similar to: Questions on dividing lists and tapply

Hello R community, I have been using R's inbuilt maximum likelihood functions, for the different methods (NR, BFGS, etc). I have figured out how to use all of them except the maxBHHH function. This one is different from the others as it requires an observation level gradient. I am using the following syntax: maxBHHH(logLik,grad=nuGradient,finalHessian="BHHH",start=prm,iterlim=2)

Hi, I have been using R for close to two years now and have grown quite comfortable with the language. I am presently trying to implement an optimization routine in R (Newton Rhapson). I have some R functions that calculate the gradient and hessian (pre requisite matrices) fairly efficiently. Now, I have to call this function iteratively until some convergance criterion is reached. I think the

Hi R community, I have a question I'm sure is very simple for most of you. I have a list, with each element being a matrix and the names of the elements are numbers (like 1,3,...). I can extract the matrices and the names individually. Now, I want to multiply each of the names to the individual list matrices (after converting to numbers of course). I could use a for loop, but the very

Hello R community, I have two questions: The first might be one of the silliest ever posted here and I apologize if I've missed some thing very obvious. It relates to using this digest. When I subscribed to the forum, I had chosen the "digest" option that bundles all mails every day into a single digest. Now, I posted a question a while ago on the logistic regression function and

Hello R community, I have a question about the logistic regression function. Specifically, when the predictor variable has not just 0's and 1's, but also fractional values (between zero and one). I get a warning when I use the "glm(formula = ... , family = binomial(link = "logit"))" which says: "In eval(expr, envir, enclos) : non-integer #successes in a binomial

Hello! I have a data frame with a factor and a numeric variable: x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200)) For each level of "factor" - I would like to divide each value of "values" by the mean of "values" that corresponds to the level of "factor" In other

hey I'd like to divide my data into four seasons. for this I made a function: Jahreszeit <- function(x) { if (x<=02 || x==12) {return("Winter") }else{ if (x>=03 && x<=05) {return("Fruehling") }else{ if (x>=06 && x<=08) {return("Sommer") }else{ if (x>=09 && x<=11) {return("Herbst") }}}}} Now, I have some

It would be cool if the default for tapply's init.value could be FUN(X[0]), so it would be 0 for FUN=sum or FUN=length, TRUE for FUN=all, -Inf for FUN=max, etc. But that would take time and would break code for which FUN did not work on length-0 objects. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jan 26, 2017 at 2:42 AM, Martin Maechler <maechler at stat.math.ethz.ch> wrote:

The following message is provided by Erik Please provide the reproducible code to do this. Generate a sample data set using the random data generating functions and show us what you'd like, we can then more easily help. ctu at bigred.unl.edu wrote: > Hi, > How about using "subset"? > x1<-tapply(subset(years, length(area)>20), function(x) length(unique(x))) >

Dear R-developers, when tapply() is invoked on factors that have empty levels, it returns NA. This behaviour is in accord with the tapply documentation, and is reasonable in many cases. However, when FUN is sum, it would also seem reasonable to return 0 instead of NA, because "the sum of an empty set is zero, by definition." I'd like to raise a discussion of the possibility of an

Dear R users, today I've got the following problem. Here you are a dataframe as example. There are some SAMPLES for which a CONCentration was recorded through TIME. The time during which the concentration was recorded is not always the same, 10 points for Sample A, 7 points for Sample B and 11 for sample C Also the initial concentration was not the same for the three samples. I would like

I have the following data.frame: index <- c("a","a","b","b","b") alpha <- c(1,2,3,4,5) beta <- c(2,3,4,5,6) table <-data.frame(index,alpha,beta) I'm now interested in getting means of alpha and beta for each of the index values and do a tapply() for each of the columns, e.g. means.alpha <- tapply(table$alpha, index,mean)

> On Jan 26, 2017 07:50, "William Dunlap via R-devel" <r-devel at r-project.org> > wrote: > It would be cool if the default for tapply's init.value could be > FUN(X[0]), so it would be 0 for FUN=sum or FUN=length, TRUE for > FUN=all, -Inf for FUN=max, etc. But that would take time and would > break code for which FUN did not work on

Is the following behaviour of tapply not disappointing? Problem with tapply occurs when dealing with na.rm when an argument additional to na.rm is sent to the applied function (here quantile). Any comment? Thank you, Philippe Lambert > x <- c(12,10,12,2,4,11,3,7,2,1,18,7,NA,NA,7,5) > fac <- gl(4,4,16) > # Works fine > tapply(x,fac,quantile,na.rm=T) $"1" 0% 25%

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What

I'm learning how to use tapply. Now I'm having a go at the following code in which dati contains almost 600 lines, Pot - numeric - are the capacities of power plants and SGruppo - text - the corresponding six technologies ("CCC", "CIC","TGC", "CSC","CPC", "TE"). .....................................................

Hei, i am trying to plot the means of two variables (d13C and d15N), by 2 grouping factors (Species and Year) that i obtained by the function tapply. I would like to plot with different colours according to the Year and show the "Species" as data labels. My data looks like this: Species d13C d13N Year "Species1" 14,4 11.5 2009 "Species2"

Consider the following: > set.seed(42) > ff <- factor(sample(c(1,3,5),42,TRUE),levels=1:5) > x <- runif(42) > tapply(x,ff,sum) 1 2 3 4 5 3.675436 NA 7.519675 NA 9.094210 I got bitten by those NAs in the result of tapply(). Effectively one is summing over the empty set, and consequently (according to what I learned as a child)

We were caught out recently attempting to use tapply to get a table of weighted means. This gives the wrong answer (or, more correctly, not the answer we were expecting), as the following example shows: R> x <- 1:10 #some data R> w <- c(1:5,5:1) #weights R> id <- rep(1:2,rep(5,2)) #id values R> weighted.mean(x[id==1],w[id==1]) #Weighted mean of x in group 1 [1] 3.666667

I'm sure I can put this together from the various 'apply's and split, but I wonder if anyone has a quick incantation: E.g. I can do tapply( data, groups, mean) but how can I do something like: tapply( list(data,weights), groups, weighted.mean ) ? (or: mapply is to sapply as ? is to tapply ) Thanks for your help. -- View this message in context: