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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

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

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 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 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

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,

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.

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]

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

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

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 like to generate some strings (a character vector) in a special way like If i have 5 variables "002.001", "003.001", "003.002", "004.001", "004.002", "004.003", "005.001", "005.002", "005.003", "005.004" so the created string vector's elements are "002.001",

Hi R users I have two variables (X and Y) X <- rnorm(100,.7,.5) Y <- rnorm (100,.3,.1) I like to know whether 1 is between each pair of X and Y or not. Thanks TM

Hi R user I need to read in some values from a computer program output. I can't change the output format because the developer of the program doesn't allow to change the format of output. There are two formats. First one looks like this if I have 10 variables, ------------------------------------------------------------------------------------------------------ [ 1]

Hi R users I have a simple question to ask x <- c("a","b","c") x [1] "a" "b" "c" I like to have "abc" instead of having "a" "b" "c" I tried to use paste and other functions related to character. Is there any function (command) that enable me to combine elements of a character vector into

Hi R users I wrote a function that generates some character strings. generate.index<-function(n.item){ for (i in 1:n.item) { for (j in ((i+1):n.item)) { cat("i",formatC(i,digits=2,flag="0"),".",formatC(j,digits=2,flag="0"),"\n",sep="") } } } I

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

Hi R-users I have two conditions. For each condition, 100 sets of 10 random numbers from N(0,1) need to be generated. Here is my question. At the begining I specify a seed number. I want to make the 100th set of the first condition and 1st set of the second conditon the same. What do I need to do ? After generating 99th set of 10 random numbers and then saving .Random.seed then using

Hi R users I like to extract lower diagonal elements of a matrix in such a way like, data[1,2], data[1,3], ...., data[5,6] are extracted from a matrix called 'data' This short script below is what I have written so far. ########################################## data <- matrix(rnorm(36,0,1),nrow=6) temp<-c() for (i in 1:(nrow(data)-1)) { for (j in (i+1):nrow(data)) {

Hi R users This looks a simple question Is there any difference between between rnorm(1000,0,1) and running rnorm(500,0,1) twice in terms of outcome ? TM