similar to: No years() function?

Hello, I've had no luck finding an R package that has the ability to estimate a Tobit model allowing for heteroskedasticity (multiplicative, for example). Am I missing something in survReg? Is there another package that I'm unaware of? Is there an add-on package that will test for heteroskedasticity? Thanks for your help. Cheers, Alan Spearot -- Alan Spearot Department of Economics

Hello, I would like to inlude the Q-Q plot by "qqmath" into a panel with other plots, say, using par(mfrow=c(1,2)). How can this be done given that "qqmath" refreshes the plotting window and there seems to be no series coming out of it? Thanks Serguei [[alternative HTML version deleted]]

Hi there, I'm writing a function that calculates the probability of different outcomes of dice rolls (e.g., the sum of the highest three rolls of five six-sided dice). I'm using the "combinations" function from the "gtools" package, which is great: it gives me a matrix with all of the possible combinations (with repetitions allowed). Now I want to count the number

Hello, I have a small statistics question, and as I'm quite new to statistics and R, I'm not sure if I'm doing things correctly. I am looking at two quantitative variables (x,y) that are correlated. When I divide the data set according to a categorical variable z, then x and y are more poorly correlated when z = A than when z = B (see attached figure). In fact x and y are two

Hello, I am trying to get the P values from the output of a summary for lm. lm <- lm(y ~ age + sex) s <- summary(lm) I thought that I might be able to get them using a combination of scan, grep and sub. But I got stuck on the first step - being able to process "s" as a text string. I could perhaps write it to file than scan it back but there is probably an easier way to do

Hi, Since "R" is a (very) generic name, I've been having some trouble searching the web for this topic. Due to this, I've just created a Google Custom Search Engine that includes several of the most relevant sites that have information on R. See it in action at: http://google.com/coop/cse?cx=018133866098353049407%3Aozv9awtetwy This is really a preliminary test. Feel free to

Hi, I'm having a weird result with the length() function: >a [... omited ...] [9994] NA "2003-12-03 16:37:00" "2002-06-26 18:43:00" [9997] "2005-07-04 04:00:00" "2007-02-16 22:09:00" "2007-02-24 15:49:00" [10000] NA > length(LastModified) [1] 9 > length(c(LastModified)) [1] 9 I was expecting to get

Hi, I have several files with data in this format: 20070102 20070102 20070106 20070201 ... The data is sorted and each line represents a date (YYYYMMDD). I would like to analyze this data using R. For instance, I would like to have a histogram by year, month or day. I've already made a simple Perl script that aggregates this data but I believe that R can be much more powerful and easy on

Just for the record, here are my steps for producing a date based histogram. Data is stored in a file where each line only has a date - 2007/02/16 >d<-readLines("filename.dat") >d<-as.Date(d, format="%Y/%m/%d") >pdf(yearly.pdf) >hist(d, "years") >dev.off() Instead of "years" you can also use "days", "weeks",

Hi, What's the best way to average dates? I though mean.POISXct would work fine but... > a [1] "2007-04-02 19:22:00 WEST" > b [1] "2007-03-17 16:23:00 WET" > class(a) [1] "POSIXt" "POSIXct" > class(b) [1] "POSIXt" "POSIXct" > mean(a,b) [1] "2007-04-02 19:22:00 WEST" > mean(b,a) [1] "2007-03-17

Hi, I'm trying to draw a 2D plot using multiple tints of red. The (simplified) setup is the following: || year | x | y || My idea is that each year is plotted with a different tint of red. Older year (lightest) -> Later year (darkest). I've managed to plot this with different scales of grays simply by doing: palette(gray(length(years):0/length(years))) before the plot and for each

Hi, I'm looking for a function to measure the dispersion of a set of values ranging from 0 to 1. This function should be 0 if all the values are evenly spaced within the interval and it should be > 0 if values are clustered. The more clustered the values are, the higher should the function be. An example: [0; 0.2; 0.4; 0.6; 0.8; 1] - function should be ~ 0 [0; 0.1; 0.1; 0.15; 1] -

Hi out there, I am trying to fit a species accumulation curve (increase in number of species known vs. sampling effort) for multiple regions and several bootstrap samples. The bootstrap samples represent different arrangements of the actual sample sequence. I fitted a series of nlme-models and everything seems OK, but since the observations are correlated I tried to include some correlation

I have an interesting lme - problem. The data is part of the Master Thesis of my friend, and she had some problems analysing this data, until one of her Jurors proposed to use linear mixed-effect models. I'm trying to help her since she has no experience with R. I'm very used to R but have very few experience with lme. The group calls of one species of parrot were recorded at many

Hi everyone, I'm interested in the analysis of spatial data, and I'm trying out several R-packages. Today I was attempting to use the package vardiag (version 0.1): >> library(vardiag) >> rs4.vo <- varobj(rs4[,2:4],trace=2) >[1] 1 >Error: couldn't find function "nlregb" so far I know "nlregb" is a S-plus function for optimization, so this

Hi, I'm pretty new to R and I've been having some problems filtering data in matrices. I have the following initial dataset: || year | name | varA || I have multiple values for "varA" for the same "year" and the same "name". Having this as the input I would like to obtain the following: || year | name | {varA mean} || Where I only have one line for each

Hi, I'm pretty new to R, I've just installed version 2.1.1 on Windows. I have a simple (it seems) doubt - how do I change the Console default Language ? It's set to Portuguese but I would like to view it in English. Thanks in advance, S??rgio Nunes

On converting character variables to ordered factors, regression result has strange names. Is it possible to obtain same variable names with and without intercept? Thanks, Naresh mydf <- data.frame(date = seq.Date(as.Date("2024-01-01"), as.Date("2024-03-31"), by = 1)) mydf[, "wday"] <- weekdays(mydf$date, abbreviate = TRUE) mydf.work <- subset(mydf, !(wday

Hi, It seems that hist() has a buggy behavior when breaking over "days". The bug can be reproduced in a few steps: > d=data.frame(date=c("2009-01-01", "2009-01-02", "2009-01-02")) > d$date=as.Date(d$date) > d$date [1] "2009-01-01" "2009-01-02" "2009-01-02" > h=hist(d$date, "days") > h$count [1] 3

I have the following xts objetct "temp" > str(temp) An ?xts? object from 2010-12-26 to 2011-03-05 containing: Data: num [1:70, 1] 2.95 0.852 -0.139 1.347 2.485 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr "t_n" Indexed by objects of class: [POSIXct,POSIXt] TZ: GMT xts Attributes: NULL > temp t_n 2010-12-26