similar to: break a vector into classes

Hi, Currently I have a density plot generated with this snippet. Is there a way I can add a line curve on top of it? I mean in one figure __BEGIN__ myhist <- hist(x col="blue", main = "Density Plot", xlab = "Exp Level", ) __END__ - Gundala Viswanath Jakarta - Indonesia

Hello everyone, I would be glad if you could help out an R-beginner here... I have a vector of categorial data like this > v <- c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4) When I do > hist(v) I get the x-axis of the histogram with floating point labels: 1.0, 1.5, 2.0, etc. Is it possible to tell R that the data consists of categories, i.e. that I only want the category names (1, 2, 3, 4) on my

Hi all, >From a vector, I want to get the unique values and the counts of these unique values in the vector. For example, x<-c(2 ,1 ,2, 1, 4 ,2 ,1, 4 ,1 ,1) xu<-unique(x) xn<-numeric(length(xu)) for (i in 1:length(xu)) {xn[i]<-length(which(x==xu[i]))} There must be a very much simpler method of doing this. Can somebody direct me to the functions that I must read in order to do

Hi all, I'm trying to superimpose a rchisq density line over a histogram with something like: hist(alnlength) lines(density(rchisq(length(alnlength), 4)),col="red") But the rchisq line won't appear anywhere, Anyone knows what I am missing here? Thanks in advance, Albert.

Hello, I got two sets of data x=(124738, 128233, 85901, 33806, ...) y=(25292, 21877, 45498, 63973, ....) When I did a t test, I got two tail p-value = 0.117, which is not significantly different. If I changed x, y to log scale, and re-do the t test, I got two tail p-value = 0.042, which is significantly different. Now I got confused which one is correct. Any help would be very appreciated.

Hi R-users, I am trying to divide a vector (say X) into equal frequency bins. If one uses the hist() function, then a histogram is plotted, but with bins of equal widths, and not with bins having the same number of data points. I have then tried the histogram() function as follows: histogram(X, nint=10, breaks=NULL, equal.widths=F) This works as I want. However, I can't extract which

Hi, Just a quick query - if I'm creating a function to produce a number of histograms per page of output (one per column from a matrix), how can I pass the column name of the matrix into the title (or indeed to form part of the x-axis label)? TIA, Laura Laura Quinn Institute of Atmospheric Science School of Earth and Environment University of Leeds Leeds LS2 9JT tel: +44 113 343 1596 fax:

Dear R users, I would like to sample from a mixture distribution p1*f(x1)+p2*f(x2). I usually sample variates from both distributions and weight them with their respective probabilities, but someone told me that was wrong. What is the correct way? Vumani

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.

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, What parameter should I feed to the hist function to draw a line, not bar graph for a histogram? Smoothed line would be great too. I've looked at curve() for normal distribution but it's not what I need. I need some curve on top of the histogram (fit to the actual data). Just as a way to illustrate, rather than bargraph. Thanks -- Regards, Anh Tran UCLA NeuroOncology Lab

I want to use R to run a reaction-time experiment: Something appears on the screen, I respond by typing something (one keystroke), the system measures the speed of my response. R would be great for this if only I didn't have to hit Enter to enter that keystroke. I am doing such experiments now but they require two actions per trial: hit keystroke, hit Enter. Is there some way that R can be

Dear all, I am trying to simulate the null distribution for the likelihood ratio test statistic for testing 1 random effect versus no random effect. The asymptotic null distribution should be a mixture of a chi-squared distribution with 0 degrees of freedom and a chi-squared distribution with 1 degree of freedom. This means that I expect a point mass of 50% on 0 for the likelihood ratio

for example: > a$tic[1:10] [1] "AIR " "ABCB " "ABXA " "ACMR " "ADCT " "ADEX " [7] "ABM " "AFCE " "AG " "ATG " Can anyone please tell me how to strip the white spaces from a$tic? Thanks, Roger [[alternative HTML version deleted]]

Thanks, I tried someting like this, but computation takes times for large matrices btransf <- function(y,X=length(y)^4) { N <- length(y) bm <- matrix(rep(1/N,N^2),N,N) for(j in 1:X){ coord <- sample(1:N,4,replace=T) d <- runif(1,0,min(bm[coord[1],coord[2]],bm[coord[3],coord[4]]))

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)

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