similar to: converting object elements to variable names and making subsequent assignments thereto

OK, what is the trick to extracting the overall p value from an lm object? It shows up in the summary(lm(model)) output but I can't seem to extract it: > test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6]))) > test2[[1]] Call: lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6]) [omitted summary output] F-statistic: 40.94 on 2 and 7 DF, p-value: 0.0001371 It does not seem

I realize this is probably pretty basic but I can't figure it out. I'm looping through an array, doing various calculations and producing a resulting data frame in each loop iteration. I need to give each data frame a different name. Although I can easily create a new character string for writing each frame to an output file, I cannot figure out how to convert such strings to

Is there any way to put an argument into an object name. For example, say I have 5 objects, model1, model2, model3, model4 and model5. I would like to make a vector of the r.squares from each model by code such as this: rsq <- summary(model1)$r.squared for(i in 2:5){ rsq <- c(rsq, summary(model%i%)$r.squared) } So I assign the first value to rsq then cycle through models 2 through

I realize this should be simple, but even after reading over the several help pages several times, I still cannot decide between the myriad "apply" functions to address it. I simply want to apply a function to all the rows (or columns) of the same index from two (or more) identically sized arrays (or data frames). For example: > a=matrix(1:50,nrow=10) >

Dear community, I've check it while working, but just to reassure myself. Let's say we have 2 models: model1 <- lm(vdep ~ log(v1) + v2 + v3 + I(v4^2) , data = mydata) model2 <- lm(vdep ~ log(v1) + v2 + v3 + v4^2, data = mydata) So in model1 you really square v4; and in model2, v4*^2 *doesn't do anything, does it? Model2 could be rewritten: model2b <- lm(vdep ~

Hi everyone, I have recently experienced a strange behavior (strange from my knowledge) in rails. In my controllers ''new'' action, I am creating a few instance variables in the following manner : @controllerModel = ControllerModel.new @model1 = Model1.all @model2 = Model2.all in my ''new'' view, I am using the @controllerModel to create the form for new and I

Dear R users, i'm using a custom function to fit ancova models to a dataset. The data are divided into 12 groups, with one dependent variable and one covariate. When plotting the data, i'd like to add separate regression lines for each group (so, 12 lines, each with their respective individual slopes). My 'model1' uses the group*covariate interaction term, and so the coefficients

I am having difficulty extracting specific results from the model object. The following code shows where I am stuck. I want to run resamplings of a data set. For each I want to extract a particular F for the contrasts. If I run a very simple model (e.g. model1 <- aov(time~group) ) I can get the relevant coefficients, for example, by using "model1$coefficients". That's fine.

Hi all, This is partly a statistical question as well as a question about R, but I am stumped! I have count data from various sites across years. (Not all of the sites in the study appear in all years). Each site has its own habitat score "habitat" that remains constant across all years. I want to know if counts declined faster on sites with high "habitat" scores. I can

Hello R-Help, ----------------------------------------------------------------------------------------------------------------------------------------- Issues (there are 2): 1) Possible bug when using lmtest::encomptest() with a linear model created using nlme::lmList() 2) Possible modification to lmtest::encomptest() to fix confusing fail when models provided are, in fact, nested. I have

Full_Name: Menelaos Stavrinides Version: 1.3. 1 OS: Windows 98 Submission from: (NULL) (193.129.76.90) When model simplification is used in glm (binomial errors) and anova is used two compare two competitive models one can use either an "F" or a "Chi" test. R always performs an F test (Although when test="Chi" the test is labeled as Chi, there isn't any

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

Within a very large matrix composed of a mix of values and NAs, e.g, matrix A: [,1] [,2] [,3] [1,] 1 NA NA [2,] 3 NA NA [3,] 3 10 17 [4,] 4 12 18 [5,] 6 16 19 [6,] 6 22 20 [7,] 5 11 NA I need to be able to consecutively number, in new columns, the non-NA values within each column (i.e. A[1,1] A[3,2] and A[3,3] would all be set to one, and

Hi, I'm trying to compare models, one of which has all parameters fixed using offsets. The log-likelihoods seem reasonble in all cases except the model in which there are no free parameters (model3 in the toy example below). Any help would be appreciated. Cheers, Jarrod x<-rnorm(100) y<-rnorm(100, 1+x) model1<-lm(y~x) logLik(model1) sum(dnorm(y, predict(model1),

I am using anova.lm to compare 3 linear models. Model 1 has 1 variable, model 2 has 2 variables and model 3 has 3 variables. All models are fitted to the same data set. anova.lm(model1,model2) gives me: Res.Df RSS Df Sum of Sq F Pr(>F) 1 135 245.38 2 134 184.36 1 61.022 44.354 6.467e-10 *** anova.lm(model1,model2,model3) gives

Hello, I have a problem concerning logistic regressions. When I add a quadratic term to my linear model, I cannot draw the line through my scatterplot anymore, which is no problem without the quadratic term. In this example my binary response variable is "incidence", the explanatory variable is "sun": > model0<-glm(incidence~1,binomial) >

Hello, I want to understand how to tell if a model is significant. For example I have vectX1 and vectY1. I seek first what model is best suited for my vectors and then I want to know if my result is significant. I'am doing like this: model1 <- lm(vectY1 ~ vectX1, data= d), model2 <- nls(vectY1 ~ a*(1-exp(-vectX1/b)) + c, data= d, start = list(a=1, b=3, c=0)) aic1 <- AIC(model1)

Hello, I have a simple theorical question about regresion... Let's suppose I have this: Model 1: Y = B0 + B1*X1 + B2*X2 + B3*X3 and Model 2: Y = B0 + B2*X2 + B3*X3 I.E. Model1 = lm(Y~X1+X2+X3) Model2 = lm(Y~X2+X3) The Ajusted R-Square for Model1 is 0.9 and the Ajusted R-Square for Model2 is 0.99, among many other significant improvements. And I want to do the anova test to choose the best

Dear All I am attempting some power analyses, based on simulated data. My experimental set up is thus: Bleach: main effect, three levels (control, med, high), Fixed. Temp: main effect, two levels (cold, hot), Fixed. Main effect interactions, six levels (fixed) For each main-effect combination I have three replicates. Within each replicate I can take varying numbers of measurements (response

Dear all, I am analysing growth data - response variable - using GAM and GAMM models, and 4 covariates: mean size, mean capture year, growth interval, having tumors vs. not The models work fine, and fit the data well, however when I try to compare models using AIC I cannot get an AIC value. This is the code for the gam model: