Hi, I believe it's lazy evaluation. See ?force
HTH, baptiste On 14 February 2010 20:32, Jyotirmoy Bhattacharya <[email protected]> wrote: > I want to use lapply and a function returning a function in order to build a > list of functions. > >> genr1 <- function(k) {function() {k}} >> l1 <- lapply(1:2,genr1) >> l1[[1]]() > [1] 2 > > This was unexpected. I had expected the answer to be 1, since that is the > value k should be bound to when genr1 is applied to the first element of > 1:2. > > By itself genr1 seems to work fine. >> genr1(5)() > [1] 5 > > I defined a slightly different higher-order function: >> genr2 <- function(k) {k;function() {k}} >> l2 <- lapply(1:2,genr2) >> l2[[1]]() > [1] 1 > > This gives the answer I expected. > > Now I am confused. The function returned by genr2 is exactly the same > function that was being returned by genr1. Why should evaluating k make a > difference? > > I am using R 2.9.2 on Ubuntu Linux. > > Jyotirmoy Bhattacharya > > [[alternative HTML version deleted]] > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
