search for: matzmoto

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 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

...ngth.of.rand.numbers-1 mat[((start):end), i]<- my.rand.num } mat } Do you (any R users) have any suggestion to this function to make this function work better or efficiently? Taka It works but I >From: "Petr Pikal" <petr.pikal at precheza.cz> >To: "Taka Matzmoto" <sell_mirage_ne at hotmail.com>,r-help at stat.math.ethz.ch >Subject: Re: [R] creating a certain type of matrix >Date: Tue, 07 Feb 2006 08:58:59 +0100 >MIME-Version: 1.0 >Received: from mail.precheza.cz ([80.188.29.243]) by >bay0-mc8-f13.bay0.hotmail.com with Microsoft S...

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

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 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

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 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

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 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)) {

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 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 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

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 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.

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

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 am trying to get cross-tabulation tables using tables. All variables used are binary ones (0 and 1). Each time I constructed cross-tabluation table using a different variable pair (e.g., variable 1 and variable 2, variable 1 and variable 3 etc) In doing so, I ran into some problems i.V2 i.V1 1 0 17 1 33 For variable 2 (i.V2) there was no one belonging to ZERO

Hi R users I like to extract (or collect) a numeric element with a name from a list. Is there any way to extract just a numeric element without the name attached to the element. For example, >mylist Mantel-Haenszel chi-squared test with continuity correction data: table(mydata[, x]) Mantel-Haenszel X-squared = 8.3832, df = 1, p-value = 0.003787 alternative hypothesis: true