similar to: duplicate values

Hi there, I've installed R somewhen in German but I'd like to have it in English. The installation of a new version does not change the language setting though I think I've chosen "English" in the installation process (this choice was available, right?). Can anybody give me a hint? (Error-Messages in German are not easy to get a solution for...) Antje

Hey hey... I'd like to create a new feature code in asterisk so when a user dials... say... *00, it would then call some other extensions and play a sound file to them. So far, this is what I have... [testing-custom] exten => *00,1,Wait(1) exten => *00,2,Playback(beep) exten => *00,3,Playback(beep) exten => *00,4,AGI(festival-script.pl|I will now attempt the call) exten =>

hei everyone, I have a weird problem: if I do a # cd /somedir; stat somesubdir |grep Modify where /somedir is some directory on an SMB-mounted filesystem I get one datetime... but I I do: # cd /somedir; stat * |grep Modify for the same directory I get a Modify time which consistently differs by 1 second. As I'm writing a perl module to detect differences in a filesystem and rely on the

Hello everyone! I need to include the confidence interval bar in a grouped barplot. I've found some options on the web, but none of them solved my problem. The question is: my barplot was created using vectors for each pair of bar and them combining them using cbind. I mean: a=c(10,15) b=c(20,24) c=c(21,23) ... hei=cbind(a,b,c) graph1=barplot(hei, beside=T,...) I've tried to include

Hi there, I am wondering if it is possible to do an xyplot with two y-axes. I'd like to print two parameters in a time series but they both have different scales. Which parameter in xyplot can I add to achieve this result? Thanks a lot for any help. Antje [[alternative HTML version deleted]]

Hi all, I'd like to know, if I can solve this with a shorter command: a <- rnorm(100) which(a > -0.5 & a < 0.5) # would give me all indices of numbers greater than -0.5 and smaller than +0.5 I have something similar with a dataframe and it produces sometimes quite long commands... I'd like to have something like: which(within.interval(a, -0.5, 0.5)) Is there anything I

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

Hello, I'm not very experienced with lattice and I was wondering whether I get get some hints from you how to create a pure heatmap (using levelplot), without any axis, title, legend, margin at all... I just want to see the coloured squares, nothing else. Any suggestions? Antje

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

I do things below: In lib/translater.rb class Translater require "rubygems" require "rmmseg" include "RMMSeg" ... def segment(text) RMMSeg::segment(text) end end in helper/xx/xxx.rb module xx require "translater.rb" def translate(text) Translater.run(text) end end when i use helper method tranlate, erros: undefined method

Hi there, as I'm not sure to understand the coloring levelplot uses, I'm looking for another easy way to create a heatmap like this: library(lattice) mat <- matrix(seq(1,5, length.out = 12), nrow = 3) mat[1,2] <- 3.5 my.at <- seq(0.5,5.5, length.out = 6) my.col.regions <- rainbow(5) graph <- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions = my.col.regions)

Hi there, assume that you have data with different sampling like d1 <- rnorm(100) d2 <- rnorm(150) now, I'd like to create two boxplots in one graph but each plot located at the sampling number at the x-axis. This, I can do with "at" l <- list(d1,d2) boxplot(l, at=c(length(d1), length(d2)), xlim=c(0,200) ) but both plots are very thin and I'd like to increase the

Dear list, I'd like to access the elements of a list within another list with the help of a variable. dict <- list( "24" = c(1,2,3,6,12,24,48,72,96,120,144,168,720), "168" = c(1,12,24,48,72,96,120,144,168,336,504,672), "8760" = c(1,24,168,730,4380,8760) ) dict$"24" works fine, but a <- "24" dict$a returns

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hi there, I have a question, which I thought is very easy to solve, but somehow I can't find a solution. Probably someone could help me quickly? Here it is: I have two matrices: a [,1] [,2] [,3] [1,] 1 4 9 [2,] 2 6 10 [3,] 3 6 11 [4,] 4 8 12 b [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 Now I want to find

Hi there, I try to rewrite some Java-code with R. It deals with reading XML files. I started with the XML package. In Java, I had a very useful method which gave me a node by using: name of the node index of appearance start point: global (false) / local (true) So, I could do something like this. setCurrentChildNode("data", 0); getValueOfElement("val",1,true); -->

Hi there, I hope this question is not as stupid as the one before ... I tried to shorten my histogram (because the distribution is quite skewed and I simply don't want to see the long tail but still use the histogram plot). How can I do something like this? (The example does not work but I don't know why...) data <- rnorm(100) # as example, of course this is not skewed... h <-