similar to: plot type any symbols?

Dear all, does anybody ever encountered the problem with diagnostic plots? x<-rnorm(100) y<-rnorm(100) plot(lm(x~y)) It gives the following message "Waiting to confirm page change..." and nothing happens. Thanks a lot! [[alternative HTML version deleted]]

Dear all, Are there any functions in R to reverse the order of the string. smth like reverse("abc") to get "cba"? Thanks a lot. [[alternative HTML version deleted]]

Dear all, Does anybody know the probable reason why <= gives false when it should give true? These two variables are of the same type, and everything works in the cycle but then it stops when they are equal. this is the output result > Rk[47] <= RB[21] [1] FALSE > Rk[47] [1] 0.002842007 > RB[21] [1] 0.002842007 Thanks a lot. [[alternative HTML version deleted]]

Dear all, might seem and easy question but I cannot figure it out. floor(100*(.58)) [1] 57 where is the trick here? And how can I end up with the right answer? Thanks a lot everybody for your help. Trafim [[alternative HTML version deleted]]

Dear all, I would like to ask you if there is a possibility in R to give the names to graphs which are not const. For example, How to name each plot, or to add notes like a=x[i], b=y[i] in this cycle x<-c(1,2,3,4,5,6) y<-c(7,8,9,10,11,12) par(mfrow=c(2,3)) for (i in 1:6){ plot(x,y) } Thank you for your time and help! [[alternative HTML version deleted]]

Dear all, I would like to ask how to extract the p-value for the whole model from summary(lm). This didn't help a lot summary.lm summary(lm(speed~dist, cars)) Thanks a lot! [[alternative HTML version deleted]]

Dear all, does anyone know how to define the structure of the required samples using function createDataPartition, meaning proportions of different types of variable in the partition? Smth like this for iris data: createDataPartition(y = c(setosa = .5, virginica = .3, versicolor = .2), times = 10, p = .7, list = FALSE) Thanks a lot for your help. Regards, Trafim [[alternative HTML version

Dear all, I have some error trying to bootstrap from a matrix. The error message is "Error in sample(n, n * R, replace = TRUE) : element 2 is empty; the part of the args list of '*' being evaluated was: (n, R)" vv <- c(0.5,3.2,5.4,1.1,1.4,1.2,2.3,2.0) Reg <- matrix(data=vv, nrow = 4, ncol = 2) bootcoeff <- function(x){ coefficients(lm(x[,1]~x[,2]))[2]+1 }

Dear all, seems that easy question but cannot find the function for that. How to get the elements of the minor diagonal of the matrix? Thanks a lot. [[alternative HTML version deleted]]

Dear all, I cannot find to explicitly get the R-squared or adjusted R-squared from summary(lm()) Thanks a lot! [[alternative HTML version deleted]]

Dear all, My question how is it possible to define a matrix A with 10 rows 1 column, so that its elements are vectors of undefined length. I need to have a possibility later to add elements like A[1,1] <- c(A[1,1],3,4,5) Thanks a lot for the help! [[alternative HTML version deleted]]

Dear all, Please, look at the following code: attach(geyser) f1 <- kde2d(duration, waiting, n = 5) a <- 0 for (i in 1:5){ for (j in 1:5){ a <- a + f1$z[i,j] } } As far as I understood from Help kde2d returns matrix elements of which are values of joint probability mass function Pr(X=x,Y=y) therefore, sum of its elements should sum to 1. Which is not the case from my check. Where is

Hi everybody, well, I definitely don't understand anything. Why the hist function with freq=FALSE gives such a strange result??? R <- c(-1.10, 0.79, -1.17, -0.53, -0.26, -0.22, 0.29, -0.26, -0.26, 0.39) hist(R, freq=FALSE, breaks=10) Thanks everybody [[alternative HTML version deleted]]

Dear all, I have the following question. Is it possible in R to call for the p-value directly in a model like this a <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) b <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) summary(lm(a~(b+0))) or even in this a <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) b <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)

Dear all, I have an error in the simple prediction function for lm(). Maybe someone experienced the same? xma <- matrix(data = 0, nrow = 100, ncol = 2) xma[, 1] <- rnorm(100) xma[, 2] <- rchisq(100, df = 3) m1 <- lm(xma[, 1] ~ xma[, 2]) predict(m1, as.data.frame(seq(-13, 13, 0.5))) Thanks a lot, Trafim [[alternative HTML version deleted]]

Hi everybody, Does anyone know what problem may be with this test. I am applying 5 different normality tests and use p-values for them, but for some reason S-W gives me NA, while sample size is 100. Any ideas? Thanks a lot! [[alternative HTML version deleted]]

Dear all, I have a simple question for which I cannot find the answer. I need to make an easy plot, but for the x axis I need to be able to specify by myself the division of x axis from x[,1] either every single observation, or every 5th, or 10th or 20th x <- matrix(data=NA, nrow=100, ncol=2) x[,1]<-seq(1,100,1) x[,2]<-rnorm(100) Thanks a lot for your help!!! [[alternative HTML

Hi! I have a dataset with three columns -the first two refer to x and y coordinates, the last one are odds ratios. I'd like to plot the data with x and y coordinates and the odds ratio shown as a fourfold plot, which I prefer to do using the stars function. Unfortunately the stars option in symbols is not as cool as the stars function on its own, and now i can't figure out how to do it!

Dear all, I have a question regarding performing test if the data fits chi-squared distribution. For example, using ks.test() I found in the examples how to fit it to gamma or weibull x<-rnorm(100) ks.test(x, "pweibull", shape=2,scale=1) for the gamma, pgamma can be used But I cannot find the value of this second parameter for the chi-squared distribution. Maybe someone

Hi everybody, I would like to ask if it is possible using pdf function or some other to print plots in cycle such that every new plot is on new page. like this pdf(file="D:/Plot.pdf",width = 9.25,height=9.25, family="Helvetica",pointsize=8,bg="white") for (i in 1:10){ x <- seq(1,40,1) y <- 2*x+1+5*rnorm(length(x)) hist(y,freq = FALSE) plot(density(y)) }