search for: newdat

...quasipoisson, data = beva) summary(gm6) library('VIM') b_dist_min_new <- as.numeric(prepare(beva$dist_min, scaling="classical", transformation="logarithm")). My first steps for the solution are following: range(b_dist_min_new) x <- seq(-1.496, 1.839, by=0.01) newdat <- data.frame(b_dist_min_new=x) y <- predict(gm6, newdata=newdat, type="response") plot(x,y, type="l", ylim=c(0,15), lty=2, xlab="Distance [scaled log.]", ylab="Number of used plant", las=1) ilogit<-function(x) exp(x)/(1 + exp(x)) logit <-functio...

...yone could help me specify the model correctly. I am trying to specify the response variable, /yseed/, as # of successes out of total observations... but I suspect that given the error below, that is not correct. Also, Newsect should be a factor, whereas Newdist is continuous. Thanks, John Newdat<-data.frame(Newtree=rep(1:3, each=20), Newsect=rep(c("a","b"), each=10), Newdist=rep(1:5, 2), y=rpois(60,2), tot=rep(c(14,12,10,8,6), 12)) yseed<-cbind(Newdat$y, Newdat$tot) mod<-glmmBUGS(yseed~Newsect + Newdist, effects="Newtree", famil...

...(even though it uses S4 classes) but am not sure whether this violates any assumptions made by the Sobel test. Also, I wish to include a matrix of predictors as opposed to a single vector "pred" - trivial to implement but is it kosher? sobel.lme<-function (pred, med, out, grpid) { NEWDAT <- data.frame(pred = pred, med = med, out = out, grpid=grpid) NEWDAT <- na.exclude(NEWDAT) model1 <- lme(out ~ pred, random=~1|grpid,data = NEWDAT) model2 <- lme(out ~ pred + med, random=~1|grpid, data = NEWDAT) model3 <- lme(med ~ pred, random=~1|grpid, data = NEWDAT...

...want to save the predicted values from that model and show the log curve in a plot ; predicted~log(x) mod<-lme(B~log(x)+as.factor(y), random=~1|cohort/Study, weights=varFixed(~I(SE^2)), na.action=na.omit, data=subset(meta), control = lmeControl(sigma = 1, apVar = FALSE)) summary(mod) newdat <- data.frame(x=seq(min(meta$x), max(meta$x),,118)) # I have 118 observations. #How do I add the factor variable to my newdat? newdat$pred <- predict(mod, newdat,level = 0,type="response") plot(B ~ x, data=meta) lines(B ~ x, data=newdat) Can you please assist me ? Thank you! Al...

...rly_met_data$date <- as.POSIXct(hourly_met_data$date) works exactly as it is supposed to (e.g. Min/max of that field are calculated correctly). However, when I perform this command to convert a date/time to just a date (e.g. I'm removing the hour portion of the date/time): hourly_met_data$newdate <- trunc(hourly_met_data$date,"day") I get: Error in `$<-.data.frame`(`*tmp*`, "newdate", value = list(sec = 0, min = 0, : replacement has 9 rows, data has 252020 If I simply set: newdate <- trunc(hourly_met_data$date,"day") And then print the data (e...

...ode for reproduction, below the results. ------------------------------------------------------------------------ --- version library( MASS ) data( housing ) hnames <- lapply( housing[,-5], levels ) house.plr <- polr( Sat ~ Infl + Type + Cont, data=housing, weights=Freq ) summary( house.plr ) newdat <- expand.grid( hnames[-1] )[1:5,] cbind( newdat, predict( house.plr, newdat, type="probs" ) ) # Baseline probs: diff( c(0,tigol( c(-0.4961,0.6907) ), 1) ) # First level of Infl: diff( c(0,tigol( c(-0.4961,0.6907) + 0.5663922 ), 1) ) # But the change of sign for eta is needed to reprod...

...that model and show the log curve > in a plot ; predicted~log(x) > > mod<-lme(B~log(x)+as.factor(y), random=~1|cohort/Study, > weights=varFixed(~I(SE^2)), na.action=na.omit, data=subset(meta), > control = lmeControl(sigma = 1, apVar = FALSE)) > summary(mod) > > newdat <- data.frame(x=seq(min(meta$x), max(meta$x),,118)) # I have 118 > observations. #How do I add the factor variable to my newdat? > newdat$pred <- predict(mod, newdat,level = 0,type="response") > > plot(B ~ x, data=meta) > lines(B ~ x, data=newdat) > > Can you p...

...ression model but it does not work. My Model: y = Worktime + Vacation + Illnes + Bankholidays My modelmatrix is of dimension 28x4 Then I want to make use of the function predict because there confidence.intervals are include. My idea was: mod <- lm(y~Worktime+Vacation+Illnes+Bankholidays) newdate=data.frame(x=c(324,123,0.9,0.1)) predict(y,newdate) But I always get the message: 'newdata' had 1 rows but variable(s) found have 28 rows What can I do? Yours, Maja -- Psssst! Schon vom neuen GMX MultiMessenger geh?rt? Der kanns mit allen: http://www.gmx.net/de/go/multimessenger

...# Random Effects model1<-rma(g,var.g,mods=~mid,method="REML") # covariate model model2<-rma(g,var.g,mods=~mod,method="REML") # factor model model3<-rma(g,var.g,mods=~mid+mod,method="REML") # multiple # example matrix for predicting against model3 newdat<-expand.grid(c(1,2,3,4,5),c(1,2,3)) predict(model1,newmods=c(1,2,3,4,5)) predict(model2,newmods=c(1,2,3)) predict(model3,newmods=newdat) -- Andrew Beckerman Sent with Sparrow (http://bit.ly/sigsprw) [[alternative HTML version deleted]]

...ve >> in a plot ; predicted~log(x) >> >> mod<-lme(B~log(x)+as.factor(y), random=~1|cohort/Study, >> weights=varFixed(~I(SE^2)), na.action=na.omit, data=subset(meta), >> control = lmeControl(sigma = 1, apVar = FALSE)) >> summary(mod) >> >> newdat <- data.frame(x=seq(min(meta$x), max(meta$x),,118)) # I have 118 >> observations. #How do I add the factor variable to my newdat? >> newdat$pred <- predict(mod, newdat,level = 0,type="response") >> >> plot(B ~ x, data=meta) >> lines(B ~ x, data=newdat)...

...each chunk is going through a loop and being spit out in a list sub3<-subset(ex.dat, c=="3") sub2<-subset(ex.dat, c=="2") sub1<-subset(ex.dat, c=="1") #put it back together; in my case, pull out the parts of the list generated by the loop that hold the data newdat<-rbind(sub3, sub2, sub1) #rebuild it such that it can be re-organized into it's original order rn<-as.numeric(row.names(newdat)) new2<-data.frame(newdat, rn) new3 = new2[do.call(order, new2["rn"]), ] ###################### It's those last three lines of code that I&...

...725649 30 7 1.49001721222926 30 8 2.00587718501980 30 9 0.450576585429981 30 10 2.87120375367357 30 11 2.25575058079324 80 12 2.03471288724508 80 13 2.67432066972984 80 14 1.74102136279177 80 15 2.29827581276955 80 and now: newvar<-(rnorm(15,4)) newdat<-as.data.frame(cbind(newvar, dat$vecs)) newdat R returns: newvar V2 1 4.300788 1 2 5.295951 1 3 5.099849 1 4 3.211045 1 5 3.703554 1 6 3.693826 2 7 5.314679 2 8 4.222270 2 9 3.534515 2 10 4.037401 2 11 4.476808 3 12 4.842449 3 13 3.109677 3 14 4.752961 3 15 4.4452...

...nt of a lexical decision task). To ascertain that the data was normally distributed, I used *shapiro.test *for each participant (see commands below), but only one out of 21 returns a p value above p.0 05. > f = function(dfr) return(shapiro.test(dfr$Target.RTinv)$p.value) > p = as.vector(by(newdat, newdat$Subject, f)) > names(p) = levels(newdat$Subject) > names(p[p < 0.05]) Removing a few outliers per subject doesn't make a difference, and "aggressive" removal of outliers (done by subject, for each of the 6 conditions ) still results in non-normally distributed data b...

....gamm <- gamm( Y ~ Group + s(X, by=Group), random = list(Subject=~1) ) Y and X are numeric, Group is an unordered factor with 5 levels, and Subject is an unordered factor with ~70 levels Now the output from gamm is a list with an lme component and a gam component. If I make a data frame "newdat" like this: > newdat X Group Subject 5 g1 s1 5 g1 s2 5 g1 s3 6 g1 s1 6 g1 s2 6 g1 s3 I can get the fixed effects prediction of the smooth by...

Hi group, suppose I have a date, say May 15 2004, and I want to now what date it is 23 days before that date. The way to calculate the new date should (...) take account of leap years :) In pseudocode: olddate <- "May 15 2004" newdate <- olddate-23 I looked around in POSIXct etc..., maybe I overlooked? Thanks, Maarten -------------------------------------------------------------- Zie ook/see also: http://www.knmi.nl/maildisclaimer.html

Dear Ben, you answerd to Nancy Shackelford about Clarks 2Dt function. Since the thread ended just after your reply, I would like to ask, if you have an idea how to use this function in R I defined it the following way: function(x , p, u) { (p/(pi*u))*(1+(x^2/u))^(p+1) } and would like to fit this one to my obeservational data (count) [,1] [,2] [1,] 15 12 [2,] 45 13 [3,]

hi netters, I'm a newbie to R and there are some very simple problems puzzeled me for two days. I've a dataframe here with several columns different in modes. Two of the columns are special for me: column 1 has the mode "factor" and column 2 has the mode "numeric vectors". The values for column 1 are either "T" or "F". I wanna do two things:

I have a dataframe that contains fields such as patid, labdate, labvalue. The same patid may show up in multiple rows because of lab measurements on multiple days. Is there a simple way to obtain just the first and last record for each patient, or do I need to write some code that performs that. Thanks, Steven

Hi, I am comparing the observed and fitted values of my GAM model, which includes the explanatory variables: longitude, depth, ssh, year and month. When I compare observed and fitted values for longitude, depth and ssh it works. But when I try to do it for month and year (which are as factors in the GAM model) it doesn't work. My observed and fitted values are exactly the same.. How is that

...um 0 0 0 0 0 0 0 0 0 0 ... $ start : num 2005 2005 2005 2006 2006 ... $ stop : num 2006 2006 2006 2007 2007 ... $ risk : num 1 2 3 1 2 3 1 2 3 1 ... $ evstatus: num 0 0 0 0 0 0 0 0 0 0 ... I am modelling the survival time until the adoption of the policy as follows: mod1<-Surv(newdat$start, newdat$stop, newdat$evstatus) mymod1<-coxph(mod1 ~ news + enviro + partisan + cluster(state) + strata(evstatus), method=c("efron"), robust=TRUE) Again, following Fox, I try to construct a data frame with a hypothetical covariate profile: n<-data.frame(news=rep(c(1,4,8)), ev...