similar to: Cost-benefit/value for money analysis

Friends I'm trying fit a survival model by maximum likelihood estimation using this function: flver=function(a1,a2,b1,b2) { lver=-(sum(st*log(exp(a1*x1+a2*x2)))+sum(st*log(hheft(exp(b1*x1+b2*x2)*t,f.heft))) -(exp(a1*x1+a2*x2)/exp(b1*x1-b2*x2))*sum(-log(1-pheft(exp(b1*x1+b2*x2)*t,f.heft)))) } emv=mle(flver,start=list(a1=0,a2=0,b1=0,b2=0)) where hheft and pheft are functions defined in

Hi folks, Suppose I create the character vector charvec by > charvec<-c("a1.b1","a2.b2") > charvec [1] "a1.b1" "a2.b2" and then I use strsplit on charvec as follows: > splitlist<-strsplit(charvec,split=".",fixed=TRUE) > splitlist [[1]] [1] "a1" "b1" [[2]] [1] "a2" "b2" I was wondering

Hi I searched in the list and only found questions without answers e.g. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/19955.html : Is there a way to get the same results with lme as with aov with Error()? Can anybody reproduce the following results with lme: id<-c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5)

Dear Wiz[R]ds, I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun, and others. We are almost there. For whatever reason, I can't change the symbol from a circle to a triangle in the upright posture plots. Any ideas? I have included the problem in full. # tritiated (3H)-Norepinephrine(NE) disappearance from plasma # concentrations supine and upright # supine datasu <-

For an application, I have formulas defined in terms of a left Kronecker product of matrices, A,B, meaning A \otimes_L B = {A * B[i,j]} -- matrix on the left multiplies each element on the right. The standard kronecker() function is the right Kronecker product, A \otimes_R B = {A[i,j] * B} -- matrix on the right multiplies each element on the left. The example below shows the result of

Hi, I have a three dimensional array, e.g., my.array = array(0, dim=c(2,3,4), dimnames=list( d1=c("A1","A2"), d2=c("B1","B2","B3"), d3=c("C1","C2","C3","C4")) ) what I would like to get is then a dataframe: d1 d2 d3 value A1 B1 C1 0 A2 B1 C1 0 . . . A2 B3 C4 0 I'm sure there is one function to do

Hello all, I have a matrix whose column names look like a1 a2 b1 b2 b3 c1 c2 1 2 3 7 1 3 2 4 6 7 8 1 4 3 Now, I can have any number of a's. not just two as shown above and same goes for b's and c's. I need to extract all the a's columns and put them in another matrix, extract all b's columns and put them in some matrix

Dear R users, Basically, from the following arbitrary data set: a <- data.frame(id=c(c("A1","A2","A3","A4","A5"),c("A3","A2","A3","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),6),rep("Life",4))) > a

Dear Wiza[R]ds, I am very grateful to Duncan Murdoch for his assistance with this problem. His help was invaluable. However, the problem has become a little more complicated for me. Now, in each plot, I need to plot the observed and fitted values of a supine and upright posture experiment. Here is what I have and how far I got. # tritiated (3H)-Norepinephrine(NE) disappearance from plasma #

Hello, I get some file names by list.files(). These names are in alphabetical order. I want to change it to logical numeric order. Example: > fileNames <- c("A10", "A1", "A2", "B1", "B2", "B10") > sort(fileNames) [1] "A1" "A10" "A2" "B1" "B10" "B2" I want to have:

I have a dataset that is basically structureless. Its dimension varies from row to row and sep(s) are a mixture of tab and semi colon (;) and example is HEADER1 HEADER2 HEADER3 HEADER3 A1 B1 C1 X11;X12;X13 A2 B2 C2 X21;X22;X23;X24;X25 A3 B3 C3 A4 B4 C4 X41;X42;X43 A5 B5 C5 X51 etc., say. Note that a blank

I have two data frame A and B adn want to cross them. A has format as: a1 a2 a3 1 2 3 2 3 1 1 3 2 ... B: b1 b2 1 2 2 1 ... the combine result shall be something like a1 a2 a3 b1 b2 1 2 3 1 2 1 2 3 2 1 2 3 1 1 2 2 3 1 2 1 1 3 2 1 2 1 3 2 2 1 .... is there a function able of doing this instead of loops? Thanks, Sun

I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + a3(b1+b2+b4) + a4(b1+b2+b3) or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i * b_i would appreciate some help. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678.html Sent from the R

Hi, I have a vector of strings like: c("a1b1","a2b2","a1b2") which I want to spilt into two parts like: c("a1","a2","a2") and c("b1","b2,"b2"). So there is always a first part with a+number and a second part with b+number. Unfortunately there is no separator I could use to directly split the vectors.. Any idea

Hi, I am trying to aggregate the sum of my test data.frame as follow: testDF <- data.frame(v1 = c("a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "c", "c", "c", "c", "c", "d", "d", "d", "d",

Dear R users: I have some difficulties analizing data with mixed effects NLME and the last version of R. More concretely, I have a repeated measures design with a single group and 2 experimental factors (say A and B) and my interest is to compare additive and nonadditive models. suj rv A B 1 s1 4 a1 b1 2 s1 5 a1 b2 3 s1 7 a1 b3 4 s1 1 a2

Dear all, I'm struggling to find the best (set of?) function(s) to do repeated measures logistic regression on some data from a psychology experiment. An artificial version of the data I've got is as follows. Firstly, each participant filled in a questionnaire, the result of which is a score. > questionnaire ID Score 1 1 6 2 2 5 3 3 6 4 4 2 ...

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

I have huge list of edges with weights. a1 b1 w1 a2 b2 w2 a3 b3 w3 a1 b1 w4 a3 b1 w5 I have to convert it into 2 dim matrix b1 b2 b3 a1 max(w1,w4) 0 0 a2 0 w2 0 a3 w5 0 w3 if edges repeated take the maximum weights. How do this efficiently without using for loop? Any idea. thanks Avi [[alternative

Hi, I am new to llvm with a particular interested in the optimization area, specially on SLP. While working through the tutorial, I ran this example [1] with the hope to see SLP vectorization in action but for some reason, I do not see it on the LLVM assembly as seen below. Is there anything I am missing? I am using Clearlinux as build machine and this has clang version 9.0.0.