similar to: ts basic question

Dear R-collegues, I'm getting an error message (Error in round) when summarising a glm.nb model, and when using anova.negbin (in R 1.3.1 for windows): > m.nb <- glm.nb(tax ~ areal) > m.bn Call: glm.nb(formula = tax ~ areal, init.theta = 5.08829537115498, link = log) Coefficients: (Intercept) areal 3.03146 0.03182 Degrees of Freedom: 283 Total (i.e. Null); 282

Dear R collegues, I'm trying to understand what's AIC in R (ver. 1.3.1), and I'm getting a different answer if I look at the AIC(of the fitted model) or the aic in the summary( of the fitted model). Is this correct? Can somebody explain me the difference between the two values? or Is the AIC criterion not appropiated for Poisson models? R session: > t1 <- glm(tax ~ areal,

Dear R collegues, I'm trying to understand what's AIC in R (ver. 1.3.1), and I'm getting a different answer if I look at the AIC(of the fitted model) or the aic in the summary( of the fitted model). Is this correct? Can somebody explain me the difference between the two values? or Is the AIC criterion not appropiated for Poisson models? R session: > t1 <- glm(tax ~ areal,

Hi Max/DTrace list, > At any rate, without the -C, I can''t use #include <sys/stream.h>. > Without the #include <sys/stream.h>, I can''t use the M_DATA. > As it is, I get the following: > > # ./strrput.d 0xd595a6c0 <-- this is a vnode for a socket > ftp is using (not important here) > dtrace: failed to compile script ./strrput.d: line 7: >

Dear collegues, How can I get the name of a variable (and not the variable) within a function ? For instance, in the following function, I'd like to create a variable in the dataframe df with the same name to the variable passed in var: prova <- function( var ) { df <- as.data.frame(matrix(nr=20,nc=0)) df[["here"]] <- seq(min(var), max(var), le= 20) # df } for

I don't quite understand the difference between the two methods for performing a chi-squared test on contingency tables: summary(table()) and chisq.test() They may different results. E.g.: aa <- gl(2, 10) bb <- as.factor(c(1,2,2,2,1,2,1,2,2,2,1,2,2,2,1,1,1,2,1,1)) aa <- c(aa, aa) bb <- c(bb, bb) table(aa, bb) summary(table(aa, bb)) chisq.test(aa, bb) Could somebody give me

Hi all, I have a large file (1.8 GB) with 900,000 lines that I would like to read. Each line is a string characters. Specifically I would like to randomly select 3000 lines. For smaller files, what I'm doing is: trs <- scan("myfile", what= character(), sep = "\n") trs<- trs[sample(length(trs), 3000)] And this works OK; however my computer seems not able to handle

Dear colleagues, I'd like to reshape a datafame in a long format to a wide format, but I do not quite get what I want. Here is an example of the data I've have (dat): sp <- c("a", "a", "a", "a", "b", "b", "b", "c", "d", "d", "d", "d") tr <- c("A",

Dear all, I'm interested in analysing a reapeated measure desing where plant height (H) was measured 3 times (Time). The experimental design include 2 fixed factor (say A and B) in which A is nested in B, and a random factor (C, the plot), using the aov(). So my first idea would be something like: aov(H ~ B * A %in% B * Time + Error(id) ) where id is the factor coded for the repeated

Dear R-users, I'm sure it must be a specific function or a better way to convert matrix to x,y,z coordinates (and viceversa), than my function below (it works). Any help? m2coord <- function(m) { k <- nrow(m)*ncol(m) aa <- data.frame(r=1:k, c=1:k, v=1:k) k <- 0 for (i in 1:nrow(m)) for (j in 1:ncol(m)) { k <- k+1 aa$f[k]=i; aa$c[k]=j; aa$v[k]=m[i,j] } aa } Juli

Dear colleges, I'm trying to combine a barplot and a plot in a single figure as follows: data <- 1:6 t <- barplot(data, axes=F) par(new= T) plot(t, data, type="b") However, as you can see in the example, the dots of the second plot do not fall in the midpoint of the bars in the first. Any trick for setting the 2 plots at the same scale? I have unsuccessfully tried: plot(t,

The cells are interpreted as counts, so by scaling you're analyzing a different experiment (one with fewer observations). So the chi-squared value will change (the terms (O-E)^2/E in the statistic scale linearly ignoring rounding and "Yates' continuity correction"). The chisq.test on the original data is a test of association. Conventionally you decide ahead of time on a

Hi I'd like to write.table a dataframe, but with an specific order of columns. Is there a direct way to do it? or I have to generate a new dataframe as follows: t <- data.frame(c=1:10, b=11:20, a=letters[1:10]) t2 <- data.frame(a=t$a, b=t$b, c=t$c) write.table(t2, row.names=F) Thanks for any comment Juli -- http://www.ceam.es/pausas

Hi Is there a way to remove blank characters from the end of strings in a vector? Something like the =TRIM functions of the OpenOffice spreadsheet. E.g., a <- c("hola ", "Yes ", "hello ") # I'd like to get: c("hola", "Yes", "hello") Thanks Juli -- http://www.ceam.es/pausas

Dear all, How could I use variables in a loop that their names are in a vector? For example: aaa <- 1:10 bbb <- aaa*2 ccc <- aaa+bbb varn <- c("aaa", "bbb", "ccc") m <- rep(NA, 3) for (i in 1:length(varn)) m[i] <- mean(varn[i]) # wrong thanks in advance Juli -- "Wars do not solve problems, wars generate even more problems"

R-users, Can an offset term be included in a Poisson model? I get an error message when trying that: >r3o <- glm(tax ~ areal + offset(o), family=poisson) Error in (if (is.empty.model(mt)) glm.fit.null else glm.fit)(x = X, y = Y, : inner loop 1; can't correct step size In addition: Warning message: Step size truncated due to divergence in: (if (is.empty.model(mt))

Dear all, I have a question on glm, family binomial. I do not see significant differences between the levels of a factor (treatment) if all data for a level is 0; and replacing a 0 for a 1 (in fact reducing the difference), then I detect the significant difference that I expected. Is there a way to overcome this problem? or this is an expected behaviour ? Here is an example: s <-

Hi all,, Is it possible to set a thicker line for the box around the plots? (i.e. for the four axes) Something like lwd (lines) but for the box. Thanks juli -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Hi, Thanks for showing me how to use RODBC and odbcConnect. Now works nicely. The question I've got now is: noms <- list.files(pattern=".DBF") # removing extension names: noms <- sapply(noms, function(x) as.character(strsplit(x,".DBF")) , USE.NAMES=F) for (i in 1:length(noms)) { s <- sqlFetch(bdades, noms[i]) # etc. } But it seems that sqlFetch()

Hi, This is a very basic question, but I would like to undestand hist(). I thought that the hist( , freq=FALSE) should provide the relative frequencies (probabilities), and so they should sum 1, however: set.seed(2) ah <- hist(rnorm(100), freq=F) sum(ah$intensities) [1] 2 set.seed(2) bh <- hist(rlnorm(100), freq=F) sum(bh$intensities) [1] 0.4999996 I'm getting similar figures with