similar to: random sequences for rnorm and runif

I retrieve for a few hundred times a group of time series (10-15 ts with 10000 values each), on every group I do some calculation, graphs etc. I wonder if there is a faster method than what presented below to get an appropriate timeseries object. Making a query with RODBC for every group I get a data frame like this: > X ID DATE VALUE 14 3 2000-01-01 00:00:03 0.5726334

Hi, I?d like to ask for some help in writing a loop. My situation is the following: I have a matrix (matrix.A) containing 3 columns and 100 rows. The columns represent parameter estimates a, b, and c. The rows contain different values for these parameter estimates. Each row is unique. I want to insert these parameter estimates into a model (say, y = a + bx + cx^2) and solve for y given a

There appears to be a mild bug, or at least a deficiency, in rnorm. The bug becomes apparent when one looks at extremes of the squares of the values generated by rnorm; rnorm is not generating quite enough extreme values. The R version that I am using is 1.4.1; I never got around to installing 1.5.0, and now since 1.5.1 is about to come out .... However, checking the 1.5.0 release notes

Hello! I am performing coupling of chains in MCMC and I need the same value of seed for two chains. I will show demo of what I want: R code, which might show my example is: niter <- 3 nchain <- 2 tmpSeed <- 123 for (i in 1:niter) { # iterations for (j in 1:nchain) { # chains set.seed(tmpSeed) a <- runif(1) cat("iter:", i, "chain:", j,

This may not actually be an R/Splus problem, but it started off that way ..... ===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+=== Executive summary: ================== Simulations involving extreme value distributions seem to ``work'' when the underlying distribution is exponential(1) or exponential(2) == chi-squared_2, but NOT when the underlying distribution is

Dear Tomas,Thank you.Regarding the "unnumbered" arguments, i.e. sprintf('%f %f', 1, 2, 3). This was the case I wanted to report, here a warning can be very useful.Regarding the "numbered" arguments, that is, sprintf('%$1f %$3f', 1, 2, 3). Here, omission of an argument might be intended, for example, in an application with support for multiple languages.

Another way to avoid the problem is to not redefine variables that are arguments. E.g., > Su3 <- function(u=100, l=u, mu=0.53, sigma2=4.3^2, verbose) { if (verbose) { print(c(u, l, mu)) } uNormalized <- u/sqrt(sigma2) lNormalized <- l/sqrt(sigma2) muNormalized <- mu/sqrt(sigma2) c(uNormalized, lNormalized, muNormalized) } > Su3(verbose=TRUE)

Dear R developers, sessionInfo() below Please have a look at the following two versions of the same function: 1. Intended behavior: > Su1 = function(u=100, l=u, mu=0.53, sigma2=4.3^2) + { + print(c(u, l, mu)) # here, l is set to u?s value + u = u/sqrt(sigma2) + l = l/sqrt(sigma2) + mu = mu/sqrt(sigma2) + print(c(u, l, mu)) + } > > Su1() [1] 100.00 100.00 0.53 [1]

Dear R developers, I am wondering if this should raise an error or a warning. > sprintf('%.f, %.f', 1, 2, 3) [1] "1, 2" I am aware that R has ?numbered? sprintf arguments (sprintf('%1$.f', ?), and in that case, omissing of specific arguments may be intended. But in the usual syntax, omission of an argument is probably a mistake. Thank you for your consideration.

Dear Bill, All makes perfect sense (including the late evaluation). I actually discovered the problem by looking at old code which used your proposed solution. Still I find it strange (and, hnestly, I don?t like R?s behavior in this respect), and I am wondering why u is not being copied to L just before u is assigned a new value. Of course, this would require the R interpreter to track all these

Dear R developers, The following command produces an interaction plot with lwd=2. interaction.plot(c(1, 2, 1, 2), c(1, 1, 2, 2), 1:4, lwd=2) In the legend, however, lwd seems to be 1, which does not seem to be intended behavior. Probably the lwd is not correctly forwarded to legend: from the interaction.plot source: legend(xleg, yleg, legend = ylabs, col = col, pch = if (type %in%

Dear R-user Suppose I have the following data y=c(2,1,5,8,11,3,1,7,50,21,33,7,60) x=data.frame(y) for(i in 4:nrow(x)) x[i,] =var(x[i-3:i-1,]) I'm trying to get a new variable with the variance of the 3 previous values (just an example) and with NA in the three first positions. I know that my for() is wrong but I'm not able to find my error. Any idea? Thanks, Marlene.

Dear R developers, I would like to have R choose the limits of the y-axis semi-automatically, e.g., zero should be included, but the maximum should be chosen depending on the data. Examples: plot(1:10, 1:10) # selects min and max automatically plot(1:10, 1:10, ylim=c(1, 10)) # manual definition plot(1:10, 1:10, ylim=c(0, Inf)) # this would be a nice feature, i.e. lower y limit = 0 defined

This is a follow-up to a query that was posted regarding some problems that emerge when running anova analyses for cox models, posted by Mathias Gondan: Matthias Gondan wrote: >* Dear List,*>**>* I have tried a stratified Cox Regression, it is working fine, except for*>* the "Anova"-Tests:*>**>* Here the commands (should work out of the box):*>**>*

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Hello, One way of preventing that is to use ?force. Just put force(l) right after the commented out print and before you change 'u'. Hope this helps, Rui Barradas Citando Matthias Gondan <matthias-gondan at gmx.de>: > Dear R developers, > > sessionInfo() below > > Please have a look at the following two versions of the same function: > > 1. Intended

My interpretation of the relation between 1-way ANOVA and Wilcoxon's test (wilcox.test() in R) is the following. 1-way ANOVA is to test if two or multiple distributions are the same, assuming all the distributions are normal and have equal variances. Wilcoxon's test is to test two distributions are the same without assuming what their distributions are. In this sense, I'm wondering

Dear list, Is anyone aware of a library for sample size calculation in R, similar to NQuery? I have to give a course in this area, and I would like to enable the students playing around with this. Best wishes, Matthias

Dear list, Apparently, there is no function like sscanf in R. I have a string, "Condition: 311", and I would like to read out the number and store it to a numeric variable. Is there an easy way to do this? Best wishes, Matthias --

Dear R-Users, Is anyone aware of a significance test which allows demonstrating that one distribution dominates another? Let F(t) and G(t) be two distribution functions, the alternative hypothesis would be something like: F(t) >= G(t), for all t null hypothesis: F(t) < G(t), for some t. Best wishes, Matthias PS. This one would be ok, as well: F(t) > G(t), for all t null