similar to: Replicate

Dear All, I wonder if there is any R function to generate a sequence of vectors from existing ones. For example: x 1<- c(1,2,3) x2 <- c(4,5) x3 <- c(6,7,8) The desired output is a list of all 3*2*3 = 18 possible combinations of elements of x1,x2 and x3. One element for example is (1,4,6). Many thanks in advance, Bernard

Dear All, I have written the following programs to find a non-singular (10*10) covariance matrix. Here is the program: nitems <- 10 x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <- try(solve(sigma), TRUE) while(inherits(inverse, "try-error")) { x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <-

Dear All, How can I sort a data frame with respect to more than one variable? I know that for one variable X one may use: df[order(df$X), ] where df is the data frame containing X. Many thanks, Bernard --------------------------------- [[alternative HTML version deleted]]

Hello. I have looked at R Site Search for this problem, and it didn't give me exactly what I needed. Consider this dataset called "results". It has the following information: Student Day Subject Score Mary 1 Math Failed David 2 Science Passed Bob 4 Reading Passed Marie 4

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Hello, Let's assume I have a vector of integers : > myvector <- c(1, 2, 3, 2, 1, 3, 5) My purpose is to obtain the cumulative distribution of these numerical data, i.e. something like : value nb_occur. <=1 2 <=2 4 <=3 6 <=4 6 <=5 7 For this, I create a table with ; > mytable <- table(myvector) 1 2 3 5 2 2 2 1 However, table() returns an array

Hi, Given a test matrix, test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3,3) A) How to compute the counts of each column (excluding the NA) i.e., 3, 2, 1 B) How to compute the counts of each column (excluding the NA) that are greater than the column means ? i.e., 1, 1, 0 I could write a for loop, but hope to use better alternative. [[alternative HTML version deleted]]

Consider this: > y <- matrix(1:8, ncol=2) > is.matrix(y[-c(1,2),]) [1] TRUE > is.matrix(y[-c(1,2,3),]) [1] FALSE > is.matrix(y[-c(1,2,3,4),]) [1] TRUE It seems like an inconsistent behaviour: - with 2 or more rows we have a matrix - with 1 row we do not have a matrix and - with 0 rows we have a matrix again I just stumbled on this behaviour, because I had a problem with my

Thanks Dimitris!!! That's much clearer now. Still have a lot of work to do this weekend to understand every bit but your code will prove very useful. Cheers, Aziz -----Original Message----- From: Dimitrios Rizopoulos [mailto:Dimitris.Rizopoulos at med.kuleuven.be] Sent: May 12, 2006 4:35 PM To: Chaouch, Aziz Subject: RE: [R] Maximum likelihood estimate of bivariate

Hi all, My question is not really urgent. I can write a loop and solve the problem. But I know that I'll be in a similar situation many more times so it would be useful to find out the answer Is there a fast way to perform linear fit to all the columns of a matrix? (or in the one dimension of a multi-dimensional array.) I'm talking about many single linear fits, not about a multiple fit.

Dear R-devels, I'd like to create a plot method for a class of objects that passes the '...' argument to both plot() and legend(), e.g., x <- list(data = rnorm(1000)) class(x) <- "foo" plot.foo <- function(x, legend = FALSE, cx = "topright", cy = NULL, ...){ dx <- sort(x$data) plot(dx, dnorm(dx), type = "l", ...) if (legend)

I have a case where I would like to change multiple columns containing numbers to factors. I can change each column one at a time as in: TEMP.FACT$EXPOS01<-factor(TEMP.FACT$EXPOS01,levels=c(1,2,3),labels=c("No ne","Low Impact","MedHigh Imp")) TEMP.FACT$EXPOS02<-factor(TEMP.FACT$EXPOS02,levels=c(1,2,3),labels=c("No ne","Low

I have the following csv file: name,x,y,z category,delta,gamma,epsilon a,1,2,3 b,4,5,6 c,7,8,9 I'd like to create a numeric matrix of just the numbers in this csv dataset. I've tried the following program: sample.data <- read.csv("sample.csv") numerical.data <- as.matrix(sample.data[-1,-1]) However, print(numerical.data) returns what appears to be a matrix of

I have a vector that has 1,974 elements and each element is one of the following (B, F, N, Y). How do I recreate that vector accept in the place of N put 0 and in the place of B, F or Y put a 1? Thanks, Jacob [[alternative HTML version deleted]]

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

hi all: xlim and ylim are used to define the interval limits of a plot. I'm interested in the scale of values between this limits. suppose xlim=c(0,10) we can have e.g. 0 5 10 0 2 4 6 8 10 0 1 2 3 4 5 6 7 8 9 10 which is the parameter that allows me to modify this? thanks in advance alexandre

Hello all, Simple version of my problem: I've got a list of data frames, where each data frame has the same number of columns and the same column names. I'd like to flatten the list into one large data frame. Is there an easy way to do this? Quick example code: a <- data.frame(x=c(1,2,3),y=c(5,7,9) b <- data.frame(x=c(2,4,7,9),y=c(2,3,5,4)) z <- list(a,b) # Do

Hi all. I have a numerical function f(x), with x being a vector of generic size (say k=4), and I wanna take the numerically computed gradient, using deriv or numericDeriv (or something else). My difficulties here are that in deriv and numericDeric the function is passed as an expression, and one have to pass the list of variables involved as a char vector... So, it's a pure R programming

Hi everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323