similar to: predict.mlm: Error: Object "X" not found (PR#1049)

Hello, I've just fitted a model with multi-responses, and I get an object of class "lm" "mlm". My problem is that as soon as I invoke the predict method for a dataframe "newdata", the methods runs and give me back prediction at the fitting points but not for newdata. Does someone has an explanation for this behavior, and some ideas to make predict.mlm work

I believe there is a coding error in the first part of predict.mlm in the splines package, but perhaps someone could explain the logic to me if I'm wrong. I don't see how the second if (missing(newdata)) could ever be true. (I would show the code but I'm using email on a different machine than R today.) Paul Gilbert

predict.lm keeps row names when working from several rows in newdata, but always removes rowname from a single row. The rownames are removed by the line in predict.lm predictor <- drop(X[, piv, drop = FALSE] %*% beta[piv]) What is the reason for that decision? I usually want to retain the row names. tmp <- data.frame(x=1:4, y=c(1,3,2,5)) tmp.lm <- lm(y ~ x, data=tmp) tmp.new <-

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

Here is the first one. It concerns the handling of multiple offsets. The following lines creates a list with 3 explanatory variables and one response. > x<-seq(0,1,length=10);y<-sin(x);z<-cos(x); w<-x+y+z+rnorm(x) > data<-list(x=x,y=y,z=z,w=w) A lm is fitted with one explanatory variable and two offsets. So far, so good. >

I would like to perform multiple comparisons or post-hoc testing on the independent variable in an Anova.mlm object generated by the Anova function of the car package. I have defined a multivariate linear model and subsequently performed a repeated measures ANOVA as per the instructions in section #3 of the following comprehensive tutorial on the subject from the Gribble lab at UWO:

Hello. I am new R user, so this question is probably quite stupid, but for the life of me I cannot figure out how to get predications using multivariate linear regression analysis. Single variable predictions work fine. I am trying the following: -- Known y's for known x's1 and x's2 ys <- c(133890, 135000, 135790, 137300, 138130, 139100, 139900, 141120, 141890, 143230, 144000,

I had a look at some influence measures, and it seems to me that currently several methods handle multiple lm (mlm) objects wrongly in R. In some cases there are separate "mlm" methods, but usually "mlm" objects are handled by the same methods as univariate "lm" methods, and in some cases this fails. There are two general patterns of problems in influence measures:

Hello, I am using the function "lm" to fit several responses at the same time (100 responses). At the end of the fit, I get an object of class "mlm". I would like to know if there is a way to access to each of the 100 underlying models separately (is it a list, ... ?). Which syntax should I use to see and use the 15th model (for instance) just like it is possible for classical

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>>>>> steven pav >>>>> on Thu, 19 Jul 2018 21:51:07 -0700 writes: > It seems that confint.default returns an empty data.frame > for objects of class mlm. For example: > It seems that confint.default returns an empty data.frame for objects of > class mlm. Not quite: Note that 'mlm' objects are also 'lm' objects, and so it is

Dear R-devel'ers Below is an outline for a set of routines to improve support for multivariate linear models and "classical" repeated measurements analysis. Nothing has been coded yet, so everything is subject to change as loose ideas get confronted by the harsh realities of programming. Comments are welcome. They might even influence the implementation... -pd General

Full_Name: Yves Rosseel Version: 2.2.1 OS: Submission from: (NULL) (157.193.116.152) The code below sketches a possible implementation of a function 'mlh.mlm' which I think would be a good complement to the 'anova.mlm' function in the stats package. It tests a single linear hypothesis of the form H_0: LBT'= 0 where B is the matrix of regression coefficients; L is a matrix

hello, I'm trying to predict some values based on a linear regression model. I've created the model using one dataframe, and have the prediction values in a second data frame (call it newdata). There are 56 rows in the dataframe used to create the model and 15 in newdata. I ran predict(model1, newdata) and get the warning: 'newdata' had 15 rows but variable(s) found have 56 rows

Dear R community, I have recently discovered the package oblique.tree and I must admit that it was a nice surprise for me, since I have actually made my own version of a kind of a classifier which uses the idea of oblique splits (splits by means of hyperplanes). So I am now interested in comparing these two classifiers. But what I do not seem to understand is why the function

Maybe a useful addition to the predict functions would be to return the values of the predictor variables. It just (unless there are problems) requires an extra line. I have inserted an example below. "predict.glm" <- function (object, newdata = NULL, type = c("link", "response", "terms"), se.fit = FALSE,

I am writing a function to assess the out of sample predictive capabilities of a time series regression model. However lm.predict isn't behaving as I expect it to. What I am trying to do is give it a set of explanatory variables and have it give me a single predicted value using the lm fitted model. > model = lm(y~x) > newdata=matrix(1,1,6) > pred =

I imitated predict.glm, my thing worked, now I need to revise. It would help me very much if someone would explain predict.glm line 28, which says object$na.action <- NULL # kill this for predict.lm calls I want to know 1) why does it set the object$na.action to NULL 2) what does the comment after mean? Maybe I need a pass by value lesson too, because I can't see how changing that

We are trying to implement Breiman's one standard deviation pruning rule for tree models. I have run into a problem with predict.tree. This bit of R code, a skeleton for a cross-validation process, library (tree) dat <- data.frame (matrix (runif (3*100), nrow=100, ncol=3)) a.dat <- dat[1:50,] b.dat <- dat[51:100,] a.tree <- tree (a.dat) b.pred <- predict.tree (a.tree,

Apologies for the coming to this late... 1. By now I hope Somkiat has realized that R-bugs is not the place to report problems in contributed packages. Please direct such reports to the package maintainer. 2. This is really user error. predict() expect the newdata to be a data frame containing variables with the same names as those used in the fitting process. E.g., you fitted the model with