similar to: small syntax suggestion

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

I ran into an interesting oddity of R, if (0) { print(1); } else { print(2); } is a syntax error, while if (0) { print(1); } else { print(2); } or if (0) { print(1); } else { print(2); } is not. I presume it has to do with the duality of the newline functioning as an end of command (;) character, though it still seems a bit odd, and it took me a while to figure out

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 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 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---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---is there a function that prints a data frame to a string? cat() cannot handle lists, so I cannot write cat("your data frame is:\n", df, "\n"). regards, /iaw ---- Ivo Welch (ivo.welch@gmail.com) [[alternative HTML version deleted]]

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---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

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--- 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

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 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-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

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 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:? 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