similar to: Access Object's Objects HELP

Dear Sir, I am a new user of R and I am doing a tast, which is: find the maximum likelihood estimate of the parameter of Gaussian distribution for generated 100 numbers by using >x=rnorm(100, mean=3, sd=1). I tried to use following Maximum Likelihood function >fn<-function(x) (-50*log((sd(x))^2))-50*log(sqrt(2*pi))-(1/2*((mean(x))^2))*(sum((x-(mean(x))^2)), but it did not work. I am

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Ladies and Gentlemen, In MATLAB, I can write: for J=1:M Y(J+1)=Y(J)+ h * feval(f,T(J),Y(J)); ... In R, I can write above as: for (J in 2:M) { y = y + h * f(t,y) ... } ---- In MATLAB, I can write: for J=1:M k1 = feval(f,T(J),Y(J)); k2 = feval(f,T(J+1),Y(J)+ h * k1 How do I write k2 in R? k1 = f(t,y) k2 = ? Thanks, /oal

Dear WizaRds, Given the experimental data, csdata<-data.frame( time=c(0,1,3,9,20), conc=c(638.697,395.69,199.00,141.58,112.16) ) weighted nls is applied, wt.MM<- function(resp, time,A1,a1,A2,a2) { pred <- A1*exp(-a1*time)+A2*exp(-a2*time) (resp - pred) / sqrt(pred) } # cs.wt <- nls( ~ wt.MM(conc, time,A1,a1,A2,a2), data=csdata,

Hi all: I have a question about multi-comparison. The data is in the attachment. My purpose: Compare the predicted means of the 3 methods(a,b,c) pairwisely. I have 3 ideas: #idea1 result_aov<-aov(y~ method + x1 + x2) TukeyHSD(result_aov) diff lwr upr p adj b-a 0.845 0.5861098 1.1038902 0.0000001 c-a 0.790 0.5311098 1.0488902 0.0000002 c-b -0.055 -0.3138902

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Hi, I am encountering a difficulty I don't understand. Be patient, I'm very new to analysis of variance. If I load this data: example12_7=read.table("http://msemac.redwoods.edu/~darnold/math15/data/chapter12/example12_7.dat",header=TRUE) The run the oneway.test: oneway.test(time~drug,data=example12_7,var.equal=TRUE) I get these results: data: time and drug F = 4.1881, num

Hi r-sers, I have a data of relative frequencies for the interval of 0-20, 20-40,...380-400.  I would like the two data on the same graph using the same x-axis label.  My question is how to get a smooth curve using kernel density code if it possible for this data.   > cbind(rel_obs,rel_gen)           rel_obs rel_gen  [1,] 0.000000000  0.0000  [2,] 0.092534175  0.0712  [3,] 0.105152471  0.1092

hello, here are some problems with ks.test. Do you think these values are consistent? (2 e^{-1} means 0.2 ) data file name& sample size& D & pvalue A&10.000&0.4202&<2.2 e^{-16}\\ B&10.000&04294&<2.2 e^{-16}\\ C &10.000&0.4484&<2.2 e^{-16}\\ D&10.000&0.4569&<2.2 e^{-16}\\ E&50.000&0.4015&<2.2 e^{-16}\\

I have a fairly large set of data with the following attributes: >str(raw.data) `data.frame': 1429 obs. of 16 variables: $ TStamp :`POSIXlt', format: chr "2001-11-25 02:00:00" "2001-11-25 01:55:00" "2001-11-25 01:50:00" "2001-11-25 01:45:00" ... $ iPDT.AHU14.14: num 0.0122 0.0125 0.0120 0.0120 0.0122 ... $ iPDT.AHU14.15: num 0.0121

Hi, I have the following question about creating data frames. I want to create a data frame with 2 components: a vector and a matrix. Let me use a simple example: y <- rnorm(10) x <- matrix(rnorm(150), nrow=10) Now if I do dd <- data.frame(x=x, y=y) I get a data frame with 16 colums, but if, according to the documentation, I do dd <- data.frame(x=I(x), y=y) then str(dd)

Hello all! I have a problem to arrange data in another form. My initial data is like this: 'data.frame': 421 obs. of 58 variables: $ 01A: num NA NA NA NA NA NA NA NA NA NA ... $ 01B: num NA NA NA NA NA NA NA NA NA NA ... $ 03A: num NA NA NA NA NA NA NA NA NA NA ... $ 03B: num NA NA NA NA NA NA NA NA NA NA ... $ 05A: num NA NA NA NA NA NA NA NA NA NA ... $ 05B: num NA NA NA NA

hi all, this is my first hit on the mailing list. hope i m posting in the right place. well, i have a dell vostro 1510 laptop with broadcom wifi adapter. n it doesnt work! :( i installed the wifi drivers with ndiswrapper. iwconfig shows the following. [root at laptop ~]# iwconfig lo no wireless extensions. eth0 no wireless extensions. sit0 no wireless extensions. wlan0

Hola Ruben, Sí precisamente es lo que comentas, en matemáticas no se suele llamar bucketización (este término se emplea más en informática) sino datos agrupados. Pero la idea es la que tu mismo dices. Respecto a las gráficas que has puesto, me han aclarado mucho sobre el tema, gracias. Si realizo lo mismo, por ejemplo con nbucket=1000 sigo obteniendo un p-valor de 1. Es decir, que casi le

I am connecting from a PC to a Linux system running CentOS release 5.5 (Final) and it is extremely slow to render plots to the X11 device. This is not R's fault but I wonder if anyone can offer guidance so I can help the system administrators address the problem. I can connect to the Linux server using a NoMachine NX client for Windows or using X-Win32. I also have access to R running

I have a moderately complex graph with three panels. There are data points plotted, and fitted lines are added using a panel function, which includes "with(alt.data[[which.packet()[1]]]" statements. It all graphs out beautifully, but none of the usual tricks to get the proper legend to plot are working, i.e., using auto.key, key, etc. One message I keep getting is Error in

Hello R-users, I have been trying to fit what I think is a simple mixed-effects model using lmer (from lme4), but I've run into some difficulty that I have not been able to resolve using the existing archives or Pinheiro and Bates (2000). I am measuring populations (of birds) which change with time at a number of different sites. These sites are grouped into regions. Sites are not measured

Hi, I am trying to construct a logistic regression model from my data (104 patients and 25 events). I build a full model consisting of five predictors with the use of penalization by rms package (lrm, pentrace etc) because of events per variable issue. Then, I tried to approximate the full model by step-down technique predicting L from all of the componet variables using ordinary least squares

Hi Gyounghwa Kim, try pasting C++ code into http://llvm.org/demo/ in order to see the LLVM IR that llvm-g++ turns it into. That way you will see how this can be done. Best wishes, Duncan.

On Fri, Mar 6, 2015 at 2:06 PM, Dmitriy Serov <serov.d.p at gmail.com> wrote: > Hello. > > Asterisk 13.2. > I transfer configs from chan_sip to res_pjsip. > In chan_sip i have "match_auth_username=yes" and have nothing in pjsip. > > I have a lot of endpoints and registrations on same SIP server. And it's > problem in pjsip now. Is not it? > > I