similar to: calculating time interval distributions

I am still fairly new to R and have a fairly rudimentary question. I am trying to name a vector of coefficients retrieved from a multilevel model using the coef function. I guess the default name is "Intercept" and I cannot figure out how to rename it. I have tried the using the code below to name the column of coefficients ind.y derived from an lme model. Unfortunately, the

Hello All, Once again thanks for all of the help to date. I am climbing my R learning curve. I've got a few more questions that I hope I can get some guidance on though. I am not sure whether the etiquette is to break up multiple questions or not but I'll keep them together here for now as it may help put the questions in context despite the fact that the post may get a little long.

Hi Again, Thanks very much for your response. It seems my example got rearranged (transposed?) after I posted it. Hopefully this example will be more clear. I have one file (ex. sheet 1) that will have a column for individuals (ind) and a column for the date (date). I would like to merge this with another file (ex. sheet 2) that has both the 'ind' and date column as well as associated

Hi, I have a correlation matrix of about 3000 items, i.e., a 3000*3000 matrix. For each of the 3000 items, I want to get the top 50 items that have the highest correlation with it (excluding itself) and generate a data frame with 3 columns like ("ID", "ID2", "cor"), where ID is those 3000 items each repeat 50 times, and ID2 is the top 50 correlated items with ID,

Dear R-Help, I have an array 1260x1260, upper triangle consisting of numbers between 0 and 100, and lower triangle all NA. I can extract the index of those values say above 99 using the following code: which(myArray>=99 , ind.arr=T) which returns: row col 5475 252 253 45423 764 765 46902 777 778 34146 611 962 50681 1220 1221 Now I would like to if poss print the actual value

I have a dataframe like this (toy example): x y z "a" "a" 0 "a" "b" 1 "a" "c" 2 "b" "a" .9 "b" "b" 0 "b" "c" 1.3 "c" "a" 2.2 "c" "b" 1.1 "c" "c" 0 The observations are from a matrix like this: c 2.2 1.1 0.0 b 0.9 0.0

Dear List, I am having trouble with a tricky merging task. I have one data sheet that has dates (continuous) that radio collared individuals were monitored via telemetry. I have a different sheet containing data from instances where individuals were recaptured and associated body condition data was recorded (sheet 2). I would like to merge the two sheets by individual and date (I can do this

Hello, I am trying to plot time-series data with certain weeks highlighted using symbols. require(ggplot2) #plotting time series data timescale <- seq(as.Date("01/01/09","%m/%d/%y"), length.out=12, by=7) data.all <- data.frame( id = c(rep('111',12),rep('222',12),rep('333',12)), week=c(timescale,timescale,timescale),

Hello all and thanks in advance for any advice. I am very new to R and have searched my question but have not come up with anything quite like what I would like to do. My problem is: I have a data set for individuals (rows) and values for behaviours (columns). I would like to know the proportion of shared behaviours for all possible pairs of individuals. The sum of shared behaviours divided by

Hello list I have a problem with a dataset (see toy example below) where I am trying to find the difference between two (or more numbers) and discard those observations which fall outside a set interval. An example and further explanation: values ind 1 2655 7A5 2 3028 7A5 3 689 ABBA-1 4 1336 ABBA-1 5 1560 ABBA-1 6 2820 ABLIM1 7 3339 ABLIM1 8

Dear all, I am trying to use R to fit mixed models. Take the following example, where ind is a random effect and sample is fixed. I wanted to fit Model 1: values = ind + sample Model 2: values =ind * sample Model 3: values=ind(sample) + sample Tried to use the below for mod1, but it did not work. Can anyone give some help on this. Thanks so much. mod1 <- lme(values ~ sample +

Hi all I am trying to solve a combinatorial optimization problem. Basically, I can reduce my problem into the next problem: 1.- Given a NxN grid of points, with some values in each cell 2.- Find the combination of K points on the grid such that, the maximum mean value is obtained I took the Travel SalesMan problem example in ?optim documentation. I am not sure if I have understood correctly

Hello List I have just written a little function that takes two matrices as arguments and returns a large matrix that is composed of the two input matrices in upper-left position and lower-right position with a padding value everywhere else. (function definition and toy example below). I need nonsquare matrices and rowname() and colname() inherited appropriately. Two questions: (1) Is there a

Dear friends, Being very new to this, I was wondering if I could get some pointers and guidance to interpreting the results of performing a linear regression with ordinal predictors in R. Here is a simple, toy example: y <- c(-0.11, -0.49, -1.10, 0.08, 0.31, -1.21, -0.05, -0.40, -0.01, -0.12, 0.55, 1.34, 1.00, -0.31, -0.73, -1.68, 0.38, 1.22, -1.11, -0.20) x <-

I am trying to define subset operator for a reference class and hitting some problem i am unable to diagnose.To give an example, here is a toy class generator that is a wrapper around a list tmpGEN<-setRefClass("TMP", fields=list( namelist="list" )) tmpGEN$methods('add'=function(obj, name){ namelist[[name]]<<-obj })

Hello all and thanks in advance for any help or direction. I have co-authorship data that looks like: Paper Author Year 1 SmithKK JonesSD 2008 2 WallaceAR DarwinCA 1999 3 HawkingS 2003 I would like: Paper Author Year 1 SmithKK 2008 1 JonesSD 2008 2

Hi, Can someone please tell me how to change the column name of a specific column. How do I change the name of the column 'Species'? Thanks in advance d <- iris colnames(d) [1] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width" "Species" ind <- which(names(d)=='Species') ind [1] 5 colnames(d[ind]) [1]

Dear R users; Consider the following toy example: a <- matrix(c(2,3,4,NA,NA,5,8,NA,8,NA), 5, 2) b <- cbind(a,apply(a, 1, diff, na.rm = TRUE)) What I would like be able to get is: c <- matrix(c(2,3,4,NA,NA,5,8,NA,8,NA,3,5,-4,8,NA), 5, 3) i.e., for each row if both values (column 1 and 2) are NA then the difference must return NA, but if any of those two values is different from NA

Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time <- cumsum(rexp(1000)/10) status <- rbinom(1000, 1, 0.5) ## kaplan meier estimates fit <- survfit(Surv(time,

Perhaps this toy example will help: ## example data output <- list(1:5, 1:7, 1:4) lens <- lapply(output, length) maxlen <- max(unlist(lens)) outputmod <- lapply(output, function(x, maxl) c(x, rep(NA, maxl-length(x))), maxl=maxlen) outputmat <- do.call(cbind, outputmod) write.csv(outputmat, na='') The idea is to pad the shorter vectors with NA (missing) before converting