similar to: crosstabulation and unlist function

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

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, 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

Hello list members, I'm looking for a way to force the results of a crosstabulation to be square - that is, to include 0 values. For example: table(letters[1:4],letters[c(1:3,3)]) yields: a b c a 1 0 0 b 0 1 0 c 0 0 1 d 0 0 1 I would like to return: a b c d a 1 0 0 0 b 0 1 0 0 c 0 0 1 0 d 0 0 1 0 Any suggestions? Thanks! -- Manuel A. Morales

I want to form a 3x3 crosstabulation for the signs of two vectors (i.e. Negative, Zero, Positive). The problem is that I am simulating the data so for some iterations one of the categories is absent. Thus the resulting table shrinks to 3x2. I want it to be 3x3 with zero column corresponding to the missing category. Moreover, I have tried but failed to give the dimension names. -- Sohail Chand

how about this: tapply(vector,list(factor1,factor2),function)? -----Original Message----- From: owner-r-help at stat.math.ethz.ch [mailto:owner-r-help at stat.math.ethz.ch]On Behalf Of Murray Jorgensen Sent: 24. september 2000 02:46 To: R-help Subject: [R] Crosstabulation I can't seem to find a function in R similar to Splus crosstabs() for creating a multi-way table from factors and a

Hello, I have a dataframe (var_1, var_2, ..., var_n) and I would like to export summary statistics to Latex in the form of a table. I want specific summary statistics by crossing numerous variables 2x2 AT ONCE. In each cell I would like sometimes to have the median (Q1 - Q3), or frequency and proportion, etc. CrossTable, xtab, etc... do not allow for multiple 2 by 2 crosstabulation. The table

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 intend to apply a mixed model on a set of longitudinal data, with a negative binomial distributed dependent variable, and after following the discussions on R help list I saw that more experienced people recommended using lmer (from lme4 pack), glmmPQL (from MASS) or glmm.admb (from glmmADMB pack)     My first problem: yesterday this syntax was ok, now I get this weird message (I

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

Hello! I have dataset where one variable is 'multiresponse', like this: [1] "1 2" "1 2 3" "4" "1 4" "4 3" etc. 'responses' separated by space. observations in different 'rows' of data.frame. I can do strsplit(data$var,' ') and make a list, where multiple responses are elements of character vectors, like this: $

  Hello R-users,   I have the following question, for which my search did not really return any usable result. If I have a matrix a1, and a vector a2 like below   a1<-matrix(c(1:4),2,2) a2<-c(8,9)   is there any function like the expand.grid (or some clever calling of the function) such that it outputs a matrix or dataframe where the entire a1 matrix is repeated for each value of a2 (the

Hello, I am trying to print a crosttabulation of mean,sd,n for a continuous variable crossclassified by anoother/s grouping variables. I came up with: xtab2 <- function(x,g1,g2) { funy <- function(z) list(mean(z,na.rm=T),sd(z,na.rm=T),length(z)) aa <- by(x,list(g1,g2),funy) bb <- matrix(unlist(aa),nrow=3 ,dimnames=list(c("mean","sd","n"),

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

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

Before I reinvent the wheel, I have need for a relatively straightforward crosstabulation (2 x n) function. I know that R has table(), ftable(), xtabs(), and summary(xtabs()), but none of these produce a fully "tricked" out cross-tabulation with marginal totals, expected cell frequencies, and an array of statistics about the contingency table. Is there a more complete (something

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

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 All: I was recently working with a dataset on arsenic poisoning. Among the variables in the dataset, I used the following three variables to produce crosstabulations (variable names: FOLSTAT, GENDER, ASBIN; all three were categorical variables, FOLSTAT denoted follow up status for the subjects and had seven levels, GENDER denoted sex (two levels: male,female), and ASBIN denoted binarized