similar to: Very simple question about list components

Hi all, I have the following code: set.seed(1) x1 <- matrix(sample(1:12), ncol=3) x2 <- matrix(sample(1:12), ncol=3) x3 <- matrix(sample(1:12), ncol=3) X <- list(x1,x2,x3) tt <- matrix(round(runif(5*4),2), ncol=4) Is there a way I can construct a new list where newlist[[i]] = tt[i,] %*% X[[i]] without using a for loop? Each element of newlist will be 3 x 1 vector. Thanks -- Tina

Dear R list, I have the following data: set.seed(1) n <- 5 # number of subjects tt <- 3 # number of repeated observation per subject numco <- 2 # number of covariates x <- matrix(round(rnorm(n*numco),2), ncol=numco) # the actual covariates x > x [,1] [,2] [1,] -0.63 -0.82 [2,] 0.18 0.49 [3,] -0.84 0.74 [4,] 1.60 0.58 [5,] 0.33 -0.31 I need to form a matrix

Hi! This is my first time posting. I've read the general rules and guidelines, but please bear with me if I make some fatal error in posting. Anyway, I have a continuous response and 29 predictors made up of continuous variables and nominal and ordinal categorical variables. I'd like to do lasso on these, but I get an error. The way I am using "lars" doesn't allow for the

I a using plink on a large SNP dataset with a .map and .ped file. I want to get some sort of file say a list of all the SNPs that plink is saying that I have. ANyideas on how to do this? -- Thanks, Jim. [[alternative HTML version deleted]]

Good afternoon, I am trying to create a plot where the bottom and top axes have the same scale but different tick marks. I tried user-defined xscale.component function but it does not produce desired results. Can anybody suggest where my use of xscale.component function is incorrect? For example, the code below tries to create a plot where horizontal axes limits are c(0,10), top axis has ticks

Hi there, sorry for troubling everybody once again, I've got a problem rewriting Sarkar's function for rewriting the tick locations in a logaritmic way (s. http://lmdvr.r-forge.r-project.org/code/Chapter08.R): His example works for log 2 but I need log e (natural logarithm). My problem is that if I replace 2 with "e" (using paste()), I get the error message that the location

Dear R-helpers, let us assume, that I have the following dataset: a <- rnbinom(200, 1, 0.5) b <- (1:200) c <- (30:229) d <- rep(c("q", "r", "s", "t"), rep(50,4)) data_frame <- data.frame(a,b,c,d) In a first step I run a glm.nb (full code is given at the end of this mail) and want to predict my response variable a. In a second step, I would

I think 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,

Hi, I'm trying to build dendrograms to pass to heatmap(). The dendrograms I build plot properly, but when I pass them to heatmap() I get the error message "row dendrogram ordering gave index of wrong length" (see output log below). I looked in the code of heatmap() and saw that the error was due to a NULL return value from order.dendrogram(), which in turn got a NULL return value

Hi, I'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 <-

<<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

I'm trying to convert an S-Plus program to R.  Since I'm a SAS programmer I'm not facile is either S-Plus or R, so I need some help.  All I did was convert the underscores in S-Plus to the assignment operator <-.  Here 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),

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:

> Thomas Lumley wrote: > > On Tue, 31 May 2005, Duncan Murdoch wrote: > > > > > >>M??kinen Jussi wrote: > >> > >>>Dear All, > >>> > >>>I luckily found the following feature (or problem) when tried to > >>>apply > >>>ifelse-function to an ordered data. > >>> > >>> >

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

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*

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;

This is my reproducible example tv.ms<-structure(list(inq = structure(4:17, .Label = c("D4", "D5", "D6a", "D6b", "D6c", "D7", "D8", "F4", "F5a", "F5b", "F6a", "F6b", "F6c", "F6d", "F7a", "F7b", "F8"), class =

Hi, I would like to use sweep to "sweep out" proportions and confidence intervals for an array, however when I supply a function which returns a string (containing something like "9% (3-18%)") I get back a list instead of an array, here is a simplified example: # example showing that sweep does not return an array with same dimensions as STATS as advertised

Hi, I am trying to fit a model with lmer in R and proc glimmix in SAS. I have simplified my code but I am surprised to see I get different results from the two softwares. My R code is : lmer(y~age_cat + (1|cat),data=fic,family=binomial(link = "logit"), NaGQ=1) My SAS code is : ods output Glimmix.Glimmix.ParameterEstimates=t_estimates; proc glimmix data=tab_psi method=laplace;