similar to: create matrix from comparing two vectors

Hi all, I made a dotplot() with lattice, which comes out nice on the graphics device. I can save this as a eps using postscript() and include this in a word document. This prints nice, but does not look good on screen. If I produce a pdf, it is nice on screen, but not on paper. How can I save a graph that looks nice on paper and on screen? Bjorn Bjorn Van Campenhout Institute of Development

I have a question, which very easy to solve, but I can't find a solution. I want to convert a data frame to matrix. Here my toy example: > L3 <- c(1:3) > L10 <- c(1:6) > d <- data.frame(cbind(x=c(10,20), y=L10), fac=sample(L3, + 6, repl=TRUE)) > d x y fac 1 10 1 1 2 20 2 1 3 10 3 1 4 20 4 3 5 10 5 2 6 20 6 2 > is.data.frame(d) [1] TRUE > sapply(d,

>cbn<-as.matrix(expand.grid( rep( list(0:1), 50))) Error in rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) : invalid 'times' value In addition: Warning message: In rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) : NAs introduced by coercion But I'm only interested in cbn matrix rows where: cbn<- cbn[rowSums(cbn)==5,] Is there a way to evaluate it

I am running R on ubuntu dapper. The version that is in the ubuntu repos is 2.2.1, so I decided to upgrade by adding deb http://cran.R-project.org/bin/linux/ubuntu dapper/ to my sources.list, as advised on the web page. After sudo apt-get update and upgrade, I have version 2.3.1. However, when I try to load the library 'foreign', I get the following error: Error in dyn.load(x,

Dear R-list, *When comparing two logistic regression models with the anova CHi test, I obtain the following error: (there are no NA's in the time series). How can this be solved such that I can compare two models on the same dataset were different explanatory variables are used? l.KBDI <- glm(zna.arson2 ~ zna.KBDI,family = binomial) l.NDWI <- glm(zna.arson2 ~ zna.NDWI,family

Dear all, I would like to take out the values from one vector that are equal to the values in another vector. Example: a <- c(1,2,3,4,5,6,7,8,9) b <- c(3,10,20,5,6) b_noRepeats = c(10,20) So I would like to have the vector b without the same values as vector a. Kind regards, João Fadista [[alternative HTML version deleted]]

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

Dear list: I'm hoping to tap in to the statistical expertise in the group, especially those familiar with simulation techniques. I'm finalizing a study where I obtain standard errors from two sources. The first source is a monte carlo simulation and the other source is an analytical model I have developed that appears to recover the standard errors from the simulation. All analysis are

hi: I have matrix with dimensions(200 X 20,000). I have another file, a tab-delim file where first column variables are row names and second column variables are column names. For instance: > tmat Apple Orange Mango Grape Star A 0 0 0 0 0 O 0 0 0 0 0 M 0 0 0 0 0 G 0 0 0 0 0 S 0 0 0 0 0

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, How is it possible to extract athe elements of a list of vectors in a fixed position? suppose that I have a list of 2-element vectors, how can I extract the 2nd element of all vectors in the list? Can it be done with indexing and not by element name? Thanks carol So in this example, I want to extract 2 and 4 v = list (c(1,2), c(3, 4)) > v [[1]] [1] 1 2 [[2]] [1] 3 4

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.

hi, i am new here and wanted to know, before i start learning yet another statistical package: I want to estimate a system of equations that is non linear in the parameters, using 3SLS. However, i will probably have to constrain some of the parameters to be between, say, zero and one. Is this possible with R, and better, is this easy? If so, since i am an absolute beginner, any pointers where

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)

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

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 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.

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

Hi, I have: a <- matrix(c(0,1,0,1),nrow=2) b <- matrix(c(1,1,1,0,0,0),nrow=3) c <- 1 d <- c(1,0,1) And I would like to join them in an object 'thing' so that I can access a, b, c, or d through an index in a for loop. For example: thing[4] would return [1] 1 0 1 Note however, that I have many of these 'thing' components. So many that a command like thing