similar to: Extending %*%

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

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

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

#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 #

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

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

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 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?

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")

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

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 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

Hello. This is a follow-up to a question I posted last week. With some previous suggestions from the R-help community, I have been able to plot survival (, hazard, and density) curves using published data for Siler hazard parameters from a number of ethnographic populations. Can the function below be modified, perhaps with a "for" statement, so that multiple curves (different line

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,]

There are function which does have optimization opportunities but because of may-alias memory dependencies sometimes optimization is not effective. May be runtime checks kills the gains of optimization. For such cases aiming to do interprocedural function specialization optimization where in the clone function version no-alias assumption can be assumed and the original function version will hold

Hi, > dd <- data.frame(A=c("b","c","a"), B=3:1) > dd A B 1 b 3 2 c 2 3 a 1 > unlist(dd) A1 A2 A3 B1 B2 B3 2 3 1 3 2 1 Someone else might get something different. It all depends on the values of its 'stringsAsFactors' option: > dd2 <- data.frame(A=c("b","c","a"), B=3:1, stringsAsFactors=FALSE) > dd2

Dear R-help readers, I'm sure this problem has been answered but I can't find the solution. I have two vectors v1 <- c("a","b") v2 <- c(1,2,3) I want an easy way to produce every possible combination of v1, v2 elements Ie I want to produce c("a1","a2","a3", "b1","b2","b3") regards Desmond Desmond

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 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