search for: predicted

Hi, I've noted that not all `predict' methods are hidden in the namespace: > methods("predict") [1] predict.ar* predict.Arima* [3] predict.arima0* predict.glm [5] predict.HoltWinters* predict.lm [7] predict.loess* predict.mlm [9] predict.nls* predict.poly [11] predict.ppr* predict.prcomp* [13]

Hi, I'm trying to use the R Survival analysis on a windows 7 system. The input data format is described at the end of this mail. 1/ I tried to perform a survival analysis including stratified variables using the following formula. cox.xtab_miR=coxph(Surv(time, status) ~ miR + strata(sex,nbligne, age), data=matrix) and obtain the following error message Warning message: In fitter(X, Y,

I''m trying to write a football prediction game. I have a Matches table and a Predictions table. Each Match :has_many Predictions, and each Prediction :belongs_to a Match Each Match has (amongst other things) a Match.match_date field. Each Prediction has (amongst other things) a User_id field I''m trying to get a list of all Predictions a particular User has made which

Hello. I just downloaded R onto a new computer, and after entering library(MASS), I still get the message "Error: could not find function "predict.lda"" when I try to use the predict.lda function (even just "predict.lda()") Can anyone help me out? Thank you! Diana Connett

Dear list, yet another question on S4 method behaviour or rather its side effects: Consider the following example (see below) where a new S4 method 'predict' is defined which by doing so makes 'predict' a S4 standardGeneric. It works and simple test examples run but I'm cautious whether this is a 'silly' thing to do. Is this a dangerous thing to do, already known to

...ike some our responses better than others and wanted to check this. Subjects could rate a query/response pair on a scale from 0 (very bad response) to 10 (very good response). Here are all the judgments we received for one particular class of response to queries which we thought users would like: Predicted-Good-0, 4 Predicted-Good-1, 1 Predicted-Good-2, 11 Predicted-Good-3, 8 Predicted-Good-4, 25 Predicted-Good-5, 12 Predicted-Good-6, 21 Predicted-Good-7, 25 Predicted-Good-8, 30 Predicted-Good-9, 52 Predicted-Good-10, 189 And here are all the judgments we received for one particular class o...

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Oh yes. The "load package" under the "packages menu" in the Windows version does that. To check I typed "library(assist)" after starting R. Same behavior, ssr is found, but others like predict.ssr, and plot.ssr, give a "not found" message. Thanks for the suggestion. Mike

Hi, I would like to step-through a non-visible function. but apparently I don't know enough about namespaces to get that to work: > methods(predict) ... deleted lines ... [27] predict.rpart* predict.smooth.spline* [31] predict.survreg.penal* Non-visible functions are asterisked > debug(predict.rpart) Error: Object "predict.rpart" not found >

Hello dear r-help group I am turning for you for help with FAQ number 7.31: "Why doesn't R think these numbers are equal?" http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f *My story* is this: I wish to run many lm predictions and need to have them run fast. Using predict.lm is relatively slow, so I tried having it run faster by

R-helpers, I am using the package "drc" to fit a 4 parameter logistic model. When I use the predict function to get prediction on a new dataset, I am not getting the requested confidence or prediction intervals. Any idea what is going on? Here is code to reproduce the problem: --- library(drc) # Fit model to existing dataset in package spinach.model <- drm(SLOPE~DOSE, data =

Hello R-Users, I have been using (nnet) by Ripley to train a neural net on a test dataset, I have obtained predictions for a validtion dataset using: PP<-predict(nnetobject,validationdata) Using PP I can find the -2 log likelihood for the validation datset. However what I really want to know is how well my nueral net is doing at classifying my binary output variable. I am new to R and I

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

For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat <= newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my

Hi there: i got a problem to get the prediction from a model recently. for example if i use a survival analysis to predict the risk. i use the code like below: i found the the prediction is not equal to (coef * x + coef * sex) , could someone help me with why this happened? and can someone explain to me how this command "predict(f, type="terms")" works? is every partial

Hi I have a problem with lm and predict I have us [1] 2789.53 3128.43 3255.03 3536.68 3933.18 4220.25 4462.83 4739.48 [9] 5103.75 5484.35 5803.08 5995.93 6337.75 6657.40 7072.23 7397.65 [17] 7816.83 8304.33 8746.98 9268.43 9816.98 10127.95 10469.60 10960.75 [25] 11685.93 12433.93 13194.70 13843.83 us.p [1] 227.62 229.92 232.13 234.25 236.31 238.42 240.59 242.75 244.97

I am new to R, so I am sure I am making a simple mistake. I am including complete information in hopes someone can help me. Basically my data in R looks good, I write it to a file, and every value is off by 1. Here is my flow: > str(prediction) Factor w/ 10 levels "0","1","2","3",..: 3 1 10 10 4 8 1 4 1 4 ... - attr(*, "names")= chr

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

Dear all, the help file for predict.gbm states that "The predictions from gbm do not include the offset term. The user may add the value of the offset to the predicted value if desired." I am just not sure how exactly, especially for a Poisson model, where I believe the offset is multiplicative ? For example: library(MASS) fit1 <- glm(Claims ~ District + Group + Age + offset(log(Holders)), data = Insurance, family = poisson) head(predict(fit1, data...

...ata.frame(x = seq(min(V1), max(V1), length.out = length(V1))) pred.w.clim <- predict(lm(V2 ~ V1), new, interval="confidence") matplot(new$x, pred.w.clim, lty=c(1,2,2), type="l", col=c("black", "red", "red"), ylab="predicted y") points(V1,V2) # Create x & y equal to V1 & V2 x <- V1 y <- V2 pred.w.clim2 <- predict(lm(y ~ x), new, interval="confidence") matplot(new$x, pred.w.clim2, lty=c(1,2,2), type="l", col=c("black", "red", "red&qu...