similar to: Reshaping data with xtabs reorders my rows

Dear ALL I have this function in R: func_LN <- function(data){ med_ge <- matrix(c(rep(NA,nrow(data)*ncol(data))), nrow = nrow(data), ncol=ncol(data), byrow=TRUE) T <- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) Tdiff<- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) T1<- c(rep(NA,ncol(data)))

Hi Folks, I am dealing with data which have been presented as at each x_i, mean m_i of the y-values at x_i, sd s_i of the y-values at x_i number n_i of the y-values at x_i and I want to linearly regress y on x. There does not seem to be an option to 'lm' which can deal with such data directly, though the regression problem could be algebraically

Hello, I have a concrete statistical question: I have a sample of an univariate mixture of an unknown number (k) of normal distributions, each time with an unknown mean `m_i' and a standard deviation `k * m_i', where k is known factor constant for all the normal distributions. (The `i' is a subscript.) Is there a function in R that can estimate the number of normal distributions k

Dear all, Lets say I have several data frames as follows: > set.seed(42) > dates <- as.Date(c("2010-01-19", "2010-01-20")) > times <- c("09:30:00", "11:30:00", "13:30:00", "15:30:00") > shows <- c("Red Dwarf", "Being Human", "Doctor Who") > > df1 <- data.frame(Date = dates[1],

Dear all, I have to handle a large matrix (1000 x 10001) where in the last column i have a value that all the preceding values in the same row has to be compared to. I have made the following code : # generate a (1000 x 10001) matrix, testm # generate statistics matrix 1000 x 4: qnt <- c(0.01, 0.05) cmp_fun <- function(x) { LAST <- length(x) smpls <- x[1:(LAST-1)] real

Greetings, Sorry, the last message was sent by mistake! Here it is again: I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix

hello for exampl, i have this programme # Generating data which are right truncated library(DTDA) library(splines) library(survival) n<-25 X<-runif(n,0,1) V<-runif(n,0.75,1) for (i in 1:n){ while (X[i]>V[i]){ X[i]<-runif(1,0,1) V[i]<-runif(1,0.75,1) }} res<-lynden(X=X,U=NA, V=V, boot=TRUE) attach(res) temps = time M_i = n.event L_t = res

Dear R-help group, I would like to model directly following random effect model: Y_ik = M_ik + E_ik where M_ik ~ N(Mew_k,tau_k^2) E_ik ~ N(0,s_ik^2) i = number of study k = number of treatment --------------------------------------------------------------------------- I have practiced using the command from 'Mixed -Effects models in S and S-plus'

Dear all, ? I am implementing a stochastic utility model that will eventually make use of multinomial logit. I found that there is a package in R called mlogit. I am not sure whether I have already found the correct package or software. May I ask am I correct? ? Basically, let's say ? I have observations of n outcomes, for each outcome 1<=i<=n, they were selected by a choice from a set

Hi, I am running simulations that does multiple comparisons to control. For each simulation, I need to model 7 nls functions. I loop over 7 to do the nls using try if try fails, I break out of that loop, and go to next simulation. I get warnings on nls failures, but the simulation continues to run, except when the internal call (internal to nls) of the chol2inv fails.

Hello all. I was wondering if there is any way to adjust the denominator degrees of freedom in lme(). It seems to me that there is only one method that can be used. As has been pointed out previously on the list, the denominator degrees of freedom given by lme() do not match those given by SAS Proc Mixed or HLM5. Proc Mixed, for example, offers five different options for computing the

> On May 12, 2018, at 9:42 AM, malika yassa via R-help <r-help at r-project.org> wrote: > > > hello > for exampl, i have this programme > # Generating data which are right truncated > library(DTDA) > library(splines) > library(survival) > n<-25 > X<-runif(n,0,1) > V<-runif(n,0.75,1) > for (i in 1:n){ > while (X[i]>V[i]){ >

Wow ! so many thanks Arun and Rui works like a charm problem solved 2013/4/13 arun <smartpink111@yahoo.com> > Hi, > Try this; > library(reshape2) > res<-dcast(Input,people~place,value.var="time") > res[is.na(res)]<-0 > res > # people beach home school sport > #1 Joe 5 3 0 1 > #2 Marc 0 4 2 0 > #3 Mary

Seems like this should be easy but I'm struggling a bit. How do I rearrange a data frame to go from the first one to the second shown below ? State Date lbs TX 200701 400 TX 200702 650 TX 200703 950 TX 200704 1000 FL 200701 200 FL 200702 300 FL 200703 500 FL 200704 333 NJ 200701 409 NJ 200702 308 NJ 200703 300 NJ 200704 800 Date TX FL NJ 200701 400 200 409 200702 650

Hello! I have a dataset with NA's in some variables (factors), for example: $ P67 : Factor w/ 2 levels "-","+": NA 2 1 NA NA 2 1 1 2 NA ... I need to use 'xtabs' like xtabs(~x$P67) It works well and produces something like this: x$P67 - + 779 1318 but i want to compute NA's too, like this: x$P67 - + NA 779 1318 137 I am trying

Hi all, I cannot get xtabs to count NA's like I expect. Below is a sample session, and note that the last two calls to xtabs() yield exactly the same thing. I am running R-2.5.0 -- if there was a bug in xtabs that got fixed, I would love to know about it. If there is a bug tracker somewhere, I can file it too. It isn't my fault that the server is so ridiculously outdated, but I would

Dear list, I am trying to make a contingency table with xtabs but I am getting a 0 where I expect a 'NA'. Here is a simple example: options(stringsAsFactors = FALSE) rn <- LETTERS[1:4] df1 <- data.frame(r07 = rep(rn, each=4), r08 = rep(rn, 4), value = 1:16) xtabs(value ~ r07 + r08, df1) # Delete the combination [A, C] df1 <- df1[-3,] # Set 'value'

Hello I have a dataset of people spending time in places. But most people don't hang out in all the places. it looks like: > Input<-data.frame(people=c("Marc","Marc","Joe","Joe","Joe","Mary"), + place=c("school","home","home","sport","beach","school"), +

>>>>> Milan Bouchet-Valat <nalimilan at club.fr> >>>>> on Thu, 19 Jan 2017 13:58:31 +0100 writes: > Hi all, > I know this issue has been discussed a few times in the past already, > but Martin Maechler suggested in a bug report [1] that I raise it here. > > Basically, there is no (easy) way of printing NAs for all variables > when calling

Hi list, is it possible convert the xtabs result xtabs(breaks~tension+wool,data=warpbreaks) wool tension A B L 401 254 M 216 259 H 221 169 to a simple matrix or data frame? A B L 401 254 M 216 259 H 221 169 Thanks a lot! Gianandrea -- View this message in context: http://www.nabble.com/From-xtabs-to-matrix-tp24304588p24304588.html