similar to: reading very large files

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-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-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 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, 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))

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'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

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 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 <-

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

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, 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

Hi, For computing correlation among variables in a matrix, I use cor( ), but for computing the p-values I'm using cor.test in the following way: cor.p <- function(X) { res <- matrix(0, ncol(X), ncol(X)) for (i in 1:ncol(X)) for (j in 1:ncol(X)) res[i, j]<- cor.test(X[, i], X[, j])$p.value rownames(res) <- colnames(res) <- colnames(X) res } I'm just wondering if there is a

Dear all, I've seen that there are some maps, or at least costlines in R. The oz package is the map of Autralia and in the fields package there is a US map and a world map. This world map allows to select smaller sections such Europe: library(fields) world(xlim=c(-10,18),ylim=c(36,60)) However, at this scale the map is quite crude (and several important big islands are messing). My

Is there a way of removing all rows with missing values from a data frame? I usually use subset(x, var1!="NA") and repeat for each variable. It would be nice to be able to do it in one fell swoop. Also, surprisingly, it doesn't always work. Sometimes I'm left with an empty set even though not all rows have missing values for the variable. Cheers, mikkel Mikkel Grum,