similar to: lm( y ~ A/x ) ... how do I extract the coefficients by factor?

I am trying to show some techniques to my graduate regression class. The textbook mentioned using bootstrap samples of regression coefficients for assessing variability. I decided to show them reasonably effective ways of doing the resampling. The following is a function I wrote to create bootstrap samples of coefficients from a fitted linear regression model. bsCoefSample <- ##

This is a follow-up to the thread ending with: http://rubyforge.org/pipermail/vpim-talk/2008/000120.html I too am in search of some ruby parser for icalendar which properly handles timezones on the datetimes in the icalendar RFC. As I understand it there are actually three types of times. 1) UTC times with a string form of yyyymmddThhmmssZ note the trailing Z indicates zulu time aka utc. 2)

You may recall that I was recently constructing a function to bootstrap the coefficients in a linear regression model. In S-PLUS I was using the model.matrix function applied to the fitted model, then taking the QR decomposition of that. I discovered that it was in fact easier to accomplish the bootstrapping in R because the QR decomposition of the model matrix is stored with the fitted model.

Full_Name: Ben Bolker Version: 2.2.1 OS: Windows XP and 2000 Submission from: (NULL) (128.227.60.124) The following code, using confint() to try to get confidence intervals on an nls object that has been fitted with algorithm="port" reliably crashes R 2.2.0 and 2.2.1 with the latest version of MASS on a Windows 2000 and a Windows XP machine here. I *think* earlier versions of MASS

I can not tell from the docs or from the mailing list what is the state of timezone support in the iCalendar package? If I want to parse an iCalendar file that has non utc dstarts and dends will it convert those times to UTC or otherwise allow me to do that? When I tried to parse an iCalendar input file started off with something like this: BEGIN:VCALENDAR METHOD:REQUEST

I''m looking at the docs for 1.0.2 and saw this example... # Now, you can make timezones like this cal = Calendar.new cal.timezone do timezone_id "America/Chicago" daylight do timezone_offset_from "-0600" timezone_offset_to "-0500" timezone_name "CDT" dtstart

Since you only have 3 predictors, each categorical with a small number of categories, you can use expand.grid to make a data.frame containing all possible combinations and give that the predict method for your model to get all possible predictions. Something like the following untested code. newdata <- expand.grid( Humidity = levels(Humidity), #(High, Medium,Low)

Does anyone know how to extract residuals in lmer? Here's the error I get: > crop.lme=lmer(response~variety*irrigation*pesticide+(1|rep)+(1|rep: pesticide)+(1|rep:pesticide:irrigation), crop.data) > qqnorm(crop.lme) Error in qqnorm.default(crop.lme) : y is empty or has only NAs > resid(crop.lme) NULL Thanks! --Jake

I need the output to have groups and the probability any given record in that group then has of being in the response class. Just like my email in the beginning i need the output that looks like if A and if B and if C then %77 it will be D. The examples you provided are just simply not similar. They are different and would take interpretation to get what i need. On Apr 14, 2016 1:26 AM,

Hi there, I really tried hard to understand and find my own solution, but now I think I have to ask for your help. I already developed some script code for my problem but I doubt that it is correct. I have the following problem: Image you develop a logistic regression model with a binary outcome Y (0/1) with possible preditors (X1,X2,X3......). The development of the final model would be

Hi all, I am trying to run Weibull PH model in R. Assume in the data set I have x1 a continuous variable and x2 a categorical variable with two classes (0= sick and 1= healthy). I fit the model in the following way. Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull") My questions are 1. Is it Weibull PH model or Weibull AFT model? Call: survreg(formula = Surv(time, delta) ~ x1

Dear all R-users, I have a GARCH related query. Suppose I fit a GARCH(1,1) model on a dataframe dat >garch1 = garch(dat) >summary(garch1) Call: garch(x = dat) Model: GARCH(1,1) Residuals: Min 1Q Median 3Q Max -4.7278 -0.3240 0.0000 0.3107 12.3981 Coefficient(s): Estimate Std. Error t value Pr(>|t|) a0 1.212e-04 2.053e-06 59.05 <2e-16 *** a1

Dear all R-users, I have a GARCH related query. Suppose I fit a GARCH(1,1) model on a dataframe dat >garch1 = garch(dat) >summary(garch1) Call: garch(x = dat) Model: GARCH(1,1) Residuals: Min 1Q Median 3Q Max -4.7278 -0.3240 0.0000 0.3107 12.3981 Coefficient(s): Estimate Std. Error t value Pr(>|t|) a0 1.212e-04 2.053e-06 59.05 <2e-16 *** a1

Hi All , Can anybody tell me if there's any way to get the summarized anova values.Now i will explain what i mean , when i say "*summarized*". Below you can see the anova table of recmeanC1 with rest* all* i.e from recmeanC2 to i15(predictors),as shown in table. Df Sum Squares Mean Square F value Significance [Pr(>F)] recmeanC2 1 89.272 89.272

Dear list members, I am trying to estimate parameters of the AR(1)-GARCH(1,1) model. I have one additional dummy variable for the AR(1) part. First I wanted to do it using garchFit function (everything would be then estimated in one step) however in the fGarch library I didn't find a way to include an additional variable. That would be the formula but, as said, I think it is impossible to add

Dear Professor Bernhard, Sorry for take your time, but I found something strange that I am not able to explain/understand. Suppose that I compute the ADF test by using the criterion="BIC" to select the lags: summary(CADFtest(y, max.lag.y = 20, type = "drift", criterion="BIC")) Suppose that 2 lags are selected. Next, if I set the lags to 2: summary(CADFtest(y,

Good morning, I am using R to try to model the proportion of burned area in Portugal. The dependent variable is the proportion. The family used is binomial and the epsilon would be binary. I am not able to find the package to be used when the proportion (%) has to be used in glm. Could someone help me? I am using normal commands of glm.. for example: glm_5<- glm(formula=p~Precipitation,

Hi! I'm running an anova on a lm and I get the message Error: invalid dimnames given for data frame Any idea where it comes from? Many thanks. Here is the fit and the attempted anova: > fit Call: lm(formula = as.formula(paste(chemical, "a~", cmd, sep = "")), singular.ok = TRUE) Coefficients: (Intercept) Odom Displace Weight

Gracias a todos por las recomendaciones. Ya me he puesto en contacto con el autor y le explicado donde esta el error. Es tan fácil como que han cambiado la estructura de los objetos tipos lmer desde la versión 2.15.3 con lo que su paquete dejo de funcionar. Por suerte la nueva estructura conserva por lo menos la parte necesaria para hacer sus funciones. Se localizar el fallo y como arreglarlo, lo

Javier, Creo que aquí aplica la ley de Linus que dice: "Dado un número suficientemente elevado de ojos, todos los errores se convierten en obvios". La persona que revisa y encuentra un error no necesariamente tiene que ser la misma que la que lo escribe. Una motivación muy importante al compartir un código es la de recibir los beneficios del control de calidad por parte de tus pares.