similar to: Simulate phi-coefficient

Newsgroup members, I appreciate the help on this topic. David Duffy provided a solution (below) that was quite helpful, and came close to what I needed. It did a great job creating two vectors of dichotomous variables with a known correlation (what I referred to as a phi-coefficient). My situation is a bit more complicated and I'm not sure it is easily solved. The problem is that I must

Newsgroup members, Does anyone have a clever way to simulate a correlation matrix such that each column contains dichotomous variables (0,1) and where each column has different prevalence rates. For instance, I would like to simulate the following correlation matrix: > CORMAT[1:4,1:4] PUREPT PTCUT2 PHQCUT2T ALCCUTT2 PUREPT 1.0000000 0.5141552 0.1913139 0.1917923 PTCUT2

Hello dear R people, for my MSc thesis I need to program some functions, and some of them simply do not work. In the following example, I made sure both vectors have the same length (10), but R gives me the following error: Error in if (vector1[i] == vector2[j]) { : missing value where TRUE/FALSE needed I googled for possible solutions, but I did not find a good explanation for this...

Hi All, I am counting the number of occurrences of the terms listed in one vector in another vector. My code runs: for( i in 1:length(vector3)){ vector3[i] = sum(1*is.element(vector2, vector1[i])) } where vector1 = vector containing the terms whose occurrences I want to count vector2 = made up of a number of repetitions of all the elements of vector1 vector3 = a vector of NAs that is

Dear all, I was encountering with a typical Matching problem and was wondering whether R can help me to solve it directly. Let say, I have 2 vectors of equal length: vector1 <- LETTERS[1:6] vector2 <- letters[1:6] Now I need to match these 2 vectors with all possible ways like: (A,B,C,D,E) & (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) & (b,a,c,d,e), however there

R Version 2.2.0 Platform: Windows When I use barplot but select a ylim value greater than zero, the graph is distorted. The bars extend below the bottom of the graph. For instance the command produces a problematic graph. barplot(c(200,300,250,350),ylim=c(150,400)) Any help would be appreciated. Paul [[alternative HTML version deleted]]

We have been using Least Angle Regression (lars) to help identify predictors in models where the outcome is continuous. To do so we have been relying on the lars package. Theoretically, it should be possible to use the lars procedure within a general linear model (glm) framework - we are particular interested in a logistic regression model. Does anyone have examples of using lars with logistic

hi is there a faster way to do this? i <- 1 for(x in vector1) for(y in vector2) { m[[i]] <- (x - y) i <- i + 1 } regards soren -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in

Dear Statisticians--- This is not even an R question, so please forgive me. I have so much ignorance in this matter that I do not know where to begin. I hope someone can point me to documentation and/or a sample. I want to compute a covariance as quickly as non-humanly possible on an Intel core processor (up to SSE4) under linux. Alas, I have no idea how to engage CPU vectorization. Do I need

Hi, I have many files containing one column of data. I like to use the scan function to parse the data. Next I like to bind to a large vector. I try this like: count<-1 files <- list.files() # all files in the working directory for(i in files) { tmp <- scan(i) assign(files[count], tmp) count<-count+1 } This part works! Now I like to plot the data in a boxplot.

Hola Jose, si CONCATENAR significa APILAR, es decir, concantenar verticalmente, por decirlo de algun modo, podrias hacerlo con rbind(): nuevovector <- rbind(vector1,vector2) Si ademas quieres que cada valor de los vectores originales sea identificado en el nuevovector, puedes usar: nuevovector <- stack(vector1,vector2) en este ultimo caso se agrega una columna adicional tipo factor, con

Hello, My name is Ick Hoon Jin and I am Ph. D. student in Texas A & M Univ.. When I run the C embedded in R in the Linux system, I confront the following error after 6,000 iteration. By googling I found this error is from the problem in C. *** caught segfault *** address (nil), cause 'memory not mapped' My C code is following:

Hello everyone, I have two vectors having only one element different: vector1 vector2 vector1 TWC TWC TWC VFC TWX NA VIA/B VFC VFC

Hi All, I am trying to match two character arrays (email lists) using either pmatch(), match() or charmatch() functions. However the function is "missing" some matches due to differences in the cases of some letters between the two arrays. Is there any way to disable case sensitivity or is there an entirely better way to match two character arrays that have identical entries but written

Recently, I was working with some lagged designs where a vector of observations at one time was used to predict a vector of observations at another time using a lag 1 design. In the work, I noticed a lot of negative correlations, so I ran a simple simulation with 2 matched points. The crude simulation example below shows that the correlation can be -1 or +1, but interestingly if you do this

Recent changes to read.spss() in the foreign package return a dataframe containing additional attributes. For example, >TEMP<-read.spss(choose.files(), to.data.frame=T,use.value.labels=F) > str(TEMP) `data.frame': 780 obs. of 8 variables: $ EXPOS01: atomic 1 1 2 1 2 3 2 4 2 1 ... ..- attr(*, "value.labels")= Named num 5 4 3 2 1 .. ..- attr(*,

Dear all, meanwhile i found out how to handle the coordinate thing and plot the lines like i like. the remaining problem is, i need something like names.arg does in barplot for my plot. my plot connects several dots with several lines. and now i like characters as the names of the intersects of the x-axis. vector1=c(a,b) vector2=c(c,d) # this vector makes sure the yellow line starts at

Does anyone know of another version of the Mcnemar test that provides: 1. Odds Ratios 2. 95% Confidence intervals of the Odds Ratios 3. Sample probability 4. 95% Confidence intervals of the sample probability Obviously the Odds Ratios and Sample probabilities are easy to calculate from the contingency table, but I would appreciate any help on how to calculate the confidence

Dear all, I hope you can forgive me my stupid questions, but I am a very new R user (; So, this is my question: I have two matrices, those are: matrix1 <- matrix(cbind(vector1, vector2), 1,2, dimnames = list(c("values"), c("T value", "p value"))) matrix2 <- matrix(dcbind,2,6,dimnames =

I have a case where I would like to change multiple columns containing numbers to factors. I can change each column one at a time as in: TEMP.FACT$EXPOS01<-factor(TEMP.FACT$EXPOS01,levels=c(1,2,3),labels=c("No ne","Low Impact","MedHigh Imp")) TEMP.FACT$EXPOS02<-factor(TEMP.FACT$EXPOS02,levels=c(1,2,3),labels=c("No ne","Low