search for: predvar

There is a bug in `predict' whereby the order of variables sometimes gets re-arranged compared to the original fit, and then disaster results. Specifically, the 'variables' and 'predvars' attributes of a 'terms' object get out of synch. This only happens when the terms in the original formula get re-ordered during fitting: test> scrunge.data_ data.frame( contin=1:10, discrete=factor( rep( c( 'cat', 'dog'), 5)), resp=runif( 10)) test> lm.ok_ lm( r...

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light on the rationale for non-preservation? Terry T. Simple e...

...; c("Not_reliable_at_all","Less_reliable","Somehow_reliable","Very reliable")) > > First model > object1 <- polr(Inc ~ Adopt ,Data,Hess = T,method = "logistic") > > Marginal effect for object1: > ocME(object1) > Error in eval(predvars, data, env) : > numeric 'envir' arg not of length one > > Second model > object2 <- polr(Inc ~ Adopt + Training ,Data,Hess = T,method = "logistic" ) > > Marginal effect for object2: > ocME(object2) > effect.Not_reliable_at_all effect.Less_reliable eff...

...Here's the behavior I assert is a bug: The output from delete.response on a terms object alters the formula by removing the dependent variable. It removes the response from the "variables" attribute and it changes the response attribute from 1 to 0. The response is removed from "predvars" But it leaves the name of the dependent variable first in the in "dataClasses". It caused an unexpected behavior in my code, so (as usual) the bug may be mine, but in my heart, I believe it belongs to delete.response. To illustrate, here's a terms object from a regression....

Hi all, Very surprising (to me!) and mystifying result from predict.glm(): the predictions vary depending on whether or not I use ns() or splines::ns(). Reprex follows: library(splines) set.seed(12345) dat <- data.frame(claim = rbinom(1000, 1, 0.5)) mns <- c(3.4, 3.6) sds <- c(0.24, 0.35) dat$wind <- exp(rnorm(nrow(dat), mean = mns[dat$claim + 1], sd = sds[dat$claim + 1])) dat <-

Dear Oscar, and any other R-project person, Can you please help me to figure out the meaning of the following error message in red ? Error in eval(predvars, data, env) : numeric 'envir' arg not of length one I computed ordered logit models using 'polr' in R (I just followed the guidance a handout I found on princeton.edu about logit, probit and multinomial logit models) . The summary results are obtained without any problem. *Howev...

....labels")= chr "x" .. .. .. ..- attr(*, "order")= int 1 .. .. .. ..- attr(*, "intercept")= int 1 .. .. .. ..- attr(*, "response")= int 1 .. .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> .. .. .. ..- attr(*, "predvars")= language list(y, x) .. .. .. ..- attr(*, "dataClasses")= Named chr [1:2] "factor" "numeric" .. .. .. .. ..- attr(*, "names")= chr [1:2] "y" "x" ..$ x : num [1:1615, 1:2] 1 1 1 1 1 1 1 1 1 1 ... .. ..- attr(*...

...e to predict value in different points. I have tried to use predict function but it doesn't work. I have used predict function like this: newdata<-seq(from=0.1, to=0.32,by=0.02) data<-predict(fm,newdata) where fm is a regression model. The predict function return me that: Error in eval(predvars, data, env) : numeric argument 'envir' doesn't have unitary length where is the error? -- View this message in context: http://www.nabble.com/predict-tp24719362p24719362.html Sent from the R help mailing list archive at Nabble.com.

...data.frame(x=rnorm(100),y=rnorm(100)) } Now, if I want to do create the five models, I do not know how to set this created data.frames I have tried: for (i in 1:5){ assign(paste0("model",i),lm(y~x,data=paste0("variable",i))) } but it does not working. Error in eval(predvars, data, env) : invalid 'envir' argument of type 'character' Can you please help me to find the correct way? Thanks! Jes?s [[alternative HTML version deleted]]

Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it works. I mean the name of the newdata is x, why it does not work though? 2. Because the prediction is conducted for log(y). I need to get the expected value of y, which is LN distributi...

...number, or 0 to exit 1: tfun(tdata) 2: survexp.test(zed ~ 1, data = mydata) 3: eval(m, parent.frame()) 4: eval(expr, envir, enclos) 5: model.frame(formula = "zed ~ 1+age+sex+year", data = mydata) 6: model.frame.default(formula = "zed ~ 1+age+sex+year", data = mydata) 7: eval(predvars, data, env) 8: eval(expr, envir, enclos) Selection: 2 Called from: model.frame.default(formula = "zed ~ 1+age+sex+year", data = mydata) Why does selecting "2" result in going to frame "6"? Terry Therneau

Hi there, I wrote a function that wraps MASS::boxcox as: bc <- function(vec) { lam <- boxcox(lm(vec ~ 1)) lam <- lam$x[which.max(lam$y)] (vec^lam - 1)/lam } When I invoke it as: > x <- runif(20) > bc(x) Error in eval(predvars, data, env) : object 'vec' not found I have googled, and rewrote the above function as: bc <- function(vec) { dat <<- data.frame(vec = vec) lam <- boxcox(lm(vec ~ 1, dat)) lam <- lam$x[which.max(lam$y)] rm(dat, envir = .GlobalEnv) (vec^lam - 1)/lam } It wor...

Hi there, I wrote a function that wraps MASS::boxcox as: bc <- function(vec) { lam <- boxcox(lm(vec ~ 1)) lam <- lam$x[which.max(lam$y)] (vec^lam - 1)/lam } When I invoke it as: > x <- runif(20) > bc(x) Error in eval(predvars, data, env) : object 'vec' not found I have googled, and rewrote the above function as: bc <- function(vec) { dat <<- data.frame(vec = vec) lam <- boxcox(lm(vec ~ 1, dat)) lam <- lam$x[which.max(lam$y)] rm(dat, envir = .GlobalEnv) (vec^lam - 1)/lam } It wor...

...rror, "Error in scale(xxA, center = 9.7846094491829, scale = 0.959413568556403) : object 'xxA' not found". By debugging predict, I can see that the error occurs in a call to model.frame. By debugging model frame I can see the error occurs with this command: variables <- eval(predvars, data, env); it seems likely that the error is because predvars looks like this: list(scale(xxA, center = 10.2058714830537, scale = 0.984627257169526), scale(xxB, center = 20.4491690881149, scale = 1.13765718273923)) An example case: dat <- data.frame(xxA = rnorm(20,10), xxB = rn...

...96 4.06743266 > unname(predict(m2, newdata = newdf)) > #> [1] 0.5194712 -0.5666554 -0.1731268 2.8134844 3.9295814 > > Is this a bug? The two objects m1 and m2 differ more than they should, so this looks like a problem in glm, not just in predict.glm. > attr(m1$terms, "predvars") list(claim, ns(wind, knots = c(25.4756277492997, 30.2270250736796, 35.4093171222502, 43.038645381669), Boundary.knots = c(12.9423820390783, 108.071583734075), intercept = FALSE)) > attr(m2$terms, "predvars") list(claim, splines::ns(wind, df = 5)) This appears to be due to a...

Dear All: Just fixed where is the problem I am trying to use the R function "svm" with "type~." , but I got the following error message SVM.Model1 <- svm(type ~ ., data=my.data.x1x2y, *type='C-classification'*, kernel='linear',scale=FALSE) *Error in eval(predvars, data, env) : object 'type' not found* I am wondering if I should install a specific R package(s). *Here is my codes:* feature.x1 <- c(0.25,0.5,1,1,1.5,2,2.25,2.5,2,1,3, 5,3.75, 1,3.5,4,4,5,5.5,6,6,6.5) length(feature.x1) feature.x2 <- c(2,3.5,1,2.5,1.75,2,1.75,1.5,2.5,1...

...me <- paste(TimePt,'df',sep='.') # invent a relevant name/symbol as a character string. assign(temp.name,IGF.df[IGF.df$TPt==TimePt,]) # this works. The relevant variable is now a data frame lm(b ~ Strain+BWt+PWt+PanPix, data=temp.name)) # this gives an error, namely Error in eval(predvars, data, env) : invalid 'envir' argument I think it's obvious what I want to achieve, but how is it done? I tried data=as.name(temp.name) but that also didn't work. I can't find anything relevant in "Introduction to R". Here is a secondary question: While trying to un...

...x2=x1*2 r.norm2=rnorm(200,0,20) x2=r.norm2+x2 reg1=glm(y1~x1,binomial) yhat=predict(reg1) ##prediction works fine yhat=predict(reg1,newdata=x2) ##gives error message because of "newdata" >Error in eval(predvars, data, env) : numeric 'envir' arg not of length one yhat=predict(reg1,data=x2) ##works fine ##using offset ofst=rep(0.5,100) reg1=glm(y1~x1,binomial,offset=ofst) yhat=predict(reg1) yhat1=predict(reg1,newdat...

Dear All: I am trying to use the R function "svm" with "type =C-classification" , but I got the following error message SVM.Model1 <- svm(type ~ ., data=my.data.x1x2y, *type='C-classification'*, kernel='linear',scale=FALSE) *Error in eval(predvars, data, env) : object 'type' not found* I am wondering if I should install a specific R package(s). *Here is my codes:* feature.x1 <- c(0.25,0.5,1,1,1.5,2,2.25,2.5,2,1,3, 5,3.75, 1,3.5,4,4,5,5.5,6,6,6.5) length(feature.x1) feature.x2 <- c(2,3.5,1,2.5,1.75,2,1.75,1.5,2.5,1...

...five models, I do not know how to set this created data.frames > > > I have tried: > > > for (i in 1:5){ > > > assign(paste0("model",i),lm(y~x,data=paste0("variable",i))) > > > } > > > but it does not working. > > Error in eval(predvars, data, env) : > invalid 'envir' argument of type 'character' > > > Can you please help me to find the correct way? > > > Thanks! > Jes?s > > > > > > > > > > [[alternative HTML version deleted]] > > > _________...