similar to: Readline

Hi, I've seen various posts on this question, but still can't get the code right. If I run the following code one line at a time, it works fine. If I run it together as a block, however, it doesn't wait for the input and gives an error. There must be a way to have are pause/wait for an answer, but I can't seem to find it. Thanks! J Code: choosefunction <-

Hi r-users, I would like to solve system of nonlinear equation using BBsolve function and below is my code.  I have 4 parameters and I have 4 eqns. mgf_gammasum <- function(p) { t  <- rep(NA, length(p)) mn <- 142.36 vr <- 9335.69 sk <- 0.8139635 kur <- 3.252591 rh  <- 0.896 # cumulants k1 <- p[1]*(p[2]+p[3]) k2 <- p[1]*(2*p[2]*p[3]*p[4] +p[2]^2+p[3]^2) k3 <-

Here is a drop-in replacement for the R incomplete beta function. src/math/pbeta.c It is a slightly modified version of the cephes library one from Netlib. In the few cases I tried it seems to give at least 14 digit agreement with the one in S-PLUS (its hard to get more). I'm not sure what performance is like. I'd like to know if it helps with some of the problems which have been

Dear Michael, One key is adjustment of nls optimizer tolerance. I notice it has to be higher than usual, but, I recovered your noisy "known" parameter values with an error of K1 (-7%) and k1 (-6%): #### Miller problem with Dalgaard modifications ## Linares 1/22/2004 ## Solution 1 nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=K1, k2=k2))[,2], data=C1.lsoda,

I've noticed that install.packages() [https://svn.r-project.org/R/trunk/src/library/utils/R/packages.R] and update.packages() [https://svn.r-project.org/R/trunk/src/library/utils/R/packages2.R] sometimes output to stdout and sometimes to stderr. It looks like stderr is used (e.g. via cat()) when the message is part of querying the user, e.g. update.packages <- function(lib.loc = NULL,

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Hi there, I'm looking for someone who can give me some hints how to make a nice levelplot. As an example, I have the following code: # create some example data # -------------------------------------- xl <- 4 yl <- 10 my.data <- sapply(1:xl, FUN = function(x) { rnorm( yl, mean = x) }) x_label <- rep(c("X Label 1", "X Label 2", "X Label 3", "X

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

I'm getting this error on the attached code and breaking my head but can't figure it out. Any help is much appreciated. Thanks, Vince CODE: library(deSolve) Res_DAE=function(t, y, dy, pars) { with(as.list(c(y, dy, pars)), { res1 = -dS -dES-k2*ES res2 = -dP + k2*ES eq1 = Eo-E -ES eq2 = So-S -ES -P return(list(c(res1, res2, eq1, eq2))) }) } pars <- c(Eo=0.02,

Vince, When that happens, one possible reason is that your DAE is of index > 1, which cannot be solved by daspk. The solver radau, also from deSolve can handle DAEs up to index 3, but you need to rewrite the problem in the form M*y' = f(x,y), where M is a mass matrix. If you do that for your problem, and solve it with radau, then radau complains that the "matrix is repeatedly

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

Hi there, I have a question concerning the behaviour of the colouring with levelplot. (I hope, I manage to explain) If I give the parameters "at" and "col.regions" like this: at <- c(1,2,3,4,5,6) col.regions <- c("blue","blue","blue","yellow","yellow","yellow") Which color would have the value 3.5? I would

Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice fit or not. So I have to apply them to observation data. But how does the method know about my observed

Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =

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A few day ago, I was looking for an answer to my question but didn't get one. Anybody who can help now? Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice