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Hi, Can you tell me if there are an R package or code for STAR model estimation and test misspecification. If no, how i could do this. Thanks in advance Best regards AJMI Noomen Phd student TUNISIA --------------------------------- [[alternative HTML version deleted]]

Dear all: I have this: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3 Example: m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<- rep(c(5,10,15),c(4,4,4)) data.frame(v,u) This is the result I want but I would like to learn a simpler way to do it. Any clue?

We have a wireless network that uses 802.1x authentication, in which domain joined computers use their machine credentials to connect. Windows machines do this automatically, and until recently Linux computers could join using wicd, wpa-supplicant, and a simple script that would retrieve the machine password with tdbdump. ( specifically tdbdump -k SECRETS/MACHINE_PASSWORD/DOMAIN

On Mon, Mar 19, 2018 at 12:44 PM, Julia Lawall <julia.lawall at lip6.fr> wrote: > > > On Mon, 19 Mar 2018, Arushi Singhal wrote: > > > This patch replace list_entry with list_{next/prev}_entry as it makes > > the code more clear to read. > > Done using coccinelle: > > > > @@ > > expression e1; > > identifier e3; > > type t; >

Hi, My english isn't brilliant and my problem is very difficult to describe but I try ;) My first question is: May I write loop "for" like this or similar - for (i in sth : sth[length(sth)], k in sth_else : length(sth_else) ) - I'd like to have two independent conditions in the same loop "for". My secound question depend on program below. I'd like to write every

Hi, I have a problem with the appearance of legend in my plots. If I specify the legend positions in characters like "topright"..etc, it appears, if i specify it in terms of coordinates like "-1, 1" .. etc, it does not appear. Can anyone help me with this? script - x.date <- as.Date(paste(year, month, day, sep="-")) ts1.n.e3 <- ts(data.nemr.e3[,3]) z1.n.e3

We are communicating with Asterisk via AMI. Running Asterisk version 13.18.5 on an Ubuntu box. If you look at the event response, the Result field is filled with random characters. I'm not sure what to do because that is a completely random result. It makes no sense. We send the following command to asterisk via AMI Action: AGI ActionID: C44415 Channel: SIP/192.168.40.105-00001338

This patch replace list_entry with list_{next/prev}_entry as it makes the code more clear to read. Done using coccinelle: @@ expression e1; identifier e3; type t; @@ ( - list_entry(e1->e3.next,t,e3) + list_next_entry(e1,e3) | - list_entry(e1->e3.prev,t,e3) + list_prev_entry(e1,e3) ) Signed-off-by: Arushi Singhal <arushisinghal19971997 at gmail.com> --- drivers/gpu/drm/drm_lease.c

Hi all, I have a data.frame e3. I use a function on it: n3 <- normalize.qspline(x = e3, samples = 0.07, na.rm = TRUE); The resulting data.frame has lost it's row-names and column names. This is the natural behaviour of the function I am calling, and I can't alter that at the moment. However, the data *is* in the correct order, as it was initially -- the names are just missing in

Hi, After reading https://llvm.org/docs/Vectorizers.html#vectorization-of-function-calls I decided to write the following C++ program: #include <cmath> using v4f32 = float __attribute__((__vector_size__(16))); v4f32 fct1(v4f32 x) { v4f32 y; y[0] = std::sin(x[0]); y[1] = std::sin(x[1]); y[2] = std::sin(x[2]); y[3] = std::sin(x[3]); return y; } v4f32 fct2(v4f32 x) { v4f32 y;

HI Matt, I am trying to replicate this particular problem. We are seeing more frequently where the Event: AsyncAGIExec is never being sent. The two scenarios I have seen in tests yesterday and today... We sendl an AMI action. For example, play a short file or hangup. AMI Events will indicate it did the work, but we never receive the Event: AsyncAGIExec with a result at all. Asterisk debug

Hi, I am translating some Gauss code to R. Gauss has an interesting way of handling constraints. Observe the following code snipplet: e1 = x[.,23] .eq 0; @ remove obs with Regular Hours = 0 @ e2 = x[.,12] .gt 1; @ remove obs with non-regular work status @ e3 = x[.,4] .lt 15; @ remove obs with agricultural and mining industry code (< 15)@ esum = e1 + e2 + e3; e = esum .gt 0; @

Hi, The lab in which I send my samples return the results in a format that is difficult for me to run my analysis. The lab outputs the results where each parameter is its own row and it’s not consistently in the same order (and not each sample is tested for the same suite of variables). e.g. >dataset site parameter value a e2 3 a e1 1 a e3 5 b e3

I am writing to get a better handle on a warning I am getting from a coxph analysis I am doing. I am analysing age of onset of dementia *after* the onset of parkinson disease. My data looks like: age.park age.dem age.death censor x1 x2 x3 x4 1 76 87 88 0 16 33 E3 E3 2 75 84 84 0 33 36 E3 E3 3 77 81 81 1 NA NA

Hello, I am passive reader of both R-devel and R-help mailing lists. I am sending the following comments to r-devel as it seemed more suitable. I am aware that this list uses GNU mailman for the list management. I have my options set that it sends a email digest. One thing I find is that the digest consists of emails that ordered temporarlly. For eg lets say there are two threads t1 and t2

Hi all, Why cov(y, y) only gives one value, and cov(t(y), t(y)) gives 3x3 NA matrix? Here my y is listed below and it is a 3x1 matrix. I am expecting that if I have a random vector y=[y1 y2 y3]', here " ' " denotes a transposition so that y is a column vector, where y1, y2, y3 are independent random variables... then cov(y, y) should be E[ y * y' ] - E[y] * E[y] ',

Hi there, I run two regressions: y = a1 + b1 * x + e1 y = a2 + b2 * z + e2 I want to test against the null hypothesis: b1 = b2. How do I design the test? I think I can add two equations together and divide both sides by 2: y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2). or just y = a3 + 0.5*b1 * x + 0.5*b2 * z + e3 If I run this new regression, I can test against

Hello, I have got a data frame df like this : > df e1 e2 e3 e4 1 1 11 1 21 2 2 12 2 22 3 3 13 3 23 4 4 14 4 24 5 5 15 5 25 6 6 16 6 26 7 7 17 7 27 8 8 18 8 28 9 9 19 9 29 10 10 20 10 30 where e1 ... e3 are vectors I have to select columns which median is in the interval [5;6] ( Here, for instance, e1 and e3) I would not want to use loop... (for, while) I search

I downloaded R v2.1.1 earlier this morning to compile under Fedora Core 4. It compiled without incident, but 'make check' failed. Below is the relevant part of its report. Is this a known problem? I used a locally compiled version of GCC v4.0.0 that reports [kent at d89h102 R-2.1.1]$ gcc -v Using built-in specs. Target: i686-pc-linux-gnu Configured with: ../gcc-4.0.0/configure

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20