similar to: write.table a df with specific column order

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

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

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 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 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,

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,

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))

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()

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",

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

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, 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 <-

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

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"

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, I'm trying to use reshape but I cannot quite understand how it works. Could somebody help me on this? Example, my data is something like: mydat <- data.frame(tree= 1:10, serra=rep(1:2, c(5,5)), bt01= 101:110, bt02= 201:210, bt03= 301:310, mm01= 9101:9110, mm02= 9201:9210, mm03= 9301:9310) > mydat tree serra bt01 bt02 bt03 mm01 mm02 mm03 1 1 1 101 201 301 9101 9201

Dear all, I'd like to subset a df within a function, and use select for choosing the variable. Something like (simplified example): mydf <- data.frame(a= 0:9, b= 10:19) ttt <- function(vv) { tmpdf <- subset(mydf, select= vv) mean(tmpdf$vv) } ttt(mydf$b) But this is not the correct way. Any help? Thanks in advance Juli

Dear all, I'm a subscriber of the r-help-Digest; however, currently I'm not receiving the daily message. The 2 last digest messages I've got are: V3 # 157 (2/11/02) and V3 #165 (12/11/02), and both are incomplete (the initial indice list many items but only few are of them are shown). I did resuscribe, but the problem persists. Is this just my problem (a problem of my system)? Or is

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