similar to: print.data.frame to string?

is it possible to throw a stop() that is so hard that it will escape even tryCatch? /iaw ---- Ivo Welch (ivo.welch at gmail.com)

dear R experts---I am trying to replicate a perl feature. I want to be able to embed R commands inside a character string, and have the string be printed with the command executed. my perl equivalent is my $a=10; my $teststring = "the expression, $a+1, is ::$a+1::, but add one more for ::$a+2::\n"; $teststring =~ s/::(.*?)::/$1/gee; print $teststring; of course, R does not use

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--- after many years, I am planning to give in and write my first R package. I want to combine my collection of collected useful utility routines. as my guide, I am planning to use Friedrich Leisch's "Creating R Packages: A Tutorial" from Sep 2009. Is there a newer or better tutorial? this one is 4 years old. I also plan on one change---given that the

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

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

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---Is there a multicore equivalent of by(), just like mclapply() is the multicore equivalent of lapply()? if not, is there a fast way to convert a data.table into a list based on a column that lapply and mclapply can consume? advice appreciated...as always. regards, /iaw ---- Ivo Welch (ivo.welch at gmail.com)

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 experts--- Has anyone written parallel versions of "by" (i.e., mcby) and "ave" (i.e. mcave) ? I did ask a question like this a year ago, and then the answer was no. for those who are googling the group for the answer to this question, in the meantime, the poor man's version of "by" is mclapply( split( ds, factor ), FUN ) I don't know the poor

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()

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 R experts: this is probably correct behavior, but I do want to point out that it is unexpected to someone not too well versed: > test=factor("A","B","C","A") > ifelse(test=="A", as.factor("A"), test) [1] 1 2 3 1 ok, my factor was just coerced into integers, even though I have a logical vector as my condition and factors as

I have a set of functions that I always load on startup. for example, there is my now infamous "is.defined()" function. I would like to add some documentation for these functions, so that I can do a ?is.defined inside R. The documentation tells me how to mark up Rd files is very good, but I wonder how one installs them for access by the R executable (on OSX for me). Do I drop

Dear R experts: I searched cran (and r-help) for "nash equilibrium" and "game" but nothing stuck out. has someone written a numerical nash optimizer for two players? player a has choices x1,x2,x3,... and cares about (maximizes) pa(x1,x2,x3,...,y1,y2,y3) player b has choices y1,y2,y3,..., and cares about (maximizes) pb(x1,x2,x3,...,y1,y2,y3) I can tune it to my problem, but

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

The treatment of dates seems to be a little inconsistent in R 2.11.1 (2010-05-31): [1] The choice of origins? > as.integer(as.Date("1970-01-01")) works and assumes as origin 1970-01-01. However, > as.Date(1) does not work. It requires an origin (as.Date(1, origin="1970-01-01")). If we set a default origin in the former, it should probably work when the input

quick programming questions. I want to "turn on" more errors. there are two traps I occasionally fall into. * I wonder why R thinks that a variable is always defined in a data frame. > is.defined(d) [1] FALSE > d= data.frame( x=1:5, y=1:5 ) > is.defined(d$z) [1] TRUE > is.defined(nonexisting$garbage) [1] TRUE this is a bit unfortunate

dear R wizards: I have a large data frame, a million rows, 40 columns. In this data frame, there are some (about 100,000) rows which I want to recompute (update), while I want to leave others just as is. this is based on a condition that I need to compute, based on what is in a few of the columns. what is the right R way to do this? I could subset out the rows that I want to recompute into a

Dear R-experts---is there a relatively low-pain way to get unicode characters into a plot to a pdf device? pdf(file="cardsymbols.pdf") plot( 0, xlim=c(0,5), ylim=c(0,5), type="n") text(1,1, "&spades;") text(2,2, "&hearts;") text(3,3, "&diams;") text(4,4, "&clubs;") dev.off() (these are the characters that I need the most