On Sep 8, 2011, at 3:09 PM, David Winsemius wrote: > > On Sep 8, 2011, at 3:54 PM, Marc Schwartz wrote: > >> >> On Sep 8, 2011, at 2:42 PM, David Winsemius wrote: >> >>> >>> On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote: >>> >>>> Am 08.09.2011 20:48, schrieb Marc Schwartz: >>>>> There was a post from Martin Maechler some years ago and I had to search >>>>> a bit to find it. For these sorts of issues, I typically trust his >>>>> judgement. >>>>> >>>>> The post is here: >>>>> >>>>> https://stat.ethz.ch/pipermail/r-help/2003-April/032471.html >>>>> >>>>> His solution also handles complex numbers. >>>> >>>> For those too lazy to follow >>>> He is basically creating the function is.whole: >>>> >>>> is.whole <- function(x) >>>> is.numeric(x) && floor(x)==x >>> >>> Are you sure? I thought the test would have been all.equal(x, round(x,0) ) >>> >>> My reasoning was that 1.999999999999 should be considered 2 but floor would >>> make it 1. >> >> David, I am confuzzled. Why would that be equal to 2? > > So that sqrt(3) * sqrt(3) would be a "whole number". (It is true the the > floor based wholeness criterion would make sqrt(2)*sqrt(2) > > Somehow it doesn't see "right" that only half of square roots of integers > that have been squared should pass the wholeness test: > > > is.whole <- function(a, tol = 1e-7) { > + is.eq <- function(x,y) { > + r <- all.equal(x,y, tol=tol) > + is.logical(r) && r > + } > + (is.numeric(a) && is.eq(a, floor(a))) || > + (is.complex(a) && {ri <- c(Re(a),Im(a)); is.eq(ri, floor(ri))}) > + } > > is.whole( sqrt(2)^2 ) > [1] TRUE > > is.whole( sqrt(3)^2 ) > [1] FALSE >
<snip content> OK. I suspect it may down to what assumptions one is willing to make, including the level of tolerance used for the comparison. is.whole() of course works for 2 because: > print(sqrt(2) ^ 2, 20) [1] 2.0000000000000004441 is slightly larger than 2, so: > floor(sqrt(2) ^ 2) [1] 2 works, as does: > round(sqrt(2) ^ 2, 0) [1] 2 On the other hand: > print(sqrt(3) ^ 2, 20) [1] 2.9999999999999995559 is slightly smaller than 3, so: > floor(sqrt(3) ^ 2) [1] 2 versus: > round(sqrt(3) ^ 2, 0) [1] 3 Not sure if Martin (cc'd now) might have any comments 8 plus years later relative to this issue, as I would again defer to his judgement here. The other solution proposed, using modulo division, would logically fail in both cases: > (sqrt(3) ^ 2) %% 1 == 0 [1] FALSE > (sqrt(2) ^ 2) %% 1 == 0 [1] FALSE Regards, Marc ______________________________________________ R-help@r-project.org 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.
