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Does anyone know how to graph a line according to a specified equation? I'd like to plot the following hyperbola: Y=139.35/(1+(0.174*X)) I know there's a way to do this, but I'm having a ridiculous time trying to remember how. Thanks! Kyle H. Ambert Department of Behavioral Neuroscience Oregon Health & Science University [[alternative HTML version deleted]]

...} tmp } but this seems inefficient. As I understand it, the subset function won't really solve my problem, but it seems like there must be something out there that will that I must be forgetting. Does anyone know of a way to solve this problem in an efficient way? Thanks! Kyle H. Ambert Graduate Student, Department of Medical Informatics & Clinical Epidemiology Oregon Health & Science University ambertk@ohsu.edu [[alternative HTML version deleted]]

Hello! Does anyone know of a handy way to wrap the names.arg text in a barplot? I'm creating a bar plot with rather long labels; I can adjust the margins, but I'd also like to have the text wrap to about 4cm. Thanks! Kyle H. Ambert Doctoral Candidate, Bioinformatics Oregon Health & Science University ambertk@gmail.com [[alternative HTML version deleted]]

...f anyone has suggestions for how to implement post- hoc analyses on X1, after finding a significant main effect and no X1:X2 interaction, given that X1 has three levels and each subject in the analysis has data associated with each level. Thank-you in advance for your suggestions. Kyle H. Ambert Graduate Student, Dept. Behavioral Neuroscience Oregon Health & Science University ambertk@ohsu.edu [[alternative HTML version deleted]]

hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x <- rbinom(200, 1, .6) y <- rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the

Hi there, I am having a strange problem. I am running nls on some data. #data x <- -(1:100)/10 y <- 100 + 10 * (exp(-x / 2) Using nls I fit an exponential model to this data and get a great fit summary(fit) Formula: wcorr ~ (Y0 + a * exp(m1 * -dist/100)) Parameters: Estimate Std. Error t value Pr(>|t|) Y0 -0.0001821 0.0002886 -0.631 0.528 a 0.1669675 0.0015223

Hello -- I would like to know of a more efficient way of writing the following piece of code. Thanks. options(stringsAsFactors=FALSE) orig <- c(rep('11111111',100000),rep('22222222',200000),rep('33333333',300000),rep('44444444',400000)) orig.unique <- unique(orig) system.time(df <- as.data.frame(sapply(orig.unique, function(x) ifelse(orig==x, 1, 0))))

Hallo, I have got an existing data frame and want to add a new column. The existing data frame was created like this: > df <- rbind( c("Fred", "Mary", 4), c("Fred", "Mary", 7), + c("Fred", "Mary", 9), c("Barney", "Liz", 3), + c("Barney", "Liz", 5) ) > df

I have a question about how to save the output of a logistic regression model in some format (delimited, preferably) so that I can manipulate it to make nice tables in excel. I have tried print, save, write, none seem to do what I want them to. Does anyone have any ideas? Lynn D. Disney, Ph.D., J.D., M.P.H. Research Analyst SEIU 1199 1771 E. 30th Street Cleveland, Ohio 44114 Telephone:

Hi all I want to apply different colors on a simple plot: If I type par(br="gray") before a plot it puts all the image in gray but (imagine I run a simple plot) want to let the centrall box (where the dots are plotted) in white or image in lightblue. Can anyone guide me to apply this second step (make the box where the series are plotted in different colours). Thanks in advance.

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Hi There, Using t.test to test hypothesis about which one is greater, A or B? where A={10,20,30},B={25,30,15}. My question is which of the following conclusions is right? #################hypothesis testing 1 h0: A greater than or equal to B h1: A less than B below is splus code A=c(10,20,30) B=c(25,30,15) t.test(c(10,20,30),c(25,30,15),alternative="less") output: p-value=0.3359

Hello all, I have a matrix whose column names look like a1 a2 b1 b2 b3 c1 c2 1 2 3 7 1 3 2 4 6 7 8 1 4 3 Now, I can have any number of a's. not just two as shown above and same goes for b's and c's. I need to extract all the a's columns and put them in another matrix, extract all b's columns and put them in some matrix

Hi, I have what I believe is a repeated-measures dataset that I'm trying to analyze using lme(). This is *not* homework, but an exercise in my trying to self-teach myself repeated-measure ANOVA for other *real* datasets that I have and that are extremely similar to the following design. I'm fairly sure the dataset described below would work with lme() -- but it'd be great if anybody

R gurus, I'm working on data analysis for a small project. My response variable is total vines per tree (median = 0, mean = 1.65, min = 0, max = 24). My predictors are two categorical variables (four sites and four species) and one continuous (tree diameter at breast height (DBH)). The main question I'm attempting to answer is whether or not the species identity of a tree has

I am (desperately) trying to get used to using a Mac here at my new location. (Why *anyone* would ever use anything other than Linux, except under duress as I am, totally escapes me, but that's another story.) Fortunately much of the Mac OSX is actually Unix, so a civilized person can manage to carry on ... But there are some things. (Like this <expletive deleted> mailer ... But

Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the