should lapply preserve attributes?
On Tue, 2002-03-19 at 16:31, Prof Brian D Ripley wrote:
On 19 Mar 2002, Timothy H. Keitt wrote:
Would it make sense to replace names(rval) <- names(X) with attributes(rval) <- attributes(X) ?? I can, of course, make a local function for this, but wondered if this change would be useful in general.
No, it would be positively harmful. X might be a data frame, or a time series .., but rval is only guaranteed to be the same length as X. If you know more, adjust the object returned by lapply.
Yes, nevermind. Ah, tiredness... I realized that's not what I meant after hitting "send" (and that Brian would catch it before I could reply to my own email! ;-). What I want is to apply a function to a list of objects and have the attributes of the objects copied to the resulting list items, e.g., out <- lapply(list(ts1=ts(1:10), ts2=ts(1:5)), sample) and have 'out' be a list of time series instead of attributeless vectors. Obviously the solution is to make a function that both permutes the values and copies the attributes as well. Tim
Timothy H. Keitt Department of Ecology and Evolution State University of New York at Stony Brook Stony Brook, New York 11794 USA Phone: 631-632-1101, FAX: 631-632-7626 http://life.bio.sunysb.edu/ee/keitt/ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-devel-request@stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._