similar to: AW: Tracking progress during iterations

Dear all, I am using R 2.1.0 for Windows on XP SP2. In a loop such as below, we can track the progress by a screen print display as below (in S): for (i in 1:10){ print(paste("Starting simulation run no...",i)) result<-c(result,mean(data)) #A dummy example of the simulation to be done } In S-plus environment, the statements Starting simulation run no...1 Starting simulation

Sorry, to solve your question I had tried: data(faithful) kmeans(faithful[c(1:20),1],10) Error: empty cluster: try a better set of initial centers But when I run this a second time it will be ok. It seems, that kmeans has problems to initialize good starting points, because of the random choose of these starting initial points. With kmeans(data,k,centers=c(...) the problem can be solved.

Hello, I am writing a package with a collection of several models. In order to allow users to play interactively with the models (in contrast to hacking lengthy scripts), I want to put all what is needed to run a particular model into a single list object for each model. Then there will be a collection of functions to run the model or to modify parameters, time steps, integration method ...,

Excellent. That was very helpful. Now I have full control about my NA?s :-) Thank you very much!!! Matthias > > Here is an another way > > count <- is.na(x) + is.na(y) > which( count == 1, arr.ind=TRUE ) > > 'count' gives you the number of missing values at for each > row and column. Then you can find out how many occurances of > both missing,

Dear all, I have several response variables estimated from some simulations, and I would like to identify the thresholds for trend changes. Fro example, below I forced two different response behaviours and on x is time unit. x<-1:1500 y<-x/exp(x^0.2) smaller15<-y[y<15] y<-ifelse(y<15,y+rnorm(length(smaller15)), 15+rnorm(1000-length(smaller15), 0, 0.9))

I have the following matrix: > dat [,1] [,2] [,3] [,4] foo 0.7574657 0.2104075 0.02922241 0.002705617 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 and given this:

Type e.g.: quantile(x,0.1) or Quantile(x,0.8) Which calculates the 10th and 80th quantile Matthias Templ -----Urspr??ngliche Nachricht----- Von: arinbasu at softhome.net [mailto:arinbasu at softhome.net] Gesendet: Montag, 26. April 2004 09:29 An: r-help at stat.math.ethz.ch Betreff: [R] Looking for help in calculating percentiles Hi All: I am working with a dataset on Arsenic toxicity,

Hi, I make this model: > moths.m6 <- glm(nind~metros+especie+habitat+metros:habitat+habitat:especie,family=poisson) > anova(moths.m6,test="F") Analysis of Deviance Table Model: poisson, link: log Response: nind Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL 81 488.32

Dear Hank, Last question first: really, only you can say for sure if 4e-281 and 5e-11 are small enough; it depends on the units you measure your state variables in. However, this strategy cannot get the state variables to exactly 0. Obviously, you could get closer to 0.0 faster by setting the derivatives even larger in absolute value. You may run into problems with the solver when the

Hi, I'm using Dovecot 1.2.14, and I've read PostLoginScripting on the wiki. Is there any way to make Dovecot use the same username/password for database access as userdb and passdb queries? Specifying the password with -p doesn't seem like a good idea, so I'm wondering if it can be handled by Dovecot directly. Or is it possible to track last logins with a plugin similar to

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,

Thanks to Dieter Menne and Spencer Graves I started to get my way through lsoda() Now I need to use it in with nls() to assess parameters I have a go with a basic example dy/dt = K1*conc I try to assess the value of K1 from a simulated data set with a K1 close to 2. Here is (I think) the best code that I've done so far even though it crashes when I call nls()

Hello! I'm a college undergrad desperately trying to finish up my thesis. I have a dataset on the distribution of a grassland bird from the Breeding Bird Survey. I have a very straightforward and simple version of the logistic growth model to describe changes in this bird's abundance over time. The variables are the natural birth rate, mortality, and carrying capacity. The equation itself

Hi r-users;   I have this values: eign_val <- c(137.810447,3.538721,2.995161,1.685670) alp    <- 1.6549 ;  lamda <- eign_val lamda_m <- min(lamda)   First I calculated manually: delta0 <- 1 delta1 <- alp*delta0*(4-lamda_m*(1/lamda[1]+1/lamda[2]+1/lamda[3]+1/lamda[4]))  delta1 delta2 <- (alp/2)*(delta1*(delta1/alp) + delta0*((1-lamda_m/lamda[1])^2+

On Tue, 23 Jun 2015, Jakub Jelen wrote: > > On 05/29/2015 09:12 AM, Damien Miller wrote: > > Hi, > > > > OpenSSH 6.9 is almost ready for release, so we would appreciate testing > > on as many platforms and systems as possible. This release contains > > some substantial new features and a number of bugfixes. > Tested basic configuration on Fedora 22. With

Like the idea - how''d you call it, "railix"? ;) .mitro -----Urspr?ngliche Nachricht----- Von: rails-bounces@lists.rubyonrails.org [mailto:rails-bounces@lists.rubyonrails.org] Im Auftrag von David Harrison *EXTERN* Gesendet: Freitag, 20. J?nner 2006 11:25 An: rails@lists.rubyonrails.org Betreff: Re: [Rails] Linux Suggestions Is this thread an indication that what would be

Hi, I have a density that I need to get MLEs from, which includes definite integrals both in the denominator and in the numerator of the density function. It looks like the outcome depends on the initial values given. My program is shown below: library(circular) ######################################## 4 parameters ######################################## z<-rvonmises(100,0,1)

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

Hello...i want to find the empirical rate for type 1 error using fixed trimmed mean. To make it easy, i'm referring to journal given by this website http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf. I already run the programme and there is no error in it but i got zero for the empirical rate of type 1 error. The empirical rate for the type 1 error given in the journal

I would like to define a function using symbols, but freeze the symbols at their current values at the time of definition. Both symbols referring to the global scope and symbols referring to arguments are at issue. Consider this (R 2.4.0): > k1 <- 5 > k [1] 100 > a <- function(z) function() z+k > a1 <- a(k1) > k1 <- 2 > k <- 3 > a1() [1] 5 > k <- 10 >