similar to: Question on partial effect

Displaying 20 results from an estimated 10000 matches similar to: "Question on partial effect"

2008 Nov 20
1
gam and ordination (vegan and labdsv surf and ordisurf)
I have a general question about using thin plate splines in the surf and ordisurf routines. My rudimentary knowledge of a gam is that with each predictive variable there is a different smooth for each one and then they are added together with no real interaction term (because they don't handle this well?). Now, If I have two variables that have a high D^2 score and a low GCV score (I am
2008 Jan 15
1
covariate in a glm
Hello mailing list! I would like to know, how I can introduce a covariate in a glm, I've two factors and a covariate. Thank you very much! _______________________________________________________________________________________________ Michelangelo La Spina Equipo de Protección de cultivos - Control Biológico Departamento de Biotecnología y Protección de Cultivos Instituto Murciano de
2009 Jul 08
1
Comparing GAMMs
Greetings! I am looking for advice regarding the best way to compare GAMMs. I know other model outputs return enough information for R's AIC, ANOVA, etc. commands to function, but this is not the case with GAMM unless one specifies the gam or lme portion. I know these parts of the gamm contain items that will facilitate comparisons between gamms. Is it correct to simply use these values
2013 Nov 25
4
lmer specification for random effects: contradictory reults
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
2010 Jan 28
2
Data.frame manipulation
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:
2011 Apr 08
1
Variance of random effects: survreg()
I have the following questions about the variance of the random effects in the survreg() function in the survival package: 1) How can I extract the variance of the random effects after fitting a model? For example: set.seed(1007) x <- runif(100) m <- rnorm(10, mean = 1, sd =2) mu <- rep(m, rep(10,10)) test1 <- data.frame(Time = qsurvreg(x, mean = mu, scale= 0.5, distribution =
2008 Feb 07
2
How to calculate normality of the residuals from a test in R?
En innebygd og tegnsett-uspesifisert tekst ble skilt ut... Navn: ikke tilgjengelig Nettadresse: https://stat.ethz.ch/pipermail/r-help/attachments/20080207/891f1e80/attachment.pl
2010 Feb 15
2
creating functions question
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)
2011 Oct 26
2
Error in summary.mlm: formula not subsettable
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]]
2010 Feb 20
3
aggregating using 'with' function
Hi All, I am interested in aggregating a data frame based on 2 categories--mean effect size (r) for each 'id's' 'mod1'. The 'with' function works well when aggregating on one category (e.g., based on 'id' below) but doesnt work if I try 2 categories. How can this be accomplished? # sample data id<-c(1,1,1,rep(4:12)) n<-c(10,20,13,22,28,12,12,36,19,12,
2012 Jun 12
1
Two-way linear model with interaction but without one main effect
Hi, I know that the type of model described in the subject line violates the principle of marginality and it is rare in practice, but there may be some circumstances where it has sense. Let's take this imaginary example (not homework, just a silly made-up case for illustrating the rare situation): I'm measuring the energy absorption of sports footwear in jumping. I have three models (S1,
2004 Nov 04
3
sub- and superscript in plot labels
Dear List, I need to add a subscript and a superscript to some of the ions in the labels on some plots. I have got to here but now I'm stuck: plot(1:10, xlab = expression(paste("nm SO"[4], " ", mu, "eq cm"^{-2}, " yr"^{-1}))) Which gives almost what I require. No matter what I tried, however, I could not get bot a sub script *and* a superscript
2010 Jul 09
1
output without quotes
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
2008 Dec 15
5
OT: (quasi-?) separation in a logistic GLM
Dear List, Apologies for this off-topic post but it is R-related in the sense that I am trying to understand what R is telling me with the data to hand. ROC curves have recently been used to determine a dissimilarity threshold for identifying whether two samples are from the same "type" or not. Given the bashing that ROC curves get whenever anyone asks about them on this list (and
2003 Feb 10
2
problems using lqs()
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,
2012 Jun 06
3
Sobel's test for mediation and lme4/nlme
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
2009 Jun 04
2
Import ARIMA coefficients
Hello, I need to know how to import ARIMA coefficients. I already determined the coefficients of the model with other software, but now i need to do the forecast in R. For Example: I have a time series named x and i have fitted an ARIMA(1,0,1) (with other software) AR coef = -.172295 MA coef = .960043 (i know that this is not a good model, it's just an example) I try to
2009 Oct 21
1
How to find the interception point of two linear fitted model in R?
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]) #######################
2008 Oct 16
1
lmer for two models followed by anova to compare the two models
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
2009 Jun 05
1
Bug in print.Arima and patch
Dear List, A posting to R-Help exposed this problem with the print method for objects of class Arima: > set.seed(1) > x <- arima.sim(n = 100, list(ar = 0.8897, ma = -0.2279)) > mod <- arima(x, order = c(1,0,1)) > coefs <- coef(mod) > mod2 <- arima(x, order = c(1,0,1), fixed = coefs) > mod2 Call: arima(x = x, order = c(1, 0, 1), fixed = coefs) Coefficients: Error