search for: pred2

...d onto it when the code below is run and temp is returned. I've been trying to figure this out for too long. It doesn't matter when I put the FPVAL in the return statement. It happens regardless of whether it's first or last. Thanks. f.lmmultenhanced <- function(response, pred1, pred2) { regmod <- lm(response ~ pred1 + pred2) lmsum <- summary(regmod) imbcoef<-lmsum$coefficients[2,1] retcoef<-lmsum$coefficients[3,1] imbpval<-lmsum$coefficients[2,4] retpval<-lmsum$coefficients[3,4] Fstat<-lmsu...

hi, i'm working with mgcv packages and specially gam. My exemple is: >test<-gam(B~s(pred1)+s(pred2)) >plot(test,pages=1) when ploting test, you can view pred1 vs s(pred1, edf[1] ) & pred2 vs s(pred2, edf[2] ) I would like to know if there is a way to access to those terms (s(pred1) & s(pred2)). Does someone know how? the purpose is to access to equation of smooths terms in order to...

...ted a simple new.data.frame: new.data.fame <- data.frame new.data.frame[,"JTemp"] <- 10.5 new.data.frame[,"SNied"] <- 430 Than I used predict() to predict values for "pnV22" in the following way: pred <- predict(result, data.frame) pred2 <- predict(result, new.data.frame) The results are the same, which I checked by ploting the values of pred and pred2 and by table(pred ==pred2) which is true for all values. Looking at the tree I would expect that pred2 has the same high value for all elements of the vector. Did I make a...

...using the predict() function, and thus get a point-estimate OR. But I can't see how to obtain the confidence interval for such an OR. For example: model <- glm(chd ~age.cat + male + lowed, family=binomial(logit)) pred1 <- predict(model, newdata=data.frame(age.cat=1,male=1,lowed=1)) pred2 <- predict(model, newdata=data.frame(age.cat=2,male=0,lowed=0)) OR <- exp(pred2-pred1) Thanks [[alternative HTML version deleted]]

...ot;dataframe". These indexed lists can be printed successfuly but are not agreeable to the plot() and lm() functions shown below as are their df$out references. Reading the documentation for plot and lm hasn't helped yet. Thanks in advance - Stan. > df=data.frame(out=1:4*3,pred1=1:4,pred2=1:4*2) > regression=function(tble,a,b) + { + plot.new() + plot(tble[a]~tble[b]) + lmm=lm(tble[a]~tble[b]) + abline(lmm) + anova(lmm) + } > df[1] out 1 3 2 6 3 9 4 12 > df out pred1 pred2 1 3 1 2 2 6 2 4...

...g other things, runs a linear model and returns r2. But, the number of predictor variables passed to the function changes from 1 to 3. How can I change the formula inside the function depending on the number of variables passed in? An example: get.model.fit <- function(response.dat, pred1.dat, pred2.dat = NULL, pred3.dat = NULL) { res <- lm(response.dat ~ pred1.dat + pred2.dat + pred3.dat) summary(res)$r.squared # other stuff happens here... } y <- rnorm(10) x1 <- y + runif(10) x2 <- y + runif(10) x3 <- y + runif(10) get.model.fit(y, x1, x2, x3) get.model.fit(y, x1,...

Dear all, I tried to run as.formula("x") and got an error message "Error: C stack usage 7971120 is too close to the limit" whether x exists or not. This is not the case in as.formula("y"), where "object 'y' not found" is the error message if y not exists, or "invalid formula" error or a formula depending on y. Can anyone confirm this is

I'm fairly new to R, and I'm trying to write out a population model that satisfies the following; the system consists of s species, i= 1, 2,...,s network of interactions between species is specified by a (s x s) real matrix, C[i,j] x[i] being the relative population of the "ith" species (0 =< x[i] =< 1, sum(x[i]=1) the evolution rule being considered is as follows;

...Groups Name Variance Std.Dev. vpnr (Intercept) 0.12965 0.36007 Xr.1 s(hours24) 1291.42444 35.93639 Number of obs: 97920, groups: vpnr, 114; Xr.1, 8 Fixed effects: Estimate Std. Error z value Pr(>|z|) X(Intercept) 0.3713345 0.0644469 5.762 8.32e-09 *** Xpred2 -0.0575848 0.0865231 -0.666 0.506 Xpred3 0.0003748 0.0869543 0.004 0.997 … Parametric coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.3713345 0.0149858 24.779 < 2e-16 *** pred2 -0.0575848 0.0197488 -2.916 0.0035...

...library(e1071) train<- iris[c(1:30,50:80,100:130),] test<- iris[-c(1:30,50:80,100:130),] y.train<- train$Species y.test<- test$Species obj<- tune.svm(train[,-5], y.train, gamma = 2^(-1:1), cost = 2^(2:4), probability=T) my.svm<- obj$best.model pred1<- predict(my.svm, test[,-5]) pred2<- predict(my.svm, test[,-5], probability=T) table(pred1, y.test) table(pred2, y.test) When I do this, the two different tables often come out different, as below: > table(pred1, y.test) y.test pred1 setosa versicolor virginica setosa 19 0 0 ver...

...library(e1071) train<- iris[c(1:30,50:80,100:130),] test<- iris[-c(1:30,50:80,100:130),] y.train<- train$Species y.test<- test$Species obj<- tune.svm(train[,-5], y.train, gamma = 2^(-1:1), cost = 2^(2:4), probability=T) my.svm<- obj$best.model pred1<- predict(my.svm, test[,-5]) pred2<- predict(my.svm, test[,-5], probability=T) table(pred1, y.test) table(pred2, y.test) When I do this, the two different tables often come out different, as below: > table(pred1, y.test) y.test pred1 setosa versicolor virginica setosa 19 0 0 ver...

...dels. When my sample size is reduced from 300 to 150, the function complains (length of dimnames[1] not equal to array) and does not produce any results. There are no missing values in the data. Any suggestions for a work-around? Thank you in Advance. > f.total=cph(Surv(fu,censor)~predictor+pred2+pred3,data=data,x=T,y=T,surv=T) > set.seed(6) > val.step=validate(f.total,B=155,bw=T) Backwards Step-down - Original Model No Factors Deleted Factors in Final Model [1] pred2 .Random.seed: 1 -1021164091 1170333634 in .GlobalEnv Iteration: 1 2 3 4 5 6 7 Error in fit(NULL,...

...ta = Newdata) > > # "correct" model matrix for predictions > p <- poly(Orthodont$age, 3) > mm2 <- model.matrix(~ poly(age, 3, coefs = attr(p, "coefs")) + Sex, data = Newdata) > > data.frame(pred1 = predict(fm, level = 0, newdata = Newdata), + pred2 = mm1 %*% fixef(fm), + pred3 = head(predict(fm, level = 0)), + pred4 = mm2 %*% fixef(fm)) pred1 pred2 pred3 pred4 1 18.61469 18.61469 23.13079 23.13079 2 23.23968 23.23968 24.11227 24.11227 3 29.90620 29.90620 25.59375 25.59375 4 36.19756 36.19756 27.03819 27.038...

...Dat) Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) # use matrix algebra x <- cbind(1, (Dat$Gender == "Male")) %*% coef(Model) pred1 <- exp(x)/(1 + exp(x)) # use the fitted line equation y <- coef(Model)[1L] + coef(Model)[2L] * (Dat$Gender == "Male") pred2 <- exp(y)/(1 + exp(y)) head(predict(Model, type="response")) head(pred1) |> c() head(pred2) Hope this helps, Rui Barradas -- Este e-mail foi analisado pelo software antiv?rus AVG para verificar a presen?a de v?rus. www.avg.com

...env") Error: object of type 'special' is not subsettable > as.formula("...") Error in eval(expr, envir, enclos) : '...' used in an incorrect context It may happen for the same reason that the following does not give an error: > y <- "response ~ pred1 + pred2" > as.formula("y") response ~ pred1 + pred2 and that the followings give a somewhat surprising result > f <- function(x) { y <- "foo ~ bar" ; as.formula(x) } > f("y") response ~ pred1 + pred2 <environment: 0x1e87978> The character method fo...

...ed(15) model = svm(data.frame(x, y), as.factor(c), probability=TRUE ) pred1 = predict( model, newdata = data.frame(x2, y2), probability=TRUE) probas1 = as.numeric(attr(pred1,"probabilities")[,"TRUE"]) set.seed(15) model = svm(data.frame(x, y), as.factor(c), probability=TRUE ) pred2 = predict( model, newdata = data.frame(x2, y2), probability=TRUE) probas2 = as.numeric(attr(pred2,"probabilities")[,"TRUE"]) sum(pred1 != pred2) # It should be 0 sum(probas1 != probas2) # It should be 0 plot(probas1,probas2,xlim=c(0.4,0.6),ylim=c(0.4,0.6),col="red&q...

Hi, My questions concerns the ROCR package and I hope somebody here on the list can help - or point me to some better place. When evaluating a model's performane, like this: pred1 <- predict(model, ..., type="response") pred2 <- prediction(pred1, binary_classifier_vector) perf <- performance(pred, "sens", "spec") (Where "prediction" and "performance" are ROCR-functions.) How can I then retrieve the cutoff value for the sensitivity/specificity tradeoff with regard to the d...

...~ x4+ x5+ x6+ x7, data = dat, family = binomial(link=logit))model2 <- glm (Y~ x1 + x2 +x3 , data = dat, family = binomial(link=logit))  and I would like to compare these two models based on the prediction that I get from each model: pred1 = predict(model1, test.data, type = "response")pred2 = predict(model2, test.data, type = "response") I have used ROCR package to compare them:pr1 = prediction(pred1,test.y)pf1 = performance(pr1, measure = "prec", x.measure = "rec")  plot(pf1) which cutoff this plot is based on? pr2 = prediction(pred2,test.y)pf2 = perform...

...e prediction does not seem to depend upon ### the values of the dependent variable included in the data pred1 <- predict(bot, newdata = test[, var_names], type="vector", update.tree.predictions = F); test$y <- as.factor(sample(0:1, size = dim(test)[1], replace = T)) pred2 <- predict(bot, newdata = test[, var_names], type="vector", update.tree.predictions = F); abs(mean(pred1[,1] - pred2[,1])) if (abs(mean(pred1[,1] - pred2[,1])) > 1e-3) { print("Results do differ."); } ### What is more curious is that the error messa...

....lme1) # How go force intercept = 0 ??? grd.lme0 = lme(newbone~t*treat-1,data=grd,random=~1|subject) grd$pred0 = predict(grd.lme0,level=0) summary(grd.lme0) # Gives true, all.equal(grd$pred1,grd$pred0) # Everything as expected without treat grd.lme2 = lme(newbone~t,data=grd,random=~1|subject) grd$pred2 = predict(grd.lme2,level=0) summary(grd.lme2) # Forced intercept = 0 grd.lme3 = lme(newbone~t-1,data=grd,random=~1|subject) grd$pred3 = predict(grd.lme3,level=0) summary(grd.lme3) # As expected: not equal all.equal(grd$pred2,grd$pred3) #------------------------------------------------------------...