similar to: aov with Error and lme

Dear R-community! I still have the problem reproducing the following example using lme. id<-factor(rep(rep(1:5,rep(3,5)),3)) factA <- factor(rep(c("a1","a2","a3"),rep(15,3))) factB <- factor(rep(c("B1","B2","B3"),15)) Y<-numeric(length=45) Y[ 1: 9]<-c(56,52,48,57,54,46,55,51,51) Y[10:18]<-c(58,51,50,54,53,46,54,50,49)

Dear R People: I have the following in a file: resp factA factB 39.5 low B- 38.6 high B- 27.2 low B+ 24.6 high B+ 43.1 low B- 39.5 high B- 23.2 low B+ 24.2 high B+ 45.2 low B- 33.0 high B- 24.8 low B+ 22.2 high B+ and I construct the data frame: > collard.df <- read.table("collard.txt",header=TRUE) > collard.aov <- aov(resp~factA*factB,data=collard.df) >

Hi, I run the following commands. 'A' has 3 levels and 'B' has 4 levels. Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3, B4)? > a=3 > b=4 > n=1000 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > > fr =

#I have a data set that looks like this. A bit more complicated actually with # three factor levels but these calculations need to be done on one factor at a #I then have a set of different rates that are applied #to it. #dataset cata <- c( 1,1,6,1,1,2) catb <- c( 1,2,3,4,5,6) doga <- c(3,5,3,6,4, 0) data1 <- data.frame(cata, catb, doga) rm(cata,catb,doga) data1 # start rates #

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

Hallo! I have data of the following design: NSubj were measured at Baseline (visit 1) and at 3 following time points (visit 2, visit 3, visit 4). There is or is not a treatment. Most interesting is the question if there is a difference in treatment between the results of visit 4 and baseline. (The other time points are also of interest.) The level of significance is alpha=0.0179 (because of an

On Mon, 15 Dec 2003, jw schultz <jw@pegasys.ws> wrote: > OK, first pass on TODO complete. .... This hardlink bug report is nearly 21 months old... So I took a look at it using 2.5.7. See below. > BUGS --------------------------------------------------------------- > > Fix hardlink reporting 2002/03/25 > (was: There seems

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

Read the data using scan(): # # a1 a2 a3 a4 # ------------- ------------- ------------- ------------- # b1 b2 b3 b1 b2 b3 b1 b2 b3 b1 b2 b3 # --- --- --- --- --- --- --- --- --- --- --- --- # # c1: # 4.1 4.6 3.7 4.9 5.2 4.7 5.0 6.1 5.5 3.9 4.4 3.7 # 4.3 4.9

Hello, I'm new in R and I want to do one thing that is very easy in excel, however, I cant do it in R. Suppose we have the data frame: data<- data.frame(A=c("a1","a2","a3","a4","a5")) I need to obtain another column in the same data frame (lets say B=c(b1,b2,b3,b4,b5) in the following way: b1=a1/(a1+a2+a3+a4+a5)

Hi all, I frequently encounter datasets that require me to repeat the same calculation across many variables. For example, given a dataset with total employment variables and manufacturing employment variables for the years 1990-2010, I might have to calculate manufacturing's share of total employment in each year. I find it cumbersome to have to manually define a share for each year and

I have lists of matrices stored in various ways. I'd like to extend %*% to work on these. Is this possible, or should I create my own new operator? A simplified example would be as follows: A <- list( A1, A2, A3) B <- list( B1, B2, B3) where A1,...,B3 are all matrices, and I'd like A %*% B to return list( A1 %*% B1, A2 %*% B2, A3 %*% B3) In the real case A and B are sometimes

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")

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?

Dear All, Here is a hypothetical sample (sorry for the clumsy code): A1 <- matrix(1:5, nrow=5, ncol=1) A2 <- matrix(6:10, nrow=5, ncol=1) A3 <- matrix(11:15, nrow=5, ncol=1) A4 <- matrix(16:20, nrow=5, ncol=1) A5 <- matrix(21:25, nrow=5, ncol=1) A6 <- matrix(26:30, nrow=5, ncol=1) B1 <- matrix(c(A1, A2, A3), nrow=5, ncol=3) B2 <- matrix(c(A2, A3, A4), nrow=5, ncol=3) B3

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

I have design with two repeated-measures factor, and no grouping factor. I can analyze the dataset successfully in other software, including my legacy DOS version BMDP, and R's 'aov' function. I would like to use 'Anova' in 'car' in order to obtain the sphericity tests and the H-F corrected p-values. I do not believe the data are truly deficient in rank. I

See if the following helps: > m <- outer(letters[1:5], 1:4, paste, sep="") > m [,1] [,2] [,3] [,4] [1,] "a1" "a2" "a3" "a4" [2,] "b1" "b2" "b3" "b4" [3,] "c1" "c2" "c3" "c4" [4,] "d1" "d2" "d3" "d4" [5,]

I'm doing a non linear regression with 8 parameters to be fitted: J.Tl.nls<-nls(Gw~(a1/(1+exp(-a2*Tl+a3))+a4)*(b1/(1+exp(b2*Tl-b3))+b4),data=Enveloppe, start=list(a1=0.88957,a2=0.36298,a3=10.59241,a4=0.26308, b1=0.391268,b2=1.041856,b3=0.391268,b4=0.03439)) First, I fitted my curve on my data by guessing the parameters'

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