similar to: esthetics --- extending the lm command to fixed effects?

Dear R-Help group: I have been tinkering with how I want my personal standard library functions to look like. They are not designed to be professional and heavyweight, but lightweight. There are probably dozens of little bugs, because I don't know or have not properly taken care of a variety of internal R code issues. still, I like how this ended up, and there is no learning curve, so I

dear R experts---now I have a case where I want to estimate very large regression models with many fixed effects---not just the mean type, but cross-fixed effects---years, months, locations, firms. Many millions of observations, a few thousand variables (most of these variables are interaction fixed effects). could someone please point me to packages, if any, that would help me estimate such

dear readers---I struggled with how to do nice fixed-effects regressions in large economic samples for a while. Eventually, I realized that nlme is not really what I needed (too complex), and all I really wanted is the plm package. so, I thought I would share a quick example. ################ sample code to show fixed-effects models? in R # create a sample panel data set with firms and years

dear R experts---quick question. I need to estimate a model that looks like y = (b*T+d*T^3) + (1-b-3*d*T^2)*x + (3*d*T)*x^2 + (-d)*x^3 I only have three parameters. Is nls() the right tool for the job, or is there something faster/better? /iaw ---- Ivo Welch (ivo.welch@brown.edu, ivo.welch@gmail.com) [[alternative HTML version deleted]]

dear R experts---I am using the coef() function to pick off the coefficients from an lm() object. alas, I also need the standard errors and I need them fast. I know I can do a "summary()" on the object and pick them off this way, but this computes other stuff I do not need. Or, I can compute (X' X)^(-1) s^2 myself. Has someone written a fast se() function? incidentally, I think

this is not an important question, but I wonder why lm returns an error, and whether this can be shut off. it would seem to me that returning NA's would make more sense in some cases---after all, the problem is clearly that coefficients cannot be computed. I know that I can trap the lm.fit() error---although I have always found this to be quite inconvenient---and this is easy if I have only

Dear R wizards: This must have an obvious solution, but I am stumped. I can run a linear regression giving a matrix as the independent set of variables, but if I give a data frame (which I would like to give, because it should tell the linear model the names of the variables), R does not like it. An example is: N=20; y= rnorm(N); x.m <- (matrix( nrow=N, ncol=2 )); x.m[,1]=rnorm(N);

Not a direct answer but you may find lm.fit worth experimenting with. Also try the high-performance computing task view on CRAN Cheers, Andrew -- Andrew Robinson Chief Executive Officer, CEBRA and Professor of Biosecurity, School/s of BioSciences and Mathematics & Statistics University of Melbourne, VIC 3010 Australia Tel: (+61) 0403 138 955 Email: apro at unimelb.edu.au Website:

Dear R wizards: is there a clean way to assign to elements in a list? what I would like to do, in pseudo R+perl notation is f <- function(a,b) list(a+b,a-b) (c,d) <- f(1,2) and have c be assigned 1+2 and d be assigned 1-2. right now, I use the clunky x <- f(1,2) c <- x[[1]] d <- x[[2]] rm(x) which seems awful. is there a nicer syntax? regards, /iaw ---- Ivo Welch

Thanks, and all well taken. But are my beautiful GPUs (with integrated memory architecture) really nothing more than a cooling area for the chip? On Fri, Nov 15, 2024 at 6:06?AM Martin Maechler <maechler at stat.math.ethz.ch> wrote: > >>>>> Andrew Robinson via R-help > >>>>> on Thu, 14 Nov 2024 12:45:44 +0000 writes: > > > Not a direct

>>>>> Andrew Robinson via R-help >>>>> on Thu, 14 Nov 2024 12:45:44 +0000 writes: > Not a direct answer but you may find lm.fit worth > experimenting with. Yes, lm.fit() is already faster, and .lm.fit() {added to base R by me, when a similar question was asked years ago ...} is even an order of magnitude faster in some cases. See

Dear R Wizards---is it possible to get R to show its current call stack (sys.calls()) upon an error abort? I don't use ESS for execution, and it is often not obvious how to locate how I triggered an error in an R internal function. Seeing the call stack would make this easier. (right now, I sprinkle "cat" statements everywhere, just to locate the line where the error appears.) Of

Dear R wizards: I am playing (and struggling) with the example in the constrOptim function. simple example. let's say I want to constrain my variables to be within -1 and 1. I believe I want a whole lot of constraints where ci is -1 and ui is either -1 or 1. That is, I have 2*N constraints. Should the following work? N=10 x= rep(1:N) ci= rep(-1, 2*N) ui= c(rep(1, N), rep(-1, N))

dear R wizards: X is factor with 20,000*20=800,000 observations of 20,000 factors. I.e., each factor has 20 observations. y is 800,000 normally distributed data points. I want to see how much R^2 the X factors can provide. Easy, right? > lm ( y ~ X) and > aov( y ~ X) Error: cannot allocate vector of size 3125000 Kb is this computationally infeasible? (I am not an expert, but

I am using a variant of stopifnot a lot. can I suggest that base R extends its functionality? I know how to do this for myself. this is a suggestion for beginners and students. I don't think it would break anything. first, I think it would be more useful if it had an optional character string, so users could write stopifnot( is.matrix(m), "m is not a matrix" ) this would

dear R wizards: here is the strange question for the day. It seems to me that nrow() is very slow. Let me explain what I mean: ds= data.frame( NA, x=rnorm(10000) ) ## a sample data set > system.time( { for (i in 1:10000) NA } ) ## doing nothing takes virtually no time user system elapsed 0.000 0.000 0.001 ## this is something that should take time; we need to add 10,000

dear R wizards: I am looking for a binary package distribution of R 2.9.2 for OSX . Looking at http://r.research.att.com/ , there seems to be only a binary for 2.9.0 . is the 2.9.2 version binary package available somewhere? (at this point, would it make sense to elevate the 64-bit version to a "standard recommended" rather than just a "boutique" version?) sincerely, /iaw

dear R wizards:? I am wrestling with reshape.? I have a long data set that I want to convert into a wide data set, in which rows are firms and columns are years. > summary(rin) firm fyear sim1 Min. :1004.00 Min. :1964.0 Min. : -1.00000 1st Qu.:1010.00 1st Qu.:1979.0 1st Qu.: -0.14334 Median :1016.00 Median :1986.0 Median : 0.00116 Mean

I sent this privately to ivo welch yesterday, and he thinks it might be useful to someone else as well. Since I'm on a Mac the screen device is quartz(): > quartz() > plot( c(0,1), c(0,1) ); > text( 0.5, 0.5, "\u2113" ) # and then File/Save As/ -- David. Begin forwarded message: > From: David Winsemius <dwinsemius at comcast.net> > Date: August 17,

dear R experts: has someone written a function that returns the results of by() as a data frame? ??of course, this can work only if the output of the function that is an argument to by() is a numerical vector. presumably, what is now names(byobject) would become a column in the data frame, and the by object's list elements would become columns. it's a little bit like flattening the by()