similar to: plot in for loop with "i" indexed variables does not work

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

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

Hello All, I was wondering how I can get the overall Pearson chi^2 test of model fit with its df and p value in the LTM package for the 2PL models. Thanks, -- Davood Tofighi Department of Psychology Arizona State University [[alternative HTML version deleted]]

Dear friends of lattice, I know how to position trellis plots with print(...,split,more=T) or (...position). Sometimes I wish I had something like the old "par(mfcol=2, mfrow=3)" mechanism, where the next free viewport is automatically chosen. I tried fiddling with grid-viewports, but could not find an easy solution. Did I miss something? Dieter Menne

I would like to put three figures next to each other in a figure. I have been reading the introduction to R, section 12 several times now, and I still can't make heads or tails out of it. Lets say that I have three dataframes a, b, c, and I want to plot a$V1, b$V1 and c$V1 in separate plots simply using plot(), how do I put them next to each other? I am sorry if this is a FAQ, but I cannot

Dear all, Are there any functions to convert decimals to fractions in R? I have the result: > summary(as.factor(complete.ID)) 0 0.0133333333333333 0.04 2256 488 230 0.0666666666666667 0.0933333333333333 0.106666666666667 2342 310 726 0.133333333333333

Hello, I would like to know if R provides the similar function to produce the 4-in-1 graph ("Normal Plot of Residuals", "I Chart of Residuals", "Histogram of Residuals" & "Residuals vs Fits") as MiniTab. If do not have, could anyone please tell me how to produce above-mentioned individual chart? Your help is greatly appreciated.

Suppose I have a nx2 matrix of data, X, the following code generate density estimation for each column and plot them denlist <- apply(X, 2, density) par(mfrow=c(1,2)) lapply(denlist, plot) Does anyone know how to change the main title of each density plot to "var 1", "var 2" by passing optional argument "main"? I've tried lapply(denlist, plot,

I have a matrix that I want to turn into a transformed matrix that includes the indices from the original matrix and the value. The matrix is simply real-valued and is square (and large (8k x 8k)). I want something that looks like (for the 3x3 case): i j value 1 1 1.0 1 2 0.783432 1 3 -0.123482 2 1 0.783432 2 2 1.0 2 3 0.928374 and so on.... I can do this with for loops, but there is

Hello list Does anybody know of any way to create an index of all the possible combinations of variables (factors) in a data frame? ie for 3 factors A, B & C we have A B C AB AC BC ABC which equates to columns 1, 2, 3, 1:2, (1,3), 2:3 and 1:3. I realise that a function like model.matrix does this, but how to get the seqence of the index? Any help would be greatly appreciated.

Dear R users, I have a data generated as the following, dat <- data.frame(matrix(sample(24), nrow=4)) dimnames(dat) <-list(rownames=c('g1','g2','g3','g4'), colnames=c("A_CH1","A_CH2","B_CH1","B_CH2","C_CH3","C_CH3")) » dat A_CH1 A_CH2 B_CH1 B_CH2 C_CH3 C_CH3 g1 16 24 7 9 14

Hi, Given names <- c("a", "b", "c", "d"). I would like to generate "a vs. b", "a vs. c", "a vs. d", "b vs. c", "b vs. d", "c vs. d" Thanks [[alternative HTML version deleted]]

Dear HelpeRs, Given: x <- rnorm(50) y <- rnorm(50) plot(x,y) abline(lm(x ~ y)) Is there a way to plot just a portion of the line? Say for values of x > 2.0 or x > -2.0 and x < 4.0. (Still fitting all the points.) Thank you, jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning.

Hi Sorry to ask such a well oiled question, but even with multiple google hits I don't think this has been answered very well. It's all well and good doing a sort of a data frame on multiple columns when you know in advance which columns you want to sort on, but what about when the names of the columns you wish to sort on are in a vector? At the minute I'm messing about with

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 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 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 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.