similar to: predict

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 <-

I want to predict values from an existing lm (linear model, e.g. lm.obj) result in R using a new set of predictor variables (e.g. newdata). However, it seems that because my linear models was made by calling scale() on the target predictor that predict exits with an error, "Error in scale(xxA, center = 9.7846094491829, scale = 0.959413568556403) : object 'xxA' not found". By

Hi, See below I reply your message for <https://stat.ethz.ch/pipermail/r-help/2008-April/160966.html>[R] predict.glm & newdata posted on Fri Apr 4 21:02:24 CEST 2008 You say it ##works fine but it does not: if you look at the length of yhat2, you will find 100 and not 200 as expected. In fact predict(reg1, data=x2) gives the same results as predict(reg1). So I am still looking for

Full_Name: Renaud Lancelot Version: Version 2.3.0 (2006-04-24) OS: MS Windows XP Pro SP2 Submission from: (NULL) (82.239.219.108) I think there is a bug in predict.lme, when a polynomial generated by poly() is used as an explanatory variable, and a new data.frame is used for predictions. I guess this is related to * not * using, for predictions, the coefs used in constructing the orthogonal

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1)) alg = "plinear", trace =

Hi all - I'm stumped by the following mdl <- glm(resp ~ . , data = df, family=binomial, offset = ofst) WORKS yhat <- predict(mdl) WORKS yhat <- predict(mdl,newdata = df) FAILS Error in drop(X[, piv, drop = FALSE] %*% beta[piv]) : subscript out of bounds I've tried without offset, quoting binomial. The offset variable ofst IS in df. Previous postings indicate possible

Hello, I'm new here, but will try to be as specific and complete as possible. I'm trying to use “lm“ to first estimate parameter values from a set of calibration measurements, and then later to use those estimates to calculate another set of values with “predict.lm”. First I have a calibration dataset of absorbance values measured from standard solutions with known concentration of

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

Hi, excuse me for my english, i am using R on windows and i have to do several graphs with axis labels and the axis text thicks has a specified font type, (Arial) and a specified font size. How can i do these? Thank you in advance -- View this message in context: http://www.nabble.com/graph%3A-axis-label-font-tp24463974p24463974.html Sent from the R help mailing list archive at Nabble.com.

Hi all - my first time here and am having an issue with the Predict function. I am using a tutorial as a guide, locate here: http://www.ats.ucla.edu/STAT/R/dae/mlogit.htm My code gives this error > newdata1$predicted <- predict(mlogit,newdata=newdata1,type="response") Error in `$<-.data.frame`(`*tmp*`, "predicted", value = c(0.332822934960197, : replacement has

I am fittling a spline to a variable in a regression model, I am then using the predict.glm funtion to make some predictions. When I use bs to fit the spline I don't have any problems using the predict.glm function however when I use ns I get the following error: Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : variable lengths differ (found for

Dear R-Users My problem is quite simple: I need to use a fitted model to predict the next point (that is, just one single point in a curve). The data was divided in two parts: identification (x and y - class matrix) and validation (xt and yt - class matrix). I don't use all values in x and y but only the 10 nearest points (x[b,] and y[b,]) for each regression (b is a vector with the

Also, and possibly more constructively, when you get an error like > CI.c = predict(mod2, data.frame( `plant-density` = x), interval = 'c') # fail Error in eval(predvars, data, env) : object 'plant-density' not found you should check your assumptions. Does "newdata" actually contain a columnn called "plant-density": > head(data.frame( `plant-density`

The model frame shows the response and predictors in a data frame with nicely labelled columns: fm <- lm(wt~qsec+log(hp)+sqrt(disp), data=mtcars) model.frame(fm) # ok When the left hand side consists of more than one response, those response variables still look good, inside a matrix: fm <- lm(cbind(qsec,hp,disp)~wt, data=mtcars) model.frame(fm)[[1]] # ok A problem arises when

Dear all, I have built glm models based on presences/absences and a number of predictor maps and would like to compute habitat suitability based on the modelled coefficients. I thought this is pretty straight forward and wanted to use predict() and supply the new data in a data frame, with one column for each predictor. However, I do get an error msg warning me that the number of rows for

Hi, I am experimenting with the function predict() in two versions of R and the R extension package "survival". library(survival) set.seed(123) testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10)) testfm=as.formula('Surv(otime,event)~x') testfun=function(dat,fm) { predict(coxph(fm,data=dat),type='lp',newdata=dat) } # Under R 2.11.1 and

Hi, I am experimenting with the function predict() in two versions of R and the R extension package "survival". library(survival) set.seed(123) testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10)) testfm=as.formula('Surv(otime,event)~x') testfun=function(dat,fm) { predict(coxph(fm,data=dat),type='lp',newdata=dat) } # Under R 2.11.1 and

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

I posted this one as an R bug (https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14767), but Prof. Ripley says I'm premature, and I should raise the question here. 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

"Thank you Rui. I didn't know about the check.names = FALSE argument. > Another good reminder to always read help, but I'm not sure I understood > what help to read in this case" ?data.frame , of course, which says: "check.names logical. If TRUE then the names of the variables in the data frame are checked to ensure that they are syntactically valid variable names