similar to: array vs matrix vs dataframe?

Hello! I have a problem with integrate() in my function nctspa(). Integrate produces an error message "evaluation of function gave a result of wrong length". I don't know what that means. Could anyone suggest me what is wrong with my function? These are the examples of function calls that work OK: nctspa(a=1:10,n=5) nctspa(a=1:10, n=5, mu=2, theta=3, renorm=0) This does not work:

Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-

Hi R-users,   I have this code here: library(numDeriv)   fprime <- function(z) { alp  <- 2.0165;   rho  <- 0.868;   # simplified expressions   a      <- alp-0.5   c1     <- sqrt(pi)/(gamma(alp)*(1-rho)^alp)   c2     <- sqrt(rho)/(1-rho)   t1     <- exp(-z/(1-rho))   t2     <- (z/(2*c2))^a   bes1   <- besselI(z*c2,a)   t1bes1 <- t1*bes1   c1*t1bes1*t2 }   ## Newton

UseRs, actuar is a package for Actuarial Science. A rather preliminary version (0.1-3) of the package has been available on CRAN since February 2006. We now announce the immediate availability of version 0.9-2 sporting a large number of new features. Non actuaries behold! There can be some features of interest for you, especially those related to new probability distribution and to the

UseRs, actuar is a package for Actuarial Science. A rather preliminary version (0.1-3) of the package has been available on CRAN since February 2006. We now announce the immediate availability of version 0.9-2 sporting a large number of new features. Non actuaries behold! There can be some features of interest for you, especially those related to new probability distribution and to the

I would like to build a package for the HT HG-U133+ PM arrays from affy, but I can't find any good documentation on how to go about it. Naively using makecdfenv's make.cdf.package() causes R to seg-fault. I'm unfamiliar with the CDF format as such, but I'm guessing that it's changed somewhat because the PM arrays no longer have P/A and mismatches. I'm looking to build

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

My machine has 8GB memory. I had quit all other programs that might take a lot of memory when I try the script (before I post the first message in this thread). The cdf file is of only 741 MB. It is strange to me to see the error. On Mon, Dec 28, 2009 at 2:38 AM, Wolfgang Huber <whuber at embl.de> wrote: > Dear Peng Yu > > how big is the RAM of your computer? You could try with

Hello R-Users, I'm new to R so I apologize in advance for any big mistake I might be doing. I'm trying to fit a set of samples with some probabilistic curve, and I have an important question to ask; in particular I have some data, from which I calculate manually the CDF, and then I import them into R and try to fit: I have the x values (my original samples) and the y values

Please give me some help, many thanks. I graphed a expected CDF line of a binomial distribution on a graph, And I have some observed points (observed CDF) from 4 groups fall on the smooth CDF line. I cannot really get the legend I want legend ('topleft', c('a, 'b', 'c', 'd', 'expected CDF'), col=c(1,2,3,4), pch=c(0,1,2, 3, '' ),

Hi, I need to write two inequalities depend on cumulative distribution (CDF) of beta binomial distribution where alpha and beta are unknown and need to find them. CDF of betabinomial(2,10,alpha,beta) <0.3<=CDF of betabinomial(3,10,alpha,beta) and CDF of betabinomial(5,10,alpha,beta) <0.8<=CDF of betabinomial(6,10,alpha,beta) How I can do that using r studio package? I tried to do

Hello R Helpers, Business - 64 bit windows 7, R 2.11.1 I am trying to match the character contents of one list, called 'exclude', to those of a second list, called 'dataset' dataset is a list of file names with folder locations, and looks like this when called: > dataset [1] "A/10-10-29a-13.cdf" "A/10-10-29a-14.cdf" "A/10-10-29a-15.cdf"

Hi r-users,   I hope somebody can help me to understand the error message.  Here is my code; ## Newton iteration newton_gam <- function(z) { n   <- length(z)   r   <- runif(n)   tol <- 1E-6   cdf <- vector(length=n, mode="numeric")   fprime <- vector(length=n, mode="numeric")   f   <- vector(length=n, mode="numeric")     for (i in 1:1000)   {

Folks, Normally, in a data frame, one observation counts as one observation of the distribution. Thus one can easily produce a CDF and (in Splus atleast) use cdf.compare to compare the CDF (BTW: what is the R equivalent of the SPlus cdf.compare() function, if any?) However, if each point should not count equally, how can I weight the points before comparing the distributions? I was thinking of

R version: 0.62.1 (June 15, 1998) I just very naively attempted to grab the 'bicreg' package for Bayesian model selection from the StatLib library, and get it running under R. I've hit a brick wall very quickly, and as an R novice I'm not sure where to go next. Here's what happened: > bicreg(as.matrix(hiv[,c(-17,-18)]),as.matrix(hiv[,18])) Error: invalid formula >

I have a table like the following. I want to fit Cm to Vm like this: Cm ~ Cl+Q1*b1*38.67*exp(-b1*(Vm-Vp1)*0.03867)/(1+exp(-b1*(Vm-Vp1)*0.03867))^2+Q2*b2*38.67*exp(-b2*(Vm-Vp2)*0.03867)/(1+exp(-b2*(Vm-Vp2)*0.03867))^2 I use nls, with start=list(Q1=2e-3, b1=1, Vp1=-25, Q2=3e-3, b2=1, Vp2=200). But I always get 'singlular gradient' error like this. But in SigmaPlot I can get the result. How

Dear all, I have define a discrete distribution P(y_i=x_i)=p_i, which I want to plot a CDF plot. However, I can not find a function in R to draw it for me after searching R and R-archive. I only find the one for the sample CDF instead my theoretical one. I find stepfun can do it for me, however, I want to plot some different CDF with same support x in one plot. I can not manage how to do it with

Hello, fellow Rdicts, I have the code for the program below. I need to integrate a function of "x" and "p". I use integrate to integrate over "x" and pass "p" as an additional argument. "p" is specified and given default value in the argument list. Still, integrate() cannot read "p", unless I explicitly insert a numeric value in the

Good Afternoon, I'm trying to create a cdf plot, with the following code. It works well, but I have little doubt, if you can help solve. When I create the plot, like the graph line would still not appear with point #cdf x<-table(Dataset$Apcode) View(s) hist(s) *plot(ecdf(x))* x<-1 37607 2 26625 3 5856 4 25992 5 30585 6 16064 7 9850 .. ... .. 186 52 -- View this message in

1. There is, I believe, some redundant code in the calculation of the test statistic in ks.test in the package ctest. Lines 34-37 of the code read x <- y(sort(x), ...) - (0:(n - 1))/n STATISTIC <- switch(alternative, two.sided = max(abs(c(x, x - 1/n))), greater = max(c(x, x - 1/n)), less = -min(c(x, x - 1/n))) Lines 35-37 could read