similar to: override ask = TRUE in plot.formula

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'

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)}

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

Full_Name: Corey A. Moffet Version: 1.8.1 OS: Windows 2000 and RHL 9 Submission from: (NULL) (199.133.140.156) The function model.tables.aov and perhaps other model.tables methods refer the user to use the function se.contrasts when the design is unbalanced. The function name is se.contrast (singular).

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

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

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

Is there a function/package that will allow you to test the hypothesis beta1 = x in a simple linear regression, where x is a constant? The AOV tests if beta1 in different from 0 but what if I want to know if beta1 one is different from 1? With best wishes and kind regards I am Sincerely, Corey A. Moffet Instructor Department of Range, Wildlife, and Fisheries Management Mail Stop 2125 Texas

I have a function for plotting soil texture that I am having a problem with. The function is: plot.psd <- function(sand, clay, ...) { conv.ter <- function(x, y) { x.con <- 100 - x - y*cos(1/3*pi) y.con <- y*sin(1/3*pi) data.frame(x=x.con, y=y.con) } plot(conv.ter(sand, clay), xlim = c(0,100), ylim = c(0,100*sin(1/3*pi)), axes = FALSE, xlab="",

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

Greetings List, I'm having some trouble with the use of the names function in a formula. Below is an example of something that works (i.e first line), and the second line is the same formula which doesn't. I want to be able to reference the column of the dataC table so I can run the variogram iteratively with a loop. > v<-variogram(A1~1,loc=~x+y, dataC) >

I'm sure this is a total noob question, but half an hour of searching bore no fruit: How do you select a subset of a vector by negative value? If I try > hist(sp$p[sp$r>0.01]) # all is well > hist(sp$p[sp$r<-0.01]) # this obviously causes issues Also, putting -0.01 in parentheses didn't help. Thanks in advance -- b

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

Folks, I've been running into an odd situation that occurs when I use length() function with aggregate(), but not with either one separately. Together, the results looks correct but is given an unexpected name. 'if (stringsAsFactors) factor(x) else x' instead of just 'x'. # Numbers work ok tt <- data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)

##data y1 <- matrix(c(3,1,0,1,0,1,1,0,0,0,1,0,0,0,1,1,0,1,1,1), nrow = 5, byrow = TRUE) y2 <- matrix(c (3,0,10,3,3,0,0,1,1,0,0,0,0,0,1,0,1,0,0,2,1,0,1,1,0,2,1,1,4,1), nrow = 5, byrow = TRUE) y1 <- as.data.frame(y1) y2 <- as.data.frame(y2) rownames(y1) <- rownames(y2) <- paste("site", 1:5, sep = "") colnames(y1) <-

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

Hi guys, I hope you can help me with this (probably) simple query: I have a data frame: -------------------------- a=c(1,1,1,1,1,1,2,2,2,2,2,2) b=c(1,1,1,2,3,4,1,1,2,2,3,4) c=c(400,200,300,100,500,300,200,100,500,400,200,100) data=data.frame(a=a,b=b,c=c) -------------------------- And I would like to get the following output: -------------------------- b a 1 2 3 4 1 900 100 500 300 2