similar to: lm: how to calculate rsquared of the predicted values?

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

Hi Sacha, On second thought, perhaps this is more the direction that you want ... X2 = cbind(X_train,y_train) colnames(X2)[3] = "y" regr2<-randomForest(y~x1+x2, data=X2,maxnodes=10, ntree=10) regr regr2 #Make prediction predictions= predict(regr, X_test) predictions2= predict(regr2, X_test) HTH, Eric On Tue, May 9, 2023 at 6:40?AM Eric Berger <ericjberger at gmail.com>

Hello, In my lab we use a four parameter logistic fit model for our ELISA data (absorbance values). We are currently testing the use of different solvents and need to find a way to add a correlation value (such as an R squared or something similar) so we can test different solvents in making this standard curve. We currently use the drc package and this is our script for the 4 parameter:

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'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,

Dear R-experts, Here below a toy example with some error messages, especially at the end of the code (Tuning the parameters). Your help to correct my R code would be highly appreciated. ####################################### #libraries library(lattice) library(ggplot2) library(caret) library(randomForest) ?? #Data

Here I am again with question I'll feel foolish for asking, when I see the answer. I'm trying to produce a report and here's where I get stuck: How do I get R2HTML to produce the same number format? Particularly remove the decimal places for Par and Sal. Are there better methods to produce this type of report? Thanks, L.A. R version 2.10.0 XP

Hi, In the first instance, sorry for my english futile. I'm from Brazil, and I'm starting in english. [Embarassed] Well, I use Debian Squeeze, and I installed the Wine for play Counter Strike 1.6. The game run normally, but whitout sound. In winecfg there not a option for modify the sound driver for alsa or OSS. How a picture is worth a thousand words, following a print: [Image:

I am calculating the mean of each column grouped by the variable 'id'. I do this using aggregate, data.table, and plyr. My aggregate results do not match the other two, and I am trying to figure out what is incorrect with my syntax. Any suggestions? Thanks. Here is the data. myData <- structure(list(var1 = c(31.59, 32.21, 31.78, 31.34, 31.61, 31.61, 30.59, 30.84, 30.98, 30.79, 30.79,

I'd like to superimpose a ksmooth() onto a barplot(). My data is: > d 2009-06-20 2009-06-21 2009-06-22 2009-06-23 2009-06-24 2009-06-25 2009-06-26 2009-06-27 2009-06-28 2009-06-29 2009-06-30 2009-07-01 2009-07-02 Same Breed (B) 12.64 21.08 13.52 12.51 13.71 9.91 14.24 7.18 11.81 5.92 12.04 17.96

Dear List, i am trying to produce a 3d plot using wireframe using the code: wireframe(Residuals_FD ~ Elevation * Temperature, data = data2, scales = list(arrows = FALSE), drape = TRUE, colorkey = TRUE) As you can see when the code (using the data below) is run the plot area is set-up correctly but the actual surface is missing? Any help would be greatly appreciated. Chris #data Elevation

Dear list, First my information: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day 01 svn rev 38247 language R version.string Version 2.3.1 (2006-06-01) Now my question: How is it possible that a command in an R script is not

Hi, I know that Rsquare in glm or in non-linear models is "wrong", but some people like this. How I make to estimate the Rsquare from a model ajusted with glmmPQL or another GLMM? Thanks for all Ronaldo -- A simplicidade ?? o ??ltimo degrau da sabedoria. -- Victor Hugo -- |> // | \\ [***********************************] | ( ?? ?? ) [Ronaldo Reis J??nior

Dear all, I have a question regarding the output of the coxph function. What does the 'max possible' exactly mean in the output below? Many thanks. coef exp(coef) se(coef) robust se z Pr(>|z|) smocc_zyban -0.4384 0.6451 0.8667 0.9473 -0.463 0.644 self 1.1857 3.2728 0.1405 0.1443 8.216 2.22e-16 *** smocc_zyban:self

I think you've badly misinterpreted the purpose of the R listserv with this request: https://www.stat.math.ethz.ch/mailman/listinfo/r-help says "The `main' R mailing list, for announcements about the development of R and the availability of new code, questions and answers about problems and solutions using R, enhancements and patches to the source code and documentation of R,

Hi, I have something like this: > x <- 1:10 > y2 <- 30+5*x+rnorm(x,sd=3) > y <- c(y1,y2) > x <- c(x,x) > plot(x,y) > x <- 1:10 > y1 <- 1+5*x+rnorm(x,sd=2) > y2 <- 30+5*x+rnorm(x,sd=5) > y <- c(y1,y2) > x <- c(x,x) > f <- factor(rep(c("a","b"),c(10,10))) > m <- lm(y~x+f) > anova(m) Analysis of Variance

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

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.

Hello, I'm trying to annotate my plots nicely and am running into trouble. This example contains two problems: a) the 'text()' arguments do not show the conditional behavior I'm trying to give them. I try to test for the intercept of my regression and reformat the output accordingly ('+ intercept' in the >= 0 case and '- sqrt(intercept^2)' in the other case),

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_