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...e next until I am left with only the plot of the eighth matrix at the end of the script. Thanks in advance for your help! Amy Paternostro National Center for Environmental Economiccs United States Environmental Protection Agency 1200 Pennsylvania Avenue, NW Washington, DC 20460 Paternostro.Amy at epamail.epa.gov

I don't understand this behavior: > a <- c(0, 1, 2, 3) > b <- c(1, 2, 3, 4) > ifelse (a == 0, 0, b[a]) [1] 0 2 3 1 rather than the desired 0 1 2 3. Thanks for any explanation.

Hi All, I want to ask if there is a transpose function for data frame like the procedure of transpose in SAS? Because I want to partially transpose a data frame which contains 5 columns (siteid, date, time, obs, mod), what I want to do is to put time as the column variables along with siteid, and date, and put obs and mod in the row names. specifically to transpose a data frame: siteid date

I'm having trouble figuring out how to create a function using "body<-" (). The help file for body() says that the argument should be a list of R expressions. However if I try that I get an error: > tmpfun <- function(a, b=2){} > body(tmpfun) <- list(expression(z <- a + b),expression(z^2)) Error in as.function.default(c(formals(f), value), envir) :

SAS proc rank has ties options of high and low that would allow producing ranks of the type found in the sports pages, e.g., rank (c(1,1,2,2,2,2,3)) == 1 1 3 3 3 3 7 Could R support these ties.methods?

I wanted a data frame component to be named "next", for example: > m <- data.frame (matrix (0, nrow=2, ncol=2)) > names (m) <- c("prev", "next") > m prev next 1 0 0 2 0 0 But "next" being reserved prevents $ indexing without quotes: > m$next Error: syntax error > m$"next" [1] 0 0 Although they are mostly

...atmanual zip error: Nothing to do! (try: zip -r9X C:/R/rw1041/src/gnuwin32/statmanual_1.0.zip . -i statmanual) -- Thomas M. Kincaid, Ph.D. Environmental Statistician Dynamac Corporation 200 SW 35th Street Corvallis, OR 97333 Phone: (541) 754-4479 Fax: (541) 754-4716 E-mail: Kincaid.Tom at epamail.epa.gov -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-reques...

Greetings, I have a meta-analysis problem in which I have fixed effects regression coefficients (and estimated standard errors) from identical models fit to different data sets. I would like to use these results to create pooled estimated regression coefficients and estimated standard errors for these pooled coefficients. In particular, I would like to estimate the model \beta_{i} = \mu +

Hello, I would like to write a loop to 1) run 100 linear regressions, and 2) compile the slopes of all regression into one vector. Sample input data are: y1<-rnorm(100, mean=0.01, sd=0.001) y2<-rnorm(100, mean=0.1, sd=0.01) x<-(c(10,400)) #I have gotten this far with the loop for (i in 1:100) { #create the linear model for each data set model1<-lm(c(y1[i],y2[i])~x)

I know R has a steep learning curve, but from where I stand the slope looks like a sheer cliff. I'm pawing through the available docs and have come across examples which come close to what I want but are proving difficult for me to modify for my use. Calculating simple group means is fairly straight forward: data(PlantGrowth) attach(PlantGrowth) stack(mean(unstack(PlantGrowth)))

Sam Rushing's Python extension 'calldll' will supposedly give me access to any DLL, and presumably R.DLL in particular, from Python. I have no experience manipulating DLLs as yet. I am learning Python. I can't find any simple step-by-step instructions on how to get done what I want to do. Does anyone have experience with this? Is there a better way? Of course, one can use

Folks, I've run into trouble while writing functions that I hope will create and modify a dataframe or two. To that end I've written a toy function that simply sets a couple of variables (well, tries but fails). Searching the archives, Thomas Lumley recently explained the <<- operator, showing that it was necessary for x and y to exist prior to the function call, but I haven't

Dear Experts, In glim() of S Plus, one of the returned values is "var", the estimated variance matrix of coefficients. However, in glm() of R (there is no glim() in R), "var" is not one of the returned values. Anyone know what could I get the varience matrix of coefficients in glm() in R? As a novice in R and S+, I'd appreciate your help Sincerely, Charlie Liu

Is the difference in behavior below, introduced in 1.8, inconsistent or, at least, undesirable? I couldn't find this in the NEWS. On the one hand, > a <- matrix (1:4, nrow=2) > a <- data.frame (a) > names (a) <- c("break","next") > names (a) [1] "break" "next" On the other, > a <- matrix (1:4, nrow=2) > dimnames(a)

When I run Rcmd check on a package on my Windows 2000 machine, I get a series of error messages like the following: * checking generic/method consistency ...c:\DOCUME~1\R5018~1.WOO\LOCALS~1\Temp/R utils138414013: cannot open c:DOCUME~1R5018~1.WOOLOCALS~1Temp/Rin138408157: no s uch file It looks as if a Windows style path to the temp directory is not being interpreted correctly, with backslashes

Hello, I am having a problem using sqldf. I'm trying to choose a subset of observations from a data set based on the date and maximum value of a variable by date. Here is the code I am using: test<-sqldf("select distinct * from bextuse group by sdate having bext=max(bext)",method="raw"); The result I get back is a data frame with 0 rows and 0 columns. I have tried

Dear Experts, I need to calculate the cdf of the standard normal distribution, i.e. H(x) = 1/sqrt(2*pi) integral(exp(-z^2/2) dz), where z is b/w -infi to infi. I know there should be a way to do it in R, but did not know to do it. I'd appreciate any help you could offer. Charlie Liu Graduate student intern at EPA/ECO

I found that predict.gnls failed with a wierd error message about a non-numeric argument to a binary vector in one of three nearly identical uses. Error in Inh/Ki : non-numeric argument to binary operator (Inh and Ki are arguments to the function used in the formula for the object whose predictions were requested). It turns out that the problem is in getCovariateFormula(). The final line in

Hello, I haven't been able to find an example for the second case below -- or perhaps I didn't recognize it when I saw it. Is there a value for x such that svalue(x) will return "bbb", either by itself or as part of an array? Or do I need to do something else entirely? (R2.5.1; Windows XP) > #### gWidgets test > options("guiToolkit"="tcltk") >

I am trying to use POSIX datetime objects rather than chron datetime objects but am having difficulty with POSIX in a time series. My question: Once a POSIXct vector is bound to a time series, is there a function to convert back to POSIXct? The following code demonstrates what I am trying to do. > ts(as.POSIXct(strptime(tmp,"%m/%d/%Y %H:%M:%S")),freq=1440) Time Series: Start =