similar to: non-linear estimation with many firm-specific parameters

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

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---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 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);

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 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

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 experts: sorry, I have to ask this again. I know that the answer is in section 7.2 of "S Programming," but I don't have the book (and I plan to buy the next edition---which I hope will be titled S/R programming ;-) ). I believe the following yields a standard fixed-effects estimation: fixed.effects = as.factor( as.integer( runif(100)*10 ) ) y=rnorm(100); x=rnorm(100);

As an end-user, I wonder about Revolution R. Is the relationship between Revolution R and the R community at-large a positive one? Do the former contribute to the development efforts of the latter? Is there a competitive aspect? is their forum competitive with r-help? any other thoughts? (most of all, I simply hope that they help some of the many helpful experts on this forum, who have

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 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

Hi All: I am working with a law firm that would like to offer to it's clients the ability to access their documents through a web interface. This would require a nice gui to offer the clients with a simple process for the clients to download from. It would also need to be secure (authentication & SSL), and mandatory disclaimer acceptance before logging into their account. Anyone

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))

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,

hi: could someone please point me to a function that allows me to solve general non-linear functions? > irr.in <- function(r, c1, c2, c3 ) { return(c1+c2/(1+r) + c3/(1+r)^2); } > solve.nonlinear( irr.in, -100, 60, 70 ); 0.189 If someone has written an irr function, this would be helpful, too---though not difficult to write, either. thanks for any pointers. Regards, /iaw

dear R group: I have the probably fairly common problem that I would like to have one code.R file do different things if it is invoked from a symbolic link, which should be easy to uncover. $ ln -s code.R code-0.R $ ln -s code.R code-1.R $ R CMD BATCH code-1.R what needs to be in code-1.R to put code-1.r into a character vector? help appreciated. regards, /ivo welch PS : I read

Hello, I have an unbalanced panel dataset and would like to exclude all objects that don't appear at least x times. Therefore, I would like to include a column indicating for every line how many periods are available, e.g. id, year, number 1, 2000, 3 1, 2001, 3 1, 2002, 3 2, 2001, 1 3, ..., ... This would allow me to exclude companies by setting "subset=number>=x". However, I

dear R experts---I am struggling with the requirements to prepare my files for my printers. I am printing in 2/2 format, which means cyan and black for me, which they take from my color-separated pdf files. R comes into play, because it produces all the figures that are embedded in my book (pdflatex). now, TeX has no problems producing CMYK files. However, R produces RGB files (for

dear R experts--- t <- 1:30 f <- function(t) { cat("f for", t, "\n"); return(2*t) } g <- function(t) { cat("g for", t, "\n"); return(3*t) } s <- ifelse( t%%2==0, g(t), f(t)) shows that the ifelse function actually evaluates both f() and g() for all values first, and presumably then just picks left or right results based on t%%2.