similar to: conditional less than

Dear R-help: Does anyone know of a package in R that will do Welch's modified F-test for heterogeneous error variances? Are there other statistical techniques available in R that test the equality of means when homoscedastisity is violated? 't.test' does this in the pairwise sense when var.equal = TRUE. With best wishes and kind regards I am Sincerely, Corey A. Moffet Support

Dear R-Help, Does anyone know of a way to use CGIwithR in Windows? David Firth the author of CGIwithR states "It ought to be possible to adapt it for use also in conjunction with web servers running under Windows NT ... but the author has no plans for that." I have developed some applications in R that I would like to make available through a web interface using CGIwithR but at this

The funciton c() works differently for strings and for factors: For strings: > l = c('a', 'b') > l [1] "a" "b" For factors: > l = c(factor('a'), factor('b')) > l [1] 1 1 What should be the right technique for merging factors? -- Svetlana Eden Biostatistician II School of Medicine

Hello This is a very newbie question on R syntax, but I do not find the answer.... I want to make a selection on an interval say choose Xint in the interval of temperatures 390-399 I tried this syntax Xint<- X[t>=390 && t< 400] typing >XintI get the answer numeric(0) it did not select any object! 'though I verified that there indeed are occrencies of X in this

Is there a function in R, which would return index of maximum value in a vector ? e.g. > v <- round(10*rnorm(8)) > v [1] 6 -3 -6 15 7 9 0 -19 > max(v) [1] 15 ??? index.max(v) ??? 4

Hi I wrote a function implementing an algoritm for simulation of spatial sampling plans (using geoR). I'm using a range of parameters to perform several simulation and sometimes this parameters create function errors wich cause the algoritm to stop. How can I avoid this ? Below there's a small example that I made to study the problem but I was not able to solve it. I've tried

Hi, Does anyone know how to manually configure the number of digits printed out from xtable()? For example, I'm exporting a data frame through xtable() into a LaTeX table, I only have two columns in the data frame so by default I only get two decimal places. But I'd like at least 5 decimal places. I had a look at ?xtable() but can't seem to find an example. Cheers, Kevin

Probably a stupid question, but I don't seem to be able to find the answer I'm looking for from any of the R literature. Basically I have a matrix with several thousand rows and 20 columns(weather stations) of wind direction data. I am wanting to extract a matrix which contains data for all columns conditional on column 20 having a value of _either_ less than 45 or greater than 315. (ie I

Hi everyone, How might I go about configuring R to keep add-on packages loaded from session to session? Is this undesireable for some reason? At present, I keep a file called "pckgs.txt" in my working directory with, e.g., library(gnlm) library(Hmisc) ... and then type source("add-ons.txt") every time I start a new session. I suspect there's a more elegant way to hold

I am having difficulty with plot.formula. For example: ------------------- dat <- data.frame(y = c(1,1,1,2,2,2,3,3,3,4,4,4), x1 = as.ordered(c(1,1,1,1,1,1,2,2,2,2,2,2)), x2 = as.ordered(c(1,1,1,2,2,2,1,1,1,2,2,2))) plot(y ~ x1 + x2, data = dat) # Works fine "interactively" asking: # "Hit <Return> to see next plot:" #

Hello, I have to add "Age (bar(x)=14.3) as a title on a chart. I am unable to get this to working. I have tried bquote, substitute and expression, but they are only doing a part of the job. new<- c(14.3, 18.5, 18.1, 17.7, 18, 15.9, 19.6, 17.3, 17.8, 17.5, 15.4, 16.3, 15, 17.1, 17.1, 16.4, 15.2, 16.7, 16.7, 16.9, 14.5, 16.6, 15.8, 15.2, 16.2, 15.6, 15, 17.1, 16.7, 15.6, 15, 15.8, 16.8,

Hello, I'm a newcomer to R so please forgive me if this is a silly question. It's that I have a linear regression: fm <- lm (x ~ y) and I want to test whether the slope of the regression is significantly less than 1. How can I do this in R? I'm also interested in comparing the slopes of two regressions: fm1 <- lm (x ~ y) fm2 <- lm (a ~ b) and asking if the slope of fm1 is

I know that there must be a cool way of doing this, but I can't think of it. Let's say I have an dataframe with NA's. > x <- data.frame(a = c(0,1,2,NA), b = c(0,NA,1,2), c = c(NA, 0, 1, 2)) > x a b c 1 0 0 NA 2 1 NA 0 3 2 1 1 4 NA 2 2 > I know it is easy to replace all the NA's with zeroes. > x[is.na(x)] <- 0 > x a b c 1 0 0 0 2 1 0 0 3 2 1

Dear R-help, Does any one know of a function/package in R that will test the assumption of spericity in a split-plot analysis of variance? How are people dealing with this issue in R? With best wishes and kind regards I am Sincerely, Corey A. Moffet Support Scientist University of Idaho Northwest Watershed Research Center 800 Park Blvd, Plaza IV, Suite 105 Boise, ID 83712-7716 (208) 422-0718

Dear R-Help, I have a generated file that looks like the following: ----- Begin file ----- # # Output File # float Version 2002.700000000000 int Numdays 31 int NumOFEs 1 # # Hillslope-specific variables # char HillVarNames[ 3 ] {Days In Simulation} {Hillslope: Precipitation (mm)}

I am trying to reproduce an analysis done in SAS from which Type III Sums of squares are used. How can you get Type III sums of squares in an aov similar to what SAS provides? With best wishes and kind regards I am Sincerely, Corey A. Moffet Instructor Department of Range, Wildlife, and Fisheries Management Mail Stop 2125 Texas Tech University Lubbock, Texas 79409-2125 (806) 742-2843 Voice

Hello all, I am trying to solve a problem, and my solution is rather ugly and not very general. The posts for "[R] help with gsub and grep functions" seemed relevent and gave me hope for a more refined and more general solution. The Problem: line <- "'this text has spaces' 'thisNot' 3 4 5 6 7 8 9 10" bad.line <- "'this text has spaces'

Hello, I have problem running the histogram function "hist". The area under the histogram is much lower than 1. Could anyone tell me what the problem is? Thanks, (The total number of observation is 992 (close to 1000), so the probability that 0<Y1<35 is approximately 0.277) miao rm(list=ls()) par(mfrow=c(1, 1)) Y <- cbind(matrix(35*0.5,1,277), matrix(35*1.5, 1, 146),

I have some data that was created for import into a Lotus 1-2-3 spreadsheet and on of the columns is time. The time is akin to Julian were the value 1 is mapped "01-Jan-00 12:00:00 AM" in Lotus 1-2-3. Is there a function in an R package that can convert this numeric vector to a POSIXct vector? With best wishes and kind regards I am Sincerely, Corey A. Moffet Instructor Department

Hi, I have a large amount of data that I would like to create a histogram of and plot and do things with in R. It is pretty much impossible to read the data into R, so I have written a program to bin the data and now have a list of counts in each bin. Is it possible to somehow import this into R and use hist(), so I can, for instance, plot the probability density? I have looked at the help page