similar to: extracting index list when using tapply()

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))) >

Hello, I have a question regarding estimation of the dispersion parameter (theta) for generalized linear models with the negative binomial error structure. As I understand, there are two main methods to fit glm's using the nb error structure in R: glm.nb() or glm() with the negative.binomial(theta) family. Both functions are implemented through the MASS library. Fitting the model using these

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"). .....................................................

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

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:

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"

Hello, I am using tapply to pull out data by the day of week and then perform functions (e.g. mean). I would like to have the number of values used for the calcuation for the functions, sorted by each day of week. A number of entries in any given column are NAs. I have tried the following code and simple variants with no luck. for (i in 1:length(a[1,])){ x<-tapply(a[,i],a[,1],mean,

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)

Hi all, the following problem is still beyond my R-knowledge: I have one data vector containing the signal from 4 channels that are measured subsequently and in repeating cycles (with one factor vector for cycle and one for channel identification). To extract the mean of each channel during each cycle tapply is the method of choice. However, I cannot use the whole measuring period for each

Hello, I'm trying to use tapply to find group means in a function. It works outside of a function, but I get the error message from the following code: "Error in tapply(index, cluster, mean) : arguments must have same length." Any suggestions? Thanks. eric d <- data.frame(cbind(cluster=1:2, value1=1:10, value2=11:20)) d FindClusterTraits <- function(framename, index){

The "no factor combination" case is distinguishable by 'tapply' with simplify=FALSE. > D2 <- data.frame(n = gl(3,4), L = gl(6,2, labels=LETTERS[1:6]), N=3) > D2 <- D2[-c(1,5), ] > DN <- D2; DN[1,"N"] <- NA > with(DN, tapply(N, list(n,L), FUN=sum, simplify=FALSE)) A B C D E F 1 NA 6 NULL NULL NULL NULL 2 NULL NULL 3 6

Full_Name: George Leigh Version: 1.8.1 OS: Windows 2000 Submission from: (NULL) (203.25.1.208) The following example gives the correct answer when the first argument of tapply is a numeric vector, but an incorrect answer when it is a factor. If the function used by tapply is "length", the type and contents of the first argument should make no difference, provided it has the same

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)

Hello R community, I have two questions about using R. The first is about dividing each element of a list with another similar sized list. So, if the first list has two elements and so does the second, then the result should also be a list with two elements. For example, the inputs are: list(matrix(1:6,ncol=2),matrix(1:6,ncol=2))->l1 l2<-list(1:3,2) I want to get a list, l3 with the

i use tapply and by often, but i always end up banging my head against the wall with the output. is there a simpler way to convert the output of the following tapply to a dataframe or matrix than what i have here: # setup data for tapply dt = data.frame(bucket=rep(1:4,25),val=rnorm(100)) fn = function(x) { ret = c(unname(quantile(x,probs=seq(.25,.75,.25),na.rm=T)),mean(x,na.rm=T)) } a =

Dear Everyone, I try to create a cvs-file with different results form the table function. Imagine a data-frame with two vectors a and b where b is of the class factor. I use the tapply function to count a for the different values of b. tapply(a,b,table) and I use the table function to have a look of the frequencies as a total table(a) I would like to put both results together in one txt or

Hi folk, I tried this and it works just perfectly tapply(iris[,1],iris[5],mean) but, how to obtain a single table from multiple variables? In tapply x is an atomic object so this code doesn't work tapply(iris[,1:4],iris[5],mean) Thanx and great summer holidays Gianandrea -- View this message in context: http://www.nabble.com/multiple-tapply-tp18868063p18868063.html Sent from the R help

Hi all, Apologies if this has been raised before ... R's tapply is very fast, but if X has names in this example, there seems to be a huge slow down: under 1 second compared to 151 seconds. The following timings are repeatable and are timed properly on a single user machine : > X = 1:100000 > names(X) = X > system.time(fast<<-tapply(as.vector(X), rep(1:10000,each=10), mean))

Full_Name: Peter Gedeck Version: R1.6.2 and R1.7.0 OS: Windows XP Submission from: (NULL) (194.191.169.72) Hello, I marked the bug report Private, as I don't want my email address on the web server. The problem that I found is best explained using an example. index <- 1:6 cluster <- c(1,1,1,2,2,3) tapply(index,cluster,sample) gives $"1" [1] 2 1 3 $"2" [1] 4 5

Hello I have a huge data frame with three columns 'Roof' 'Month' and 'Temp' i want to run analyses on the numerical Temp data by the factors Roof and Month, separately and together. For using more than one factor i understand i should use aggregate, but i am struggling with the tapply for single factor analysis. > tapply(Temp, INDEX = Roof, FUN = median) This works