similar to: matrix with Loop

Hello, ?can i implement this as 10% censored data where t gives me failure and x censored. Thank you p=2;b=120 n=50 set.seed(132); r<-sample(1:50,45) t<-rweibull(r,shape=p,scale=b) t set.seed(123);? cens <- sample(1:50, 5)? x<-runif(cens,shape=p,scale=b)? x Chris Guure Researcher, Institute for Mathematical Research UPM

Dear all, ?The code below is used to generate interval censored data but unfortunately there is an error with the ifelse which i am not able to rectify. ?Can somebody help correct it for me. Thank you t<-rexp(20,0.2)? v<-c(0,m,999)? y<-function(t,v){ ? z<-numeric(length(t (( ? ? s<-numeric(length(t (( ? ? ? for(i in 1:length(t)){ ? ? ? ? for(j in 1:length(v-1))? ? ? ? ? { ifelse

Dear All, Can you help me, with the code below how do I obtain the fisher information from it. Is my q<-replicate(1000,x) the right way to do simulation. thank you. x<-rweibull(100,0.8,1.5) q<-replicate(1000,x) z<-function(p){ beta<-p[1] eta<-p[2] log1<-(n*log(beta)-n*beta*log(eta)+(beta-1)*sum(log(x))-sum((x/eta)^beta)) return(-log1) } zz<-optim(c(0.5,0.5),z) zz Chris

Hello, ?When i generate data with the code below there appear NA as part of the generated data, i prefer to have zero (0) instead of NA on my data. Is there a command i can issue to replace the NA with zero (0) even if it is after generating the data?? Thank you library(survival) p1<-0.8;b<-1.5;rr<-1000 for(i in 1:rr){ r<-runif(45,min=0,max=1) t<-rweibull(45,p1,b)

Your function is giving NaN's during the optimization. The R-forge version of optimx() has functionality specifically intended to deal with this. NOTE: the CRAN version does not, and the R-forge version still has some glitches! However, I easily ran the code you supplied by changing optim to optimx in the penultimate line. Here's the final output. KKT condition testing Number of

Hello, I have being trying to estimate the parameters of the?generalized?exponential distribution. The random number generation for the GE distribution is?x<-(-log(1-U^(1/p1))/b), where U stands for uniform dist. The data i have generated to estimate the parameters is right censored and the code is given below; The problem is that, the newton-Raphson approach isnt working and i do not know what

Hello, I wish to?censor 10% of my sample units of 50 from a Weibull distribution. Below is the code for it. I will need to know whether what i have done is correct and if not, can i have any suggestion to improve it? Thank you ?p=2;b=120 n=50 r=45 t<-rweibull(r,shape=p,scale=b) meantrue<-gamma(1+(1/p))*b meantrue cen<- runif(n-r,min=0,max=meantrue) cen Chris Guure Researcher,

Hello, When i run the code below from Weibull distribution with 30% censoring by using optim i get an error form R, which states that Error in optim(start, fn = z, data = q, hessian = T) :? ? objective function in optim evaluates to length 25 not 1 can somebody?help me remove this error. Is my censoring approach correct. n=25;rr=1000 p=1.5;b=1.2 for (i in 1:rr){ q<-c(t,cen)

Hello, I want to estimate the exponential parameter by using?optim?with the following input, where t contains 40% of the data and q contains 60% of the data within an interval. In implementing the code command for optim i want it to contain both the t and q data so i can obtain the correct estimate. Is there any suggestion as to how this can be done. I have tried h<-c(t,q) but it is not working

Hello, I have a matrix (class matrix) composed of GridCell (row and column). The matrix value is the beta diversity index value between two grids. Now I would like to get the average value of each GridCell. Please kindly advise how to make the calculation. Thank you. Elaine The matrix looks like (cited from Michael Friendly) I would like to get the average value of each color. Obs

Hi is anybody aware of a function to calculate a the median for specified columns in a dataframe or matrix - so analogous to rowMeans? Thanks Max --

Hola a todos Estoy intentando hacer un análisis rápido de una serie temporal de datos diarios pero me encuentro con algunos problemas. Me gustaría en primera instancia hacer una regresión lineal pero no encuentro la forma. Tras leer los datos diarios creo un objeto de la clase zoo y sobre éste no puedo utilizar lm(). He leído algo sobre dynlm pero no encuentro la forma. Se agradece

- This avoids problematic "reloc'ed while mapped" messages and some associated corruption as well. Signed-off-by: Maarten Maathuis <madman2003 at gmail.com> --- src/gallium/drivers/nouveau/nouveau_screen.c | 21 +++++++++++++++++++++ src/gallium/drivers/nouveau/nouveau_screen.h | 3 +++ src/gallium/drivers/nouveau/nouveau_stateobj.h | 13 +++++++++++++

Hi all, I'm trying to plot multiple histograms in one plot (cross-validation values of model parameters), but I cannot seem to reduce the margins enough to fit as many of them in as I would like. I'm using split.screen to divide the window into a 5x4 grid, then plotting with hist. I've tried explicitly reducing the margins with par(mar=c(1,1,1,1)), but it doesn't seem to have

Hi I am trying to create a loop which averages replicates in my data. The original data has many rows. and consists of 40 column zz[,2:41] plus row headings in zz[,1] I am trying to average each set of values (i.e. zz[1,2:3] averaged and placed in average_value[1,2] and so on. below is my script but it seems to be stuck in an endless loop Any suggestions?? for (i in 1:length(average_value[,1])) {

--- On Fri, 15/5/09, Amit Patel <amitrhelp at yahoo.co.uk> wrote: > From: Amit Patel <amitrhelp at yahoo.co.uk> > Subject: Help with loops > To: r-help at r-project.org > Date: Friday, 15 May, 2009, 12:17 PM > Hi > I am trying to create a loop which averages replicates in > my data. > The original data has many rows. and consists of 40 column > zz[,2:41]

Thanks Dragos, I assume I will be setting those parameters during initialization of encoder right? Question is, if connection gets too lossy, how will opus adapt to it? Can it automatically shift bitrate down to minimize impact? Mark from IRC suggests that the app has to be aware of the losses and change it on the fly. Has anybody on the list tried this? Kelvin Chua On Wed, Mar 4, 2015 at 5:53

Hi, May be this helps: ?set.seed(24) ?mydata4<- as.data.frame(matrix(sample(1:100,10*38,replace=TRUE),ncol=38)) ?dim(mydata4) #[1] 10 38 ?library(matrixStats) res<-do.call(cbind,lapply(1:400, function(i) {permutation<-sample(mydata4); (rowMeans(permutation[,1:27])-rowMeans(permutation[,28:38]))/(rowSds(permutation[,1:27])+rowSds(permutation[,28:38]))} )) ?dim(res) #[1]? 10 400 A.K.

hi guys, do you have any suggestions on how to connect 2 TE410P via E1? (for simulation and testing purposes) asterisk1 --> TE410P ----> ? ---------> ? ---->TE410P -->asterisk2 -------------- next part -------------- An HTML attachment was scrubbed... URL: http://lists.digium.com/pipermail/asterisk-users/attachments/20030907/698cd499/attachment.htm

I am using libopus for my implementation. I wonder if anybody in the list have any experience on how to make libopus dynamically adjust its bitrate? On Mar 3, 2015 10:42 PM, "Benjamin Schwartz" <benjamin.m.schwartz at gmail.com> wrote: > It sounds like your software isn't adjusting the opus bitrate in response > to network conditions. For example, many WebRTC