similar to: How to replace NA with zero (0)

Hello! I got something to ask..whether you can help me with the R program...i got this for example 5x4 matrix..and i want to find: ?i) mean for each row of the matrix ii) median for each column of the matrix and i need to do this using a loop function...below is my program..u try to check it for me as the output that i got is not what i desired...thanks.. data<-rnorm(20,0,1)

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

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, 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, ?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

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,

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 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

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, If i write a function as below using log of weibull distribution i do not get the required results in estimating the parameters what do i do, please a/b * (t/b)^a-1 * exp(-t/b)^a n=500 x<-rweibull(n,2,2) z<-function(p) {(-n*log(p[1])+n*log(p[2])- (p[1]-1)*sum(log(x))+(p[1]-1)*log(p[2])+(sum(x/p[2])^(p[1]))  )} zz<-optim(c(0.5,0.5),z) zz [[alternative HTML version deleted]]

Does Surv in psm handle interval2 data? The argument list seems to indicate it does but I get an error. Thanks, Chris # code library('survival') left <- c(1, 3, 5, NA) right <-c(2, 3, NA, 4) Surv(left, right, type='interval2') survreg(Surv(left, right, type='interval2') ~ 1) library('rms') Surv(left, right, type='interval2') # error args(Surv)

Can anybody give me a neat example of interval censored data analysis codes in R? Given that suvreg(Surv(c(1,1,NA,3),c(2,NA,2,3),type="interval2")~1) works why does survreg(Surv(data[,1],data[,2],type="interval2")~1) not work where data is : T.1 T.2 Status 1 0.0000000 0.62873036 1 2 0.0000000 2.07039068 1 3 0.0000000

In splus rexp() and rweibull() are related: > set.seed(153) > rexp(1) [1] 0.0493267 > set.seed(153) > rweibull(1, shape=1) [1] 0.0493267 (you can also try shape =2, then rweibull = sqrt(rexp) ) However, in rw0.64.1 (on Win NT) they are different > .Random.seed <- 1:4 > rexp(1) [1] 1.412030 > .Random.seed <- 1:4 > rweibull(1, shape=1) [1] 2.054032 May be rweibull

- 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 +++++++++++++

hie i'm tryimg to generate two survival data using the following code (I know its ugly ) but it seems to repeat two of the variables can any one tell me whats the porblem. n=20 n1=n/2 n2=n/4 a11=1 ;a12=1.4 ;a21=16 ;a22=a12 * a21 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

Hello all, I have a installation of dovecot 1.2.16, sieve-0.1.18, managesieve-0.11.12 in dovecot-ldap.conf auth_bind=yes user_attrs = irisMailbox=home,fiMailQuotaSize=quota_rule=*:storage user_filter = (|(uid=%u)(mail=%u)) pass_attrs = userPassword=password,irisMailbox=userdb_home,fiMailQuotaSize=userdb_quota_rule=*:storage pass_filter = (|(uid=%u)(mail=%u)) I'm doing some test to

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 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

hie Im trying to calculate the power of the logrank test for different values of rho .I was just wandering whether the following programme would do it. any suggestions are welcome s=50 number=1 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count7=0; count8=0;count9=0 while(s!=0){ n=20 n1=n/2