similar to: back tick names with predict function

?s 17:38 de 30/11/2023, Robert Baer escreveu: > I am having trouble using back ticks with the R extractor function > 'predict' and an lm() model.? I'm trying too construct some nice vectors > that can be used for plotting the two types of regression intervals.? I > think it works with normal column heading names but it fails when I have > "special"

Also, and possibly more constructively, when you get an error like > CI.c = predict(mod2, data.frame( `plant-density` = x), interval = 'c') # fail Error in eval(predvars, data, env) : object 'plant-density' not found you should check your assumptions. Does "newdata" actually contain a columnn called "plant-density": > head(data.frame( `plant-density`

Sorry for the last email, sent too early. I have a small data set that has a hierarchical structure. It has both temporal (year, months) and spatial (treatment code and zone code). The following explains the data: WSZ_Code the water supply zone code (1 to 8) Treatment_Code the treatment plant which supplies each water supply zone (1 to 4)

"Thank you Rui. I didn't know about the check.names = FALSE argument. > Another good reminder to always read help, but I'm not sure I understood > what help to read in this case" ?data.frame , of course, which says: "check.names logical. If TRUE then the names of the variables in the data frame are checked to ensure that they are syntactically valid variable names

Hello, I tried to do a 'sem' analysis for data of how blueberry consumption by birds is influenced by a pollution gradient, using distance and vegetation structural and composition variables, but I got the following error message: Error in sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : S must be a square triangular or symmetric matrix This may be very

Hi all, I attempted to add an EL5.1 client to our puppet server (EL5), and after signing the client cert, got the error "Certificates were not trusted: hostname not match with the server certificate" I found the mailing list discussion and the relevant page: http://www.reductivelabs.com/trac/puppet/wiki/RubySSL-2007-006 As far as I can tell, my puppermaster''s cert CN matches

Hi I am comparing four different linear mixed effect models, derived from updating the original one. To compare these, I want to use anova(). I therefore do the following (not reproducible - just to illustration purpose!): dat <- loadSPECIES(SPECIES) subs <- expression(dead==FALSE & recTreat==FALSE) feff <- noBefore~pHarv*year # fixed effect in the model reff <-

Hi All, I'm conducting a meta-analysis and have taken a data.frame with multiple rows per study (for each effect size) and performed a weighted average of effect size for each study. This results in a reduced # of rows. I am particularly interested in simply reducing the additional variables in the data.frame to the first row of the corresponding id variable. For example:

I'm trying to get predicted probabilities out of a regression model, but am having trouble with the "newdata" option in the predict() function. Suppose I have a model with two independent variables, like this: y=rbinom(100, 1, .3) x1=rbinom(100, 1, .5) x2=rnorm(100, 3, 2) mod=glm(y ~ x1 + x2, family=binomial) I can then get the predicted probabilities for the two values of

Hello, Any advice or pointers for implementing Sobel's test for mediation in 2-level model setting? For fitting the hierarchical models, I am using "lme4" but could also revert to "nlme" since it is a relatively simple varying intercept model and they yield identical estimates. I apologize for this is an R question with an embedded statistical question. I noticed that a

When I fit a multivariate linear model, and the formula is defined outside the call to lm(), the method summary.mlm() fails. This works well: > y <- matrix(rnorm(20),nrow=10) > x <- matrix(rnorm(10)) > mod1 <- lm(y~x) > summary(mod1) ... But this does not: > f <- y~x > mod2 <- lm(f) > summary(mod2) Error en object$call$formula[[2L]] <- object$terms[[2L]]

Hi All, I am interested in printing column names without quotes and am struggling to do it properly. The tough part is that I am interested in using these column names for a function within a function (e.g., lm() within a wrapper function). Therefore, cat() doesnt seem appropriate and print() is not what I need. Ideas? # sample data mod1 <- rnorm(20, 10, 2) mod2 <- rnorm(20, 5, 1) dat

Hi All, I am interested in creating a function that will take x number of lm objects and automate the comparison of each model (using anova). Here is a simple example (the actual function will involve more than what Im presenting but is irrelevant for the example): # sample data: id<-rep(1:20) n<-c(10,20,13,22,28,12,12,36,19,12,36,75,33,121,37,14,40,16,14,20)

Hi everyone, I'm trying to print a table justified to the left, but it doesn't work. Any hints? KennArt <- data.frame(NR=c(171,172,174,175,176,177,181,411,980), TYP=c("K?rnermais", ?"Corn Cob Mix", "Zuckermais", "Mischanbau (Silo)Mais/Sonnenblumen", ?"Mais mit Bejagungsschneise in gutem landwirtschaftlichen und ?kologischen Zustand",

Dear Colleagues, I run this model: mod1 <- lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + category * subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 4102 4670 -1954 3665 3908 Random effects: Groups Name Variance

Hi All, I was wondering if someone could help me to solve this issue with lmer. In order to understand the best mixed effects model to fit my data, I compared the following options according to the procedures specified in many papers (i.e. Baayen <http://www.google.it/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDsQFjAA

I am teaching a DTrace class and a student noticed that 2 iosnoop scripts run in two different windows were producing different results. I was not able to answer why this is. Can anyone explain this. Here are the reults from the two windows: # io.d ... sched 0 <none> 1024 dad1 W 0.156 bash 1998

Dear List-members, I found a strange behaviour in the lqs function. Suppose I have the following data: y <- c(7.6, 7.7, 4.3, 5.9, 5.0, 6.5, 8.3, 8.2, 13.2, 12.6, 10.4, 10.8, 13.1, 12.3, 10.4, 10.5, 7.7, 9.5, 12.0, 12.6, 13.6, 14.1, 13.5, 11.5, 12.0, 13.0, 14.1, 15.1) x1 <- c(8.2, 7.6,, 4.6, 4.3, 5.9, 5.0, 6.5, 8.3, 10.1, 13.2, 12.6, 10.4, 10.8, 13.1, 13.3, 10.4, 10.5, 7.7, 10.0, 12.0,

The abline function can be used to draw the regression line when one passes the lm object as an argument. However, if it's an intercept-only model, it appears to use the intercept as the slope of the abline: mod <- lm(dist ~ 1, data = cars) plot(dist ~ speed, data = cars) abline(reg = mod) # nothing appears This behaves as documented, but might catch someone. Would it be an improvement

Dear All, Let have 10 pair of observations, as shown below. ###################### x <- 1:10 y <- c(1,3,2,4,5,10,13,15,19,22) plot(x,y) ###################### Two fitted? models, with ranges of [1,5] and [5,10],?can be easily fitted separately by lm function as shown below: ####################### mod1 <- lm(y[1:5] ~ x[1:5]) mod2 <- lm(y[5:10] ~ x[5:10]) #######################