similar to: Trouble receiving messages from the mailing list

Why does R CMD INSTALL work for some packages (e.g., lme4) but not others (e.g., nlme)? Thanks, Gang

I want to create a list based on the information from a data.frame, Model. So I tried the following: MyList <- list(colnames(Model)[2] = levels(Model$(colnames(Model)[2]))) but it failed with an error: Error: unexpected '=' in "list(colnames(Model)[2] =" I have the following problems with this command line: (1) I wanted to use colnames(Model)[2] as a tag for the list:

Hi experts, I just started to learn R today, and tried to work with an add-on package sem. I have a version of 2.3.1 on MacOS X 10.4.6 with sem put under /Library/Frameworks/R.framework/Versions/2.3/Resources/library However when I typed library(sem) the following error showed up: Error in library(sem) : 'sem' is not a valid package -- installed < 2.0.0? Why is this? Thank

I want to run deconvolution of a time series by an impulse or point-spread function through Wiener filter, regularized filter, Lucy-Richardson method, or any other approaches. I searched the CRAN website and the mailing list archive, but could not find any package for such a deconvolution analysis. Does anybody know an existing R function for deconvolution? TIA, Gang

Suppose I have a 4-D array X with dimensions (dx, dy, dz, dp). I want to collapse the first 3 dimensions of X to make a 2-D array Y with dimensions (dx*dy*dz, dp). Instead of awkward looping, what is a good way to do this? Is there a similar function like reshape in Matlab? Thanks, Gang

This must be very simple, but I'm stuck. I have a command line in R defined as a variable of a string of characters. How can I convert the variable so that I can execute it in R? Really appreciate any help, Gang

I do not think anyone has answered this. > I'm trying to run a simple one-way ANCOVA with the lmer > function in R package lme4, but have encountered some > conceptual problem. The data file MyData.txt is like this: > > Group Subj Cov Resp > A 1 3.90 4.05 > A 2 4.05 4.25 > A 3 4.25 3.60 > A 4 3.60 4.20 > A 5 4.20 4.05 > A 6 4.05 3.85

Suppose I have a mixed-effects model where yij is the jth sample for the ith subject: yij= beta0 + beta1(age) + beta2(age^2) + beta3(age^3) + beta4(IQ) + beta5(IQ^2) + beta6(age*IQ) + beta7(age^2*IQ) + beta8(age^3 *IQ) +random intercepti + eij In R how can I get an F test against the null hypothesis of beta6=beta7=beta8=0? In SAS I can run something like contrast age*IQ 1,

I'm trying to use the following loop to open a window multiple times to select files, but only the last window shows up. What am I missing? library(tcltk) nWin <- 6 fn <- vector('list', nWin) for (ii in nWin) { fn[[ii]] <- tclvalue( tkgetOpenFile( filetypes = "{{Files} {.1D}} {{All files} *}", title = paste('Choose number', ii, 'time

Dear r-help mailing list, this seems stupid, but I actually don't find the solution: if I have a vector of numbers x of length n, ranging, say, from -3 to 4, if I do barplot (x) all the values below 0 go downwards, and all the positive values go upward. How can I make them all begin from the minimum pointing upwards? Thanks! Gabriele Zoppoli, MD Ph.D. Fellow, Experimental and Clinical

I want to run a R program, prog.R, interactively. My question is, is there a way I can start prog.R on the shell terminal when invoking R, instead of using source() inside R? TIA, Gang

I can run one-sample t-test on an array, for example a matrix myData1, with the following apply(myData1, 2, t.test) Is there a similar fashion using apply() or something else to run 2-sample t-test with datasets from two groups, myData1 and myData2, without looping? TIA, Gang

Sorry for this dumb question. Suppose I have a named array ww defined as ww <- 1:5 names(ww) <- c("a", "b", "c", "d", "e") How can I extract the whole array of numbers without the names? ww[1:5] does not work while ww[[1]] can only extract one number at a time. Thanks, Gang

A very simple question. With a data frame like this: > n = c(2, 3, 5) > s = c("aa", "bb", "cc") > df = data.frame(n, s) I want df$s[1] or df[1,2], but how can I get rid of the extra line in the output about the factor levels: > df$s[1] [1] aa Levels: aa bb cc Thanks, Gang

I want to identify whether a variable is character(0), but get lost. For example, if I have > dd<-character(0) the following doesn't seem to serve as a good identifier: > dd==character(0) logical(0) So how to detect character(0)? Thanks, Gang

Hi, I want to store some number of outputs from running a bunch of analyses such as lm() into an array. I know how to do this with a one-dimensional array (vector) by creating myArray <- vector(mode='list', length=10) and storing each lm() result into a component of myArray. My question is, how can do this for a multiple dimensional array? It seems array() does not have such a

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

I am trying to upload into R 143 Affymetrix chips onto using R on the NIH Nimbus server. I can load 10 chips without a problem, however, when I try to load 143 I receive a error message: cannot create a vector of 523263 KB. I have expanded the memory of R as follows: R --min-vsize=10M --max-vsize=2500M --min-nsize=10M -max-nsize=50M (as specified in help in R). After running this command the

I want to pass a predefined string ww ("fa*fb+fc") to function lme so that I can run > lme(y ~ fa*fb+fc, random = ~1|subj, model) I've tried something like > lme(y ~ paste(ww), random = ~1|subj, model) and > lme(y ~ sprintf(ww), random = ~1|subj, model) but both give me the following error: Error in model.frame(formula, rownames, variables, varnames, extras,

Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between