similar to: crazy loop error.

Dear R Users. This seems like a simple task, but I'm stuck. I have a list with 3 elements: (2 vectors and 1 matrix). I wish to extract each of these data elements using index subscripts and multiply them with a vector multiplier. What I have: > betas [1] 0.01 0.01 0.01 > LData[1] $int [1] 1 1 1 1 1 1 1 1 1 1 $date [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]

Hello, I have used nlme to fit a model, the R syntax is like fmla0<-as.formula(paste("~",paste(colnames(ldata[,9:13]),collapse="+"),"-1")) > fmla1<-as.formula(paste("~",paste(colnames(ldata[,14:18]),collapse="+"),"-1")) >

Rafael Espíndola <rafael.espindola at gmail.com> writes: > Now that aliases can have any expressions, can't you use something like > > @data = private global [2 x i32] [i32 42, i32 43] > @symbol = alias getelementptr ([2 x i32]* @data, i32 0, i32 1) > > This produces > > .Ldata: > .long 42 # 0x2a > .long 43

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

I am currently puzzled by a seach path behavior. I have a library of a dozen routines getlabs(), getssn(), getecg(), ... that interface to local repositories and pull back patient information. All have a the first 6 arguments in common, and immediately call a second routine to do initial processing of these 6. The functions "joe" and "fred" below capture the relevant

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1]

Hi list, I am unsuccessfully trying to produce a serious of trellis barcharts from within a for-loop. The barcharts work outside the loop. What am I missing? Example attached. Thanks Herry #XXXXXXXXXXXXXXXXXXXXXX trellis.device(bg="white") trellis.par.get("fontsize")->fontsize fontsize$default<-16 trellis.par.set("fontsize",fontsize)

Hello again, after obtaining the keytab file I tried to use kinit keytab.file followed by the spn $ samba-tool spn list z1 z1 User CN=z1,CN=Users,DC=home,DC=lan has the following servicePrincipalName: zookeeper/ap42.home.lan $ samba-tool domain exportkeytab z1.ktab --principal=z1 $ samba-tool domain exportkeytab z1.ktab --principal=zookeeper/ap42.home.lan $ kinit -V -k -t z1.ktab

Hi Team, I am trying to very simple plot with command plot. Question : I am trying to plot (x,y) based on the value of Column e. If column e value is greater than 0 then plot(x,y) otherwise do not plot it. Data : structure(list(x = c(1, 1, 1, 2, 2, 2, 3, 3, 3), y = c(1, 2, 3, 1, 2, 3, 1, 2, 3), e = c(0, -1, -2, 1, 0, -1, 2, 1, 0)), row.names = c(NA, -9L), .Names = c("x",

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should

Dear R users, I have posted this question several days ago and received not a single suggestion. I believe I have provided sufficient information for at least some help. Here I repost the question with several modifications. I want to run nested fixed-factor Anova in R on different experiments. I have 48 levels of the main factor x1 and 242 levels of the nested factor z1, and continuous response

hello, this is my script: #1) read in data: daten<-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt', header=TRUE, sep="\t") daten<-as.matrix(daten) #2) create empty matrix: indxind<-matrix(nrow=617, ncol=617) indxind[1:20,1:19] #3) compare cells to each other, score: for (s in 3:34) { #walks though the matrix colum by colum, starting at

Hi,   I use fCopulae package to draw different graphs of univariate and bivariate skew t.  But the plots titles overlap.  I tried using cex.main, font.main to adjust the size but they still overlaps.  Here is my code: par(mfrow = c(3, 1)) mu = 0 Omega = 1 alpha1 = 0 alpha2 = 1.5 alpha3 = 2 alpha4 = 0.5 Z1 = matrix(dmvst(x, 1, mu, Omega, alpha1, df = Inf), length(x)) Z2 = matrix(dmvst(x, 1, mu,

Hi R users: I got an error that "l1" is not found. This is a sample code: f1<-function(i,l1){ print(l1[[i]]) l1[[i+1]]<<-i return(i) } z1<-list(x=100,NULL,a=c(1,1)) lapply(1:3,f1,l1=z1) My final goal is to update the value of each list component to use it in the following step. I see that print function founds l1[[1]], but the assign instruction not. Thank you

Dear R-group members, I use: platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32 status major 1 minor 4.1 year 2002 month 01 day 30 language R I try to fit a 2 compartment model. The compartments are open, connected to each other and

Dear Rusers, I would like some comments about the following results (under R-2.2.0) > m = matrix(1:6 , 2 , 3) > m [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 > z1 = m[(m[,1]==2),] > z1 [1] 2 4 6 > is.matrix(z1) [1] FALSE > z2 = m[(m[,1]==0),] > z2 [,1] [,2] [,3] > is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about

Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-

Hello, I would like to perform a multivariate regression analysis to model the relationship between m responses Y1, ... Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. Based on my readings of the R help archives and R documentation, the function lm() should be able to

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im(z2) [1] -2

Dear all, I have a dataset. I want to make barplot from this data. Zero1 <- " V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1