search for: mod2

...e 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: id<-c(1,2,2,3,3,3) es<-c(.3,.1,.3,.1,.2,.3) mod1<-c(2,4,4,1,1,1) mod2<-c("wai","other","calpas","wai","itas","other") data<-as.data.frame(cbind(id,es,mod1,mod2)) data id es mod1 mod2 1 1 0.3 2 wai 2 2 0.1 4 other 3 2 0.2 4 calpas 4 3 0.1 1 itas 5 3 0...

...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) mod1.out <- summary(model1)$tTable mod2.out <- summary(model2)$tTable mod3.out <- summary(model3)$tTable indir <- mod3.out[2, 1] * mod2.out[3, 1] effvar <- (mod3.out[2, 1])^2 * (mod2.out[3, 2])^2 + (mod2.out[3, 1])^2 * (mod3.out[2, 2])^2 serr <- sqrt(effvar) zvalue = indir/serr out <- lis...

...g x2 constant at 0 like this: p2=predict(mod, type="response", newdata=as.data.frame(cbind(x1, x2=0))) unique(p2) However, I am running regressions as part of a function I wrote, which feeds in the independent variables to the regression in matrix form, like this: dat=cbind(x1, x2) mod2=glm(y ~ dat, family=binomial) The results are the same as in mod. Yet I cannot figure out how to input information into the "newdata" option of predict() in order to generate the same predicted probabilities as above. The same code as above does not work: p2a=predict(mod2, type=...

...), X2=rep(0:1, 2))) # model with 2 effects mod <- coxph(Surv(Time) ~ X + X2, data=thedata) summary(mod) # coefficient of X2 should be 0 # coefficient of X should be log(lambda1/lambda0) log(lambda1/lambda0) predict(mod , newdata=ndf, type="expected", se.fit=TRUE) # stratified model mod2 <- coxph(Surv(Time) ~ strata(X) + X2, data=thedata) predict(mod2, newdata=ndf, type="expected" ) # no se predict(mod2, newdata=ndf, se.fit=TRUE) # type="lp" predict(mod2, type="expected", se.fit=TRUE) # prediction at old dat...

...en 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$terms[[2L]] <- as.name(ynames[i]) : objeto de tipo 'symbol' no es subconjunto I would say that the problem is in the following difference: > class(mod1$call$formula) [1] "call" > class(mod2$...

...&bvm> R.H. et al 2008) Here, dT_purs is the response variable, T and Z are the fixed effects, and subject is the random effect. Random and fixed effects are crossed.: mod0 <- lmer(dT_purs ~ T + Z + (1|subject), data = x) mod1 <- lmer(dT_purs ~ T + Z + (1 +tempo| subject), data = x) mod2 <- lmer(dT_purs ~ T + Z + (1 +tempo| subject) + (1+ Z| subject), data = x) mod3 <- lmer(dT_purs ~ T * Z + (1 +tempo| subject) + (1+ Z| subject), data = x) mod4 <- lmer(dT_purs ~ T * Z + (1| subject), data = x) anova(mod0, mod1,mod2, mod3, mod4) Data: x Models: mod0: dT_purs ~ T +...

...mple (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) r<-c(.68,.56,.23,.64,.49,-.04,.49,.33,.58,.18,-.11,.27,.26,.40,.49, .51,.40,.34,.42,.16) mod2<-factor(c(rep(c(1,2,3,4),5))) da<-data.frame(id, n, r, mod1, mod2) reg0<-lm(da$r ~ 1) reg1<-lm(da$r ~ da$mod1) reg2<-lm(da$r ~ da$mod1 + da$mod2) # This is as far as I get with the function: MRfit <- function( ...) { models <- list(...) fit<- anova(models) return(fit)...

...), `plant-density` = c(137, 107, 132, 135, 115, 103, 102, 65, > ?????????????????????? 149, 85, 173, 124, 157, 184, 112, 80, 165, 160, > 157, 119)), > class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -20L)) > > # regression model works > mod2 = lm(`cob-wt` ~ `plant-density`, data = cob) > > # x sequence for plotting CI's > # Set up x points > x = seq(min(cob$`plant-density`), max(cob$`plant-density`), length = 1000) > > # Use predict to get CIs for a plot > # Add CI for regression line (y-hat uses 'c')...

...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 <- data.frame(mod1, mod2) # collapsing the colnames to 'mod1+mod2' temp <- paste(names(dat), collapse="+") temp # this gives quotes print(temp, quote = FALSE) # no quotes but includes [1] # need the output like this (no quotes & no [1]):...

...REMLdeviance 4102 4670 -1954 3665 3908 Random effects: Groups Name Variance Std.Dev. id (Intercept) 0.11562 0.34003 Residual 0.22765 0.47713 number of obs: 2568, groups: id, 107 run this model (only difference is I've removed the interaction term): mod2 <- lmer(x~category+subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 3987 4151 -1966 3823 3931 Random effects: Groups N...

...y ~ x1 + x2, method="lms", nsamp="exact") mod1$coefficients (Intercept) x1 x2 35.5293489 0.4422742 -1.2944534 mod1$bestone [1] 12 17 27 Now, instead of using the formula, I want to provide the design matrix and the response, then: X <- cbind(1, x1, x2) mod2 <- lqs.default(X, y, intercept=F, method="lms", nsamp="exact") mod2$coefficients x1 x2 35.4217275 0.4276641 -1.2834731 mod2$bestone [1] 6 14 15 The results are not the same (?!). Furthermore, if I create the design matrix without the column of 1&...

...ndows 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 /dtrace/mod2 1024 dad1 R 8.807 bash 5184 /usr/bin/ls 8192 dad1 R 10.332 ls 5184 /dtrace/mod2/examples 1024 dad1 R 0.259 fsflush 3 /var/tmp/dtrace-1b 8192 dad1 W 0.278 # io.d ......

...224, 257, 200, 190, 208, 224 ), `plant-density` = c(137, 107, 132, 135, 115, 103, 102, 65, ?????????????????????? 149, 85, 173, 124, 157, 184, 112, 80, 165, 160, 157, 119)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -20L)) # regression model works mod2 = lm(`cob-wt` ~ `plant-density`, data = cob) # x sequence for plotting CI's # Set up x points x = seq(min(cob$`plant-density`), max(cob$`plant-density`), length = 1000) # Use predict to get CIs for a plot # Add CI for regression line (y-hat uses 'c') # usual trick is to assign x to ac...

...ns, 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]) ####################### I'm looking for the interception points between these two straight lines from these fitted models, mod1 and mod2. Could someone advice me the way to do it? Thank you Fir?

My OQO mod2s just arrived! All 4 of them; portable testbed here it comes!!! I hope.... The units weigh in at ~.75lbs, and measure aprox. 5 5/8 x 3 3/8 x 1 1/8" A little larger when you push up the 5" screen to reveal the keyboard. WiFi, Bluetooth, HDMI, and a single USB built in; an adapter for...

...data = cars) abline(reg = mod) # nothing appears This behaves as documented, but might catch someone. Would it be an improvement if this situation was detected so as to plot the appropriate horizontal line, i.e. abline(a = coef(mod), b = 0) ? Would it also be an improvement if for a model like mod2 <- lm(dist ~ 1 + offset(speed), data = cars) abline(reg = mod2) would be equivalent to abline(a = coef(mod2), b = 1) ? For models through the origin, the current function works fine, but one might even consider models through the origin and having the independent variable in an offset() to be...

Hi , I know this question has been asked twice in the past but to my knowldege, it still hasn't been solved. I am doing a zero inflated binomial model using the VGAM package, I need to obtain p values for my Tvalues in the vglm output. code is as follows > mod2=vglm(dmat~Season+Diel+Tidal.phase+Tidal.cycle,zibinomial, data=mp1) > summary(mod2) Call: vglm(formula = dmat ~ Season + Diel + Tidal.phase + Tidal.cycle, family = zibinomial, data = mp1) Pearson Residuals: Min 1Q Median 3Q Max logit(phi) -3.6496 0.273...

...y) abline(v=-1,col=2,lty=2) mod<-lm(y~x+x*(x>-1)) summary(mod) yy<-predict(mod) lines(x[order(x)],yy[order(x)],col=2,lwd=2) #--lme #grouping factor, unbalanced g<-as.character(c(1:200)) id<-sample(g,size=1000,replace=T, prob=sample(0:1,200,rep=T)) table(id) #unbalanced mod2<-lme(y~x+x*(x>-1),random=~x|id, data=data.frame(x,y,id)) summary(mod2) newframe<-data.frame( #fictious id id="fictious", x) newframe[1:5,] #predictions yy2<-predict(mod2,level=0, newdata=newframe) lines(x[order(x)],yy2[order(x)],col="blue",lwd=2) # add va...

...ars that CL2_Free is the most significantly correlated so i've decided to include that in the final model. Positively skewed I'm thinking of doing a mixed effects model with random intercepts as the treatment code and zones within the treatment cose and random slopes as the seasons. mod2 <- lme(tthm ~ cl2free, random= ~ seasons| treatcode/loc_code) but that doesn't work. these seems to work good: mod2 <- lme(tthm ~ cl2free, random= ~ 1| loc_code, data=new.data, method="ML") mod3 <- lme(tthm ~ cl2free, random= ~ 1| treatcode/loc_code, d...

Dear R-list, I would like to use the lattice library to show several groups on the same graph. Here's my example : ## the data f1 <- factor(c("mod1","mod2","mod3"),levels=c("mod1","mod2","mod3")) f1 <- rep(f1,3) f2 <- factor(rep(c("g1","g2","g3"),each=3),levels=c("g1","g2","g3")) df <- data.frame(val=c(4,3,2,5,4,3,6,5,4), x=rep(c(1,2,3),3...