similar to: expand.grid using a matrix and a vector as input

Hello R-users, Checking the archives, I recently came across this topic: "export tables to Excel files" (http://r.789695.n4.nabble.com/export-tables-to-Excel-files-td1565679.html#a1565679), and the following interesting references have been proposed: http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows

Hi list, Using the lmList function from ?nlme package? I get the following error message: ?Error in !unlist(lapply(sum.lst, is.null)) : invalid argument type? # this is the syntax used # fm_cream <-lmList(MULTDV~TIME|cod, data=zz, na.action=na.omit) # summary(fm_cream) When I generate a simple example the function works, but on the real data I get that message. I imagine that it is a

Hi List, I have the following problem: I am using the multilevel package and make.univ function for available in the package and then xyplot from lattice and I want to know how could I be able to use the ?coefficient? for the straight line that passes the data ? Example from help: library(multilevel) data(univbct) #a data set already in univariate or stacked form for job satisfaction

Good morning! I have 4 questions which trouble me: 1. I want to test the hypothesis that the 2 proportions (the mean of a binomial) which come from 2 different samples are equal. I want to use the following function z= (p1-p2)/ sqrt((p1(1-p1)/n1)+(p2(1-p2)/n2)) which is one of the standard formulas for this case. Is there such a function in R? p1=the proportion from the first sample n1=the

Hi R users! I am trying to learn some multilevel analysis, but unfortunately i am now very confused. The reason: http://www.ats.ucla.edu/stat/hlm/seminars/hlm_mlm/mlm_hlm_seminar.htm http://www.ats.ucla.edu/stat/sas/seminars/sas_mlm/mlm_sas_seminar.htm and MlmSoftRev. pdf from mlmRev package. >From what i see, the first two links seem to declare the level one variable as a random part (i

Dear R users, It may be very simple but it is being difficult for me. I'd like to calculate the difference in percent between to measures. My data looks like this: set.seed(123) df1 <- data.frame(measure = rep(c("A1", "A2", "A3"), each=3), water = sample(c(100:200), 9), tide = sample(c(-10:+10), 9)) df1 # What I want to calculate is:

Dear Group, I have a following dataset: > a A B C D 1 22 3 31 40 2 26 31 36 32 3 3 7 49 16 4 24 40 27 26 5 20 45 47 0 6 34 43 11 18 7 48 48 24 2 8 3 16 39 48 9 20 49 7 21 10 17 36 47 10 > dput(a) structure(list(A = c(22L, 26L, 3L, 24L, 20L, 34L, 48L, 3L, 20L, 17L), B = c(3L, 31L, 7L, 40L, 45L, 43L, 48L, 16L, 49L, 36L), C = c(31L, 36L, 49L, 27L, 47L, 11L, 24L,

Dear list, I would like to expand a DF with all the missing levels of a variable. a <- c(2,2,3,4,5,6,7,8,9) a.cut <- cut(a, breaks=c(0,2,6,9,12), right=FALSE ) (x <- data.frame(a, a.cut)) # In 'x' the level "[0,2)" is "missing". AddMissingLevel <- function(xdf) { xfac <- factor( c("[0,2)", "[2,6)", "[6,9)",

Hello everyone, I need to fit a bivariate normal regression model to a dataset where the same covariate (say, X) influences two separate but correlated responses (say, Y1 and Y2). So, the bivariate model would look like : Y1 = a1 + b1*X + e1 Y2 = a2 + b2*X + e2 where e1 and e2 are error terms which can be correlated. Is there any package in R which can help me fit this model ? Any help will be

Hi R users I want to use the lme package for a multilevel analysis on the following example: > math<-c(2, 3,2, 5, 6 ,7 , 7) > sex<-c(1, 2, 1, 2, 2, 2, 1) > school_A<-c(1,1,1,2,2,2,2) > school_B<-c(10,10,10,20,20,20,20) > mydata<-data.frame(math, sex, school_A, school_B) > mydata School_A and school_B are two different school characteristics, math is an

Hello R-users, My toy example: aa<-c(1:5) bb<-c(NA,2,NA,4,5) cc<-c(1,2,NA,4,NA) dd<-c("A","B","B","A","C") df<-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B,

Hi List, Can anyone tell me how could i put the "BACK" button in the following code, just under the "AAA" menu? I want this button to go back to the previous page, and since it has nothing to do with the "1" and "2" buttons, i want it somehow separated from these two buttons, but i don't know how. I searched the web for some examples but my results

Dear R helpers Suppose x  <- c(1:3) y  <- matrix(1:12, ncol = 3, nrow = 4) > y      [,1] [,2] [,3] [1,]    1    5    9 [2,]    2    6   10 [3,]    3    7   11 [4,]    4    8   12 I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of y by 2nd element of x i.e. 2 an so on. Thus the resultant matrix should be like > z      [,1]   [,2]    [,3] [1,]    1   

Hi everyone After the following setup sector=2 # Define Number of Sectors sectors=LETTERS[seq( from = 1, to = sector )] # Name sectors No_ent=round(3/runif(sector)) # Number of entities per sector #Tot_No_ent=sum(No_ent) Goal is to get a List like (A1, A2, A3, B1, B2, B3, B4) where A is denoting an industrial sector and then a numbered sequence of companies within the sector. The step

Dear All, I would appreciate any help with the following: given the vector 'x' x <- c("Ass1", "Ass.s1", "Ass2", "Ass.s2") I would like to pick up the positions where the character string contains "Ass" but does not contain "Ass.s", so for 'x' that would be positions 1 and 3. I guess this could be programmed around

Hi List, I have a problem with using Latex and Sweave for creating a document. So I downloaded the Sweave manual from http://www.statistik.lmu.de/~leisch/Sweave/Sweave-manual.pdf and i have tried to replicate the example on pages 4-5, but i encounter the following problem: IT DOES NOT WORK. I can run Sweave, create the .tex file and then when building the .tex file in TeXnikCenter i get all

Good afternoon! I'm trying to use the Survey package for a stratified sample which has 4 criteria on which the stratification is based. I would like to get the corrected weights and for every element i get a weight of 1 E.g: tipping design <- svydesign (id=~1, strata= ~regiune + size_loc + age_rec_hhh + size_hh, data= tabel) and then weights(design) gives

Hi, I have this lame question. I want to convert a list (each with varies in length) to matrix with same row length by eliminating vectors outside the needed range. For example: l<-list(NULL) l[[1]]=1,2,3.7 l[[2]]=3,4,5,6,3 l[[3]]=4,2,5,7 l[[4]]=2,4,6,3,2 l[[5]]=3,5,7,2 #so say I want to only have 4 rows and 5 column in my matrix (or data.frame) and eliminating the 5th index value in l[[2]]

Hi R users! I have the following problem: how appropriate is my aov model under the violation of anova assumptions? Example: a<-c(1,1,1,1,1,1,1,1,1,1,2,2,2,3,3,3,3,3,3,3) b<-c(101,1010,200,300,400, 202, 121, 234, 55,555,66,76,88,34,239, 30, 40, 50,50,60) z<-data.frame(a, b) fligner.test(z$b, factor(z$a)) aov(z$b~factor(z$a))->ll TukeyHSD(ll) Now from the aov i found that my model

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives