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...UE)))) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] FALSE 4 3 5 5 4 5 4 3 4 4 TRUE 1 2 0 0 1 0 1 2 1 1 > Thanks for your thoughts. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richstat at earthlink.net

...linear regression. Then calculate the antilog of the residulas and plot these as your trendline. If 2) then take natural logs of y and x and follow the rest of the procedure described in 1). Hope this helps. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richstat at earthlink.net ----- Original Message ----- From: "chrisli1223" <chrisli at austwaterenv.com.au> To: <r-help at r-project.org> Sent: Thursday, January 07, 2010 10:33 PM Subject: [R] R exponential regression > > Hi all, > > I have a dataset which consists of 2...

...ase R. Can I be sure I am only working with base R? Or do I need to maintain two installations of R, one for use with GLP/GCP and one for when I want to use contributed packages? I hope this is clear. Thanks, Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richstat at earthlink.net

Hi! I'm comparing the differences in contaminant concentration between 2 different groups of people ( N=36, N=37). When using a simple linear regression model I found no differences between groups, but when evaluating the diagnostic plots of the residuals I found my independent variable to have deviations from normality (even after log transformation). Therefore I have used bootstrap on

Hi all, I have a dataset which consists of 2 columns. I'd like to plot them on a x-y scatter plot and fit an exponential trendline. I'd like R to determine the equation for the trendline and display it on the graph. Since I am new to R (and statistics), any advice on how to achieve this will be greatly appreciated. Many thanks, Chris -- View this message in context:

Hi, I'm not used to thinking along these lines, and wanted to ask your advice: Suppose you have a sample of around 100, consisting of patients according to doctors, in which patients and doctors are given a questionnaire with categorical responses. Each patient somehow has roughly 3 doctors, or 3 rows of data. The goal is to assess by category of each question or DV the agreement between

Hi! I have some data that looks like this up down percentaje uew_21 20 14 58.82 uew_20_5 27 40 40.29 uew_20 8 13 38.09 uew_19_5 17 42 28.81 So I have 4 experimental conditions and I am counting number of animals in the up and down compartment and the calculating the percentage, I want to know which one of the conditions is different from each other. If the data wouldn't be percentage

Hi All, I downloaded the R program (version 2.6.2) in last Jan 2008. I now want to upgrade the program to its latest version, but I don't want to go through the process of deleting the existing version and downloading the new version. This is because my existing R program has numerous packages that I downloaded for my research work. I want to upgrade my R program with those packages in it. Is

I am working on a project that requires me to do very large factorial evaluations. On R the built in factorial function and the one I created both are not able to do factorials over 170. The first gives an error and mine return Inf. Is there a way to have R do these larger calculations (the calculator in accessories can do 10000 factorial and Maple can do even larger) -- View this message in

I have a few hundred files of formatted data. Unfortunately most of them end with a spurious CONTROL-Z. I want to rewrite the files without the spurious character. Here's what I've come up with so far, but my code is unsafe because it assumes without justification that the last row of df contains a control character (and some NAs to fill up the record). options(warn=-1) #turn off

I'm sure there is a more general way to ask this question but how do you use the elements of a character vector as names of objects in an expression? For example, say you have: a = c(1,3,5,7) b = c(2,4,6,8) n=c("a","b") and you want to use the names a and b in a function (e.g. sum) sum(eval(as.name(n[1])),eval(as.name(n[2]))) works but what is a simpler way to