similar to: as.data.frame() methods for model objects

Following Duncan Murdoch's off-list comments (thanks again!), here is a more complete/flexible version: as.data.frame.lm <- function(x, ..., level = 0.95, exp = FALSE) { cf <- x |> summary() |> stats::coef() ci <- stats::confint(x, level = level) if (exp) { cf[, "Estimate"] <- exp(cf[, "Estimate"]) ci <- exp(ci) } df <-

>>>>> SOEIRO Thomas via R-devel >>>>> on Fri, 17 Jan 2025 14:19:31 +0000 writes: > Following Duncan Murdoch's off-list comments (thanks again!), here is a more complete/flexible version: > > as.data.frame.lm <- function(x, ..., level = 0.95, exp = FALSE) { > cf <- x |> summary() |> stats::coef() > ci <- stats::confint(x,

Hi! I'm failing to understand the value of the intercept value in a multiple linear regression with categorical values. Taking the "warpbreaks" data set as an example, when I do: > lm(breaks ~ wool, data=warpbreaks) Call: lm(formula = breaks ~ wool, data = warpbreaks) Coefficients: (Intercept) woolB 31.037 -5.778 I'm able to understand that the value of

Hello, R-i-zens. I'm working on an data set with a factorial ANOVA that has a significant interaction. I'm interested in seeing whether the simple effects are different from 0, and I'm pondering how to do this. So, I have my.anova<-lm(response ~ trtA*trtB) The output for which gives me a number of coefficients and whether they are different from 0. However, I want the

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after

Does anyone know how could I get the significant codes from mixed model output fitted with a GEE package? The output I got is the following: GEE: GENERALIZED LINEAR MODELS FOR DEPENDENT DATA gee S-function, version 4.13 modified 98/01/27 (1998) Model: Link: Logit Variance to Mean Relation: Binomial Correlation Structure: Exchangeable Call: gee(formula = bru

Hello, I am performing a BACI analysis with ANOVA using the following glm: fit1<-glm(log(Cucs_m+1)~(BA*Otter)+BA+Otter+ID+Primary, data=b1) The summary(aov(fit1)) shows significance in the interaction; however, now I would like to determine what combinations of BA and Otter are significantly different (each factor has two levels). ID and PRIMARY substrates are categorical and included in

Hi A quick question. Standard errors reported by gee/yags differs from the ones in geeglm (geepack). require(gee) require(geepack) require(yags) mm <- gee(breaks ~ tension, id=wool, data=warpbreaks, corstr="exchangeable") mm2 <- geeglm(breaks ~ tension, id=wool, data=warpbreaks, corstr="exchangeable", std.err = "san.se") mm3 <- yags(breaks ~

I'm interested in restricting the pairwise comparisons of interaction effects in a multi-way factorial ANOVA, because I find comparisons of interactions between all different variables different to interpret. For example (supposing a p<0.10 cutoff just to be able to use this example): > summary(fm1 <- aov(breaks ~ wool*tension, data = warpbreaks)) Df Sum Sq Mean Sq F

Hello, I have a data set with 9700 records, and 7 parameters. The data were collected for a survey of forest communities. Sample plots (1009) and species (139) are included in this data set. I need to determine how species are accumulated as new plots are considered. Basically, I want to develop a species area curve. I've included the first 20 records from the data set. Point

Hello, I would like to do a large number of e.g. 1000 paired ttest using the by-function. But instead of using only the data within the 1000 groups, R caclulates 1000 times the ttest for the full data set(The same happens with Wilcoxon test). However, the by-function works fine with the lme function. Did I just miss something or is it really not working? If not, is there any other possibility to

Hi, I've been trying to figure out a special set of covariance matrices that causes some symmetric zero elements in the inverse covariance matrix but am having trouble figuring out if that is possible. Say, for example, matrix a is a 4x4 covariance matrix with equal variance and zero covariance elements, i.e. [,1] [,2] [,3] [,4] [1,] 4 0 0 0 [2,] 0 4

Dear useRs, I'm working on multiple comparison design on two factor (2 3 levels) ANOVA. Each of the tests I have tried (Tukey, multcomp package) seem to do only with one factor at a time. fm1 <- aov(breaks ~ wool * tension, data = warpbreaks) tHSD <- TukeyHSD(fm1, "tension", ordered = FALSE) $tension diff lwr upr p adj M-L -10.000000 -19.35342

Hi, I am seeking help with an error when running the example from R Documentation for TukeyHSD. The error occurs with any example I run, from any text book or website. thank you... > plot(TukeyHSD(fm1, "tension")). Error in plot(confint(as.glht(x)), ylim = c(0.5, n.contrasts + 0.5), ...) : error in evaluating the argument 'x' in selecting a method for function

Hi all!

Hi all, I frequently encounter datasets that require me to repeat the same calculation across many variables. For example, given a dataset with total employment variables and manufacturing employment variables for the years 1990-2010, I might have to calculate manufacturing's share of total employment in each year. I find it cumbersome to have to manually define a share for each year and