similar to: annotation problems (conditional text())

I estimate two competing simple regression models, A and B where the LHS is the same in both cases but the predictor is different ( I handle the intercept issue based on other postings I have seen ). I estimate the two models on a weekly basis over 24 weeks. So, I end up with 24 RSquaredAs and 24 RsquaredBs, so essentally 2 time series of Rsquareds. This doesn't have to be necessarily thought

Dear all, I have a question about the knowing for which row I have the max value of one of my variables. I calculated the Rsquared for different columns and made a list to gather them. I unlisted this list to create a vector with this values. I want to know for which column I have the max value of Rsquared. The columns were always named in the same way. They always start with results4$depth_

Hello, I'm with a conceptual doubt regarding Rsquared of both lm() and postResample(library caret). I've got a multiple regression linear model (lets say mlr) with anR² value of 67.52%. Then I use this model pro make predictions with predict() function using the same data as input , that is, use the generated model to predict the value associated with data that I used as input. Next, if

hello mailing list! i still consider myself an R beginner, so please bear with me if my questions seems strange. i'm in the field of biology, and have done consecutive hydraulic conductivity measurements in three parallels ("Sample"), resulting in three sets of conductivity values ("PLC" for percent loss of conductivity, relative to 100%) at multiple pressures

I calculate an anova test in the following way: expdata<-read.table("/home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP", header=TRUE)

Hello, I seem unable to construct a legend which contains a substitution as well as math symbols. I'm trying to do the following: strain2 <- "YJG48" legend.txt <- c( substitute( strain * %==% * "YJG45, rpn10" * %Delta%, list(strain=strain2) ), "Verhulst/Logistic", "Malthus" ) legend( 100,2.5, legend.txt, cex=0.75,

Hi Mark, I don't know wether you recived a sufficient reply or not, so here are my comments to your problem. Supressing the constant term in a regression model will probably lead to a violation of the classical assumptions for this model. From the OLS normal equations (in matrix notation) (1) (X'X)b=X'y and the definition of the OLS residuals (2) e = y-Xb you get - by

Hello, I'm new to this and am trying to teach myself some R by plotting biological data. The growth curve in question is supposed to be fitted to the Verhulst equation, which may be transcribed as follows: f(x)=a/(1+((a-0.008)/0.008)*exp(-(b*x))) - for a known population density (0.008) at t(0). I am trying to rework the example from "An Introduction to R" (p. 72) for my case and

Dear all, Can anyone please shed some light onto how to do this? This will give me all "intensity" columsn in my data frame: intensityindeces <- grep("^Intensity",names(dataframe),value=TRUE) This will give me the maximum intensity for the first row: intensityone <- max(dataframe[1,intensityindeces]) What I'm now looking for is how to dfo this for the whole data

Hello, I'm doing some linear regression: >lm<-lm(osas~alp,data) >summary(lm) However, the Rsquared in the output of summary() is not the same as the "standard" Rsquared calculated by spreadsheets, and outlined in statistical guidebooks, being SSR/SSTO. The output says "multiple Rsquared", but it is no multiple regression... What's the difference? Thanks,

Hello R-ologists, Imagine you have a list "list" like so: >list [[1]] [1] "IPI00776145.1" "IPI00776187.1" [[2]] [1] "Something" "IPI00807764.1" "IPI00807887.1" [[3]] [1] "IPI00807764.1" [[4]] [1] "Somethingelse" What I need to achieve is a filtered list "list2" like so: >list2 [[1]] [1]

Dear R-users first, thanks to Paul Murrell for fixing my problem "What line is labeled in persp()". It works great now (in the development version). One more question (I don't know if it's related to the "old" problem): the annotation on the z-axis overwrites the numbers at the tickmarks and sometimes, if numbers are pretty long, the numbers themselfe overlap the

Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I

Hi all, I'm to inexperienced to come up with the matrix solution elusively appearing on the horizon for the following problem and would appreciate if you could give me a nudge ... I have two vectors a, and b and need to find the closest match for each value of a in b. How to do that efficiently? Thanks, Joh

Hi, I'm using glm function to do logistic regression and now I want to know if it exists a kind of R-squared with this function in order to check the model. Thank you.

Hi, I know that if i want to see the SPLUS or R-source code of a function,i should give the command without the brackets.For example:  rsquared.lmRobMM function(x) {  str0 <- "Initial S-estimate"  str1 <- "Final M-estimate"  if(x$est == "final") {   z <- x$r.squared   attr(z, "info") <- str1  }  if(x$est == "initial") {   z <-

Dear r-helpers, I have some multiplot figures that I would like to annotate. To each plot frame I would like to add the associated correlation coefficients. To do so, I believe I need to understand the usage of `expression' better. The following is a good start. petstring <- expression(paste(r[rho*p],"=" )) mtext(petstring, line=0.3, adj=0.1, cex=0.8) My question is,

Hi, I have zlib compressed strings (example is attached) and would like to decompress them using memDecompress ... I try this: > connection <- file("compressed.txt","r") > compressed <- readLines(connection) > memDecompress(as.raw(compressed),type="g") Error in memDecompress(as.raw(compressed), type = "g") : internal error -3 in

All, Happy World Cup and Wimbledon. This morning finds me with the first of my many daily questions. I am running a ks.test on residuals obtained from a regression model. I use this code: > ks.test(Year5.lm$residuals,pnorm) and obtain this output One-sample Kolmogorov-Smirnov test data: Year5.lm$residuals D = 0.7196, p-value < 2.2e-16 alternative hypothesis: two.sided Warning

Hello, Please help - I'm blanking on this ... I have a matrix like this: [,1] [,2] [1,] 1 2 [2,] 1 3 [3,] 2 3 and would like to have a list of vectors, where a vector contains the entries in a matrix row ... Can somebody nudge me to the place I need to go? Thanks, Joh