similar to: how to evaluate character vector within pnorm()

Hi, I have the following problem. Calling wireframe within a loop results into an empty window(s) #generate some data temp = expand.grid(A = 1:3,B = 1:3) temp = cbind(temp, y1 = rnorm(9)) temp = cbind(temp, y2 = runif(9)) #plot y1 and y2 in two different windows for(i in 1:2) { wireframe(y1 ~ A*B, temp, shade =T) windows() wireframe(y2 ~ A*B, temp, shade =T) } #However, calling it twice

#Hi, # #somebody knows how to remove the outer box around a wireframe and reduce the height # # test = data.frame(expand.grid(c(1:10), c(1:10))) z = test[,1] + test[,2] test = cbind(test, z) names(test) = c("x", "y", "z") require(lattice) wireframe(z ~ x*y, data = test, par.box = c(col = "transparent") ) #not this one but the remaining outer box.

I have a generalized linear model to solve. I used package "geepack". When I use the correlation structure "unstructured", I get a messeage that- R GUI front-end has stopped working. Why this happens? What is the solution? The r codes are as follows: a<-read.table("d:/bmt.txt",header=T")

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

I stumbled across this and I am wondering if this is unexpected behavior or if I am missing something. > pnorm(-1.0e+307, log.p=TRUE) [1] -Inf > pnorm(-1.0e+308, log.p=TRUE) [1] NaN Warning message: In pnorm(q, mean, sd, lower.tail, log.p) : NaNs produced > pnorm(-1.0e+309, log.p=TRUE) [1] -Inf I don't know C and am not that skilled with R, so it would be hard for me to look into

Hello, I want to import R-output via Rserve to Java, especially for the function ctree from the package party. Rserve is working properly. Yet, I only get the predictions with the Java code try{ RConnection c = new RConnection(); ... c.voidEval("modell <- ctree(...)"); REXP y = c.eval("nodes(modell,1)[[1]]$prediction"); ...

As to the function"pnorm",the default degree of freedom(df) is infinite. I wanna know how to set the df as I want. Help on pnorm doesn't have df setting.The only choice are:"mean, sd, lower.tail, log.p",but no df. For instance: sample size=6 df=6-1=5 t value=9.143 I wanna to the corresponding p value by using function "pnorm". How can I do it? Thanks a lot

Hi, I've read in a csv file (test.csv) which gives me the following table: Hin1 Hin2 Hin3 Hin4 Hin5 Hin6 HAI1 9534.83 4001.74 157.16 3736.93 484.60 59.25 HAI2 13272.48 1519.88 36.35 33.64 46.68 82.11 HAI3 12587.71 5686.94 656.62 572.29 351.60 136.91 HAI4 15240.81 10031.57 426.73 275.29 561.30 302.38 HAI5 15878.32 10517.14 18.93 22.00 16.91

Dear R people The function below should be decreasing, convex, and tend to zero when x tends to infinity. curve((1-pnorm(x))/dnorm(x),from=0, to=9) >From the plot we see that for x between 8.0 and 8.3 the function is fluctuating. As far as I understand, this is due to the function pnorm() not being sufficiently accurate in the tails. I am using pnorm() in a way that has probably not been

Hi, I would have thought that these two constructions would produce the same result but they do not. Resp <- rbinom(10, 1, 0.5) Stim <- rep(0:1, 5) mm <- model.matrix(~ Stim) Xb <- mm %*% c(0, 1) ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) pnorm(as.vector(Xb), lower.tail = Resp, log.p = TRUE) > ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) [1] -0.6931472 -1.8410216

Dear R list, I calculated a two-sided p values according to 2*(1-pnorm(8.104474)), which gives 4.440892e-16. However, it appears to be 5.30E-16 by a colleague and 5.2974E-16 from SAS. I tried to get around with mvtnorm package but it turns out to be using pnorm for univariate case. I should have missed some earlier discussions, but for the moment is there any short answer for a higher

Full_Name: James Michael Rath Version: all (I think) OS: doesn't matter Submission from: (NULL) (129.116.226.162) The code for pnorm in R was adapted from a Fortran library published in the ACM TOMS journal. The published version had a typographical error, though, which was pointed out in a second article published three years after the original. The error was that a macro/variable named

Full_Name: Julian Faraway Version: R-Release (June 15) OS: Linux (redhat) Submission from: (NULL) (141.211.66.172) R fails to compile on the current released version. Some constants are undefined in pnorm.c. It appears that adding #include "nmath.h" to pnorm.c solves this problem. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list

Hi, I've been looking for Image Processing packages. Thresholding, Edge Filters, Dct, Segmentation, Restoration. I'm aware, that Octave, Matlab etc. would be a good address but then I'm missing the "statistical power" of R. Does anybody know of packages, projects etc. Comments on wether the use of R for such matters is useful are welcome. Greetings Thomas Kaliwe

Dear R users Is there any way to increase the decimal accuracy for the normal probability distribution? When one needs an accurate p-value for instance this is provided by pnorm(10,lower.tail=F) [1] 7.619853e-24 However, what happens when instead of a P[X<x], a more accurate P[X>=x] is the objective. Thank you in advance for your responses. Dimitris [[alternative HTML version

What would be the R formulae for a two-sided test? I have a formula for a one-sided test: powertest <- function(a,m0,m1,n,s){ t1 = -qnorm(1-a) num = abs(m0-m1) * sqrt(n) t2 = num/s pow = pnorm(t1 + t2) } Would you pls let me know if you know of? Thank you, ej

I have some questions about the use of lme(). Below, I constructed a minimal dataset to explain what difficulties I experience: # two participants subj <- factor(c(1, 1, 1, 1, 2, 2, 2, 2)) # within-subjects factor Word Type wtype <- factor(c("nw", "w", "nw", "w", "nw", "w", "nw", "w")) # within-subjects factor

--=-YFjXKq8/D/t1qWmIzQ9D Content-Type: text/plain Content-Transfer-Encoding: quoted-printable I was going over the source in src/nmath/pnorm.c and noticed a little bug in pnorm_both (in R 1.7.0). The else-if on line 205 covers the entire real line. Seems you want an &&, not an ||. Doesn't make a big difference (you still get a 0 or 1 from extreme starting values) but your log

Dear useRs, What is an efficient way to randomly sample from clustered data such that I get equal representation from each cluster? For example, let's say I want to randomly sample two cases from each cluster created by the "id" variable in the following data frame: > id<-c(rep("100", 4),rep("101", 3), rep("102", 6), rep("103",

Hi, I'm sure there's an easy approach to this issue, I'm just not seeing it. I have a data frame of the following form: Date class subclass count 8/1/2009 A X 1 8/1/2009 B X 2 8/1/2009 A Y 9 8/1/2009 B Y 3 8/2/2009 A X 1 8/2/2009 B X 5 8/2/2009 A Y