > 1) for (i in 2:length(v)) ans <- ans | v[i] ans }

...o USA (where the sun will soon set into the mountains) ************************************************************************** *** arithmetic.c Mon Nov 10 19:35:01 1997 --- /net/aster/R/src/main/arithmetic.c Wed Apr 1 16:33:27 1998 *************** *** 428,433 **** --- 428,443 ---- else ans = allocVector(INTSXP, n); + /* copy attributes from longest argument */ + if (n1 > n2) + copyMostAttrib(s1, ans); + else if (n1 == n2) { + copyMostAttrib(s2, ans); + copyMostAttrib(s1, ans); + } + else + copyMostAttrib...