search for: takas

Hi R users Having both a faster CPU and more memory will boost computing power. I was wondering if only adding more memory (1GB -> 2GB) will significantly reduce R computation time? Taka, _________________________________________________________________ Get FREE company branded e-mail accounts and business Web site from Microsoft Office Live

Dear R-users I have four simple linear models, which are all in the form of a*X+b The estimated parameter vectors are a <- c(1,2,3,4) b <- c(4,3,2,1) My goal is to draw a plot where x-axis is X (range from -100 to 100) and y-axis is the sum of all four linear models X <- seq(-100,100, length=10000) plot(X, sum of the four linear functions) I started with a function for summing

Hello, I''m trying to setup postgreSQL onto rails on linux and when I type bundle install this is the error that i get. I guess its not finding the files Gem::Installer::ExtensionBuildError: ERROR: Failed to build gem native extension. /home/taka/.rvm/rubies/ruby-1.9.3-p194/bin/ruby extconf.rb checking for bzlib.h... no checking for BZ2_bzWriteOpen() in -lbz2... no ***

Dear R-users I need to read some text files produced by some other software. I used readLines (with n = -1 ) command and then tried to find some numbers I liked to extract or some line numbers I like to know. The problem is that there is no empty line at the end of the text files. R gave me this warning below; In addition: Warning message: incomplete final line found by readLines on

Dear R-users >prob <- c(0.5,0.4,0.3,0.1,0.0) >cal <- prob * log(prob,base=2) >cal [1] -0.5000000 -0.5287712 -0.5210897 -0.3321928 NaN Is there any way to change NaN to zero ? I did come up with this by applying Ripley's relpy to my previous question cal <-prob*log(pmax(prob,0.00000001),base=2) Any suggestion ? Thank you Taka

Dear R-users, I am looking for a way to scan a directory and then make a string vector consisting of the file names in the directory. For example, under c:\temp\ there are 4 files a.txt, b.txt, c.txt, and d.txt I would like to have a string vector c("a.txt","b.txt","c.txt","d.txt") How do I do that? Thanks Taka,

Hi R users Here is what I got with help from Petr Pikal (Thanks Petr Pikal). I modified Petr Pikal's code to a little to meet my purpose. I created a function to generate a matrix generate.matrix<-function(n.variable) { mat<-matrix(0,n.variable,(n.variable/2)/5+1) #matrix of zeroes dd<-dim(mat) # actual dimensions mat[1:(dd[1]/2),1]<-1 #put 1 in first half of first column

Hi R users Which version of S does the current R version use? Venables and Ripley's book (2000) says R is version 3 of S language. Is there any change since 2000 ? I searched R-help archive for using C or C++ with R. Although there are some postings, I am looking for up-to-date answers. Which one ( C and C++) works better with R? What complier is the best suited for R ( gcc or visual

Hi R-users I have a data set. There are 10 products and the numbers of people who ranked the products. The format of the data set is productID rank1 rank2 rank3 rank4 rank5 rank6 rank7 rank8 rank9 rank10 ------------------------------------------------------------------------------------------------------- 1 10 2 3 3 6 4 2 5

Dear R-users I wrote a small program for assigning a membership Here is my script sample.size <- 60 x <- rnorm(sample.size, 0, 1) y <- rnorm(sample.size, 0, 1) x.mean <- mean(x) y.mean <- mean(y) membership <- numeric(sample.size) for (i in 1:sample.size) { if ((x[i] < x.mean) && (y[i] < y.mean)) { membership[i]

Hi RUsers, I am wonder if I can search observations whose IDs matches any of the values in another vector, such as in MySQL. While I am learing MySQL for future database management, I appreciate if anyone could give me a hint. Suppose I have one 5*1 vector containing observation IDs and frequencies, and one 3*1 vector containing observation IDs. observation<-c(1,2,3,4,5) ID<-c(1,3,4)

Dear R-users I have 5 dependent variables (y1 to y5) and one independent variable (x) and 3 conditioning variables (m, n, and 0). Each of the conditioning variables has 2 levels. I created 2*4 panel plots. xyplot(y1+y2+y3+y4+y5 ~ x | m*n*o,layout = c(4,2)) I would like to reorder the 8 panels. I tried to use index.cond (e.g., index.cond = list(c(1,3,2,4,5,7,6,8)) but it didn't work out.

Hello I am trying to boot/install VM on XenClient. I would like to learn the way to install VM. Please teach the example of the operation and config file. (Especially, configuration file of method of displaying screen.) thanks. ========================================= [My Executed History] (1) I made a config file for VM(Windows XP:below). (2) I executed the following commands. # xenvm

Dear R-user I have 100 lists. Each list has several components. For example, >data1 $a [1] 1 2 $b [1] 3 4 $c [1] 5 There are data1, data2,...., data100. All lists have the same number and the same name of components. Is there any function I can use for applying to only a specific component across 100 lists? (e.g., taking mean of $c acorss 100 lists) or do I need to write my own

Dear R users, I would like to extract df and Mean Sq values. I tried several things (e.g., str(model1), names(model1)) but I can't find a way to extract these values. I also tried to search using RSiteSearch. Any help will be appreciated. Thanks Taka, model1<-aov(dv~iv.1*iv.2*iv.3) summary(model1) Df Sum Sq Mean Sq iv.1 1 3.200 3.200

Cze??. Aktualnie mam do czynienia z klastrem pocztowym, w kt?rym pliki wiadomo?ci pocztowych u?ytkownik?w s? przechowywane na raid software-owym udost?pnianym po NFS. Serwer?w NFS jest kilka, ka?dy udost?pnia podzbi?r wiadomo?ci dla odpowiedniego serwera z postfix-em, kt?ry jest klientem NFS. Dane z jakiego serwera NFS host frontend z postfix-em powinien montowa? katalogi z poczt? s? w MySQL.

Hi I bought a new laptop HP dv9500 just a week ago and installed a Ubuntu gutsy ribbon on this laptop. I wanted to install Fedora but there are more threads on Ubuntu, so I decided to install Ubuntu. After hours of struggle in configuring x server/graphic card stuff, I installed R for Gutsy ribbon using "sudo apt-get install g77 r-core'. Now when I tried to install 'sp' package

Hi everyone, I need help. I want to have a "uniform" kind distribution. When I used sample function I got almost twice many zeros compared to other numbers. What's wrong with my command ? temp <-sample(0:12, 2000, replace=T,prob=(rep(1/13,13))) hist(temp) Thanks in advance, Taka,

Hi R users I like to generate a certain type of matrix. If there are 10 variables, the matrix will have nrow=10, ncol=((10/2))/5+1. so the resulting matrix's dimension 10 by 2. If there are 50 variables the dimension of the resulting matrix will be 50 by 6. The arrangement of elements of this matrix is important to me and I can't figure out how to arrange elements. If I have 20

Hi R users I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and one continuous variable (e.g., Y1). I combined these variables using data.frame() mydata <- data.frame(X1,X2,X3,X4,X5,X6,X7,Y1) after that, I sorted this data.frame rank.by.Y1<-order(mydata[,8]) sorted.mydata<-mydata[rank.by.Y1,] after that, I replaced Y1's values with values ranging from 1