Ah, my bad, you're right of course.
sum(abs(diff(diff( sort(x))))) < eps
for some reasonable eps then, would do as a oneliner, or
all(abs(diff(diff(sort(x)))) < eps)
or
max(abs(diff(diff(sort(x))))) < eps
-Mark
Op vr 31 aug. 2018 om 16:14 schreef Iñaki Ucar <iu...@fedoraproject.org>:
> El vie., 31 ago. 2018 a las 16:00, Mark van der Loo
> (<mark.vander...@gmail.com>) escribió:
> >
> > how about
> >
> > is_evenly_spaced <- function(x,...) all.equal(diff(sort(x)),...)
>
> This doesn't work, because
>
> 1. all.equal does *not* return FALSE. Use of isTRUE or identical(.,
> TRUE) is required if you want a boolean.
> 2. all.equal compares two objects, not elements in a vector.
>
> Iñaki
>
> >
> > (use ellipsis to set tolerance if necessary)
> >
> >
> > Op vr 31 aug. 2018 om 15:46 schreef Emil Bode <emil.b...@dans.knaw.nl>:
> >>
> >> Agreed that's it's rounding error, and all.equal would be the way to go.
> >> I wouldn't call it a bug, it's simply part of working with floating
> point numbers, any language has the same issue.
> >>
> >> And while we're at it, I think the function can be a lot shorter:
> >> .is_continous_evenly_spaced <- function(n){
> >> length(n)>1 && isTRUE(all.equal(n[order(n)], seq(from=min(n),
> to=max(n), length.out = length(n))))
> >> }
> >>
> >> Cheers, Emil
> >>
> >> El vie., 31 ago. 2018 a las 15:10, Felix Ernst
> >> (<felix.gm.er...@outlook.com>) escribió:
> >> >
> >> > Dear all,
> >> >
> >> > I a bit unsure, whether this qualifies as a bug, but it is
> definitly a strange behaviour. That why I wanted to discuss it.
> >> >
> >> > With the following function, I want to test for evenly space
> numbers, starting from anywhere.
> >> >
> >> > .is_continous_evenly_spaced <- function(n){
> >> > if(length(n) < 2) return(FALSE)
> >> > n <- n[order(n)]
> >> > n <- n - min(n)
> >> > step <- n[2] - n[1]
> >> > test <- seq(from = min(n), to = max(n), by = step)
> >> > if(length(n) == length(test) &&
> >> > all(n == test)){
> >> > return(TRUE)
> >> > }
> >> > return(FALSE)
> >> > }
> >> >
> >> > > .is_continous_evenly_spaced(c(1,2,3,4))
> >> > [1] TRUE
> >> > > .is_continous_evenly_spaced(c(1,3,4,5))
> >> > [1] FALSE
> >> > > .is_continous_evenly_spaced(c(1,1.1,1.2,1.3))
> >> > [1] FALSE
> >> >
> >> > I expect the result for 1 and 2, but not for 3. Upon
> Investigation it turns out, that n == test is TRUE for every pair, but not
> for the pair of 0.2.
> >> >
> >> > The types reported are always double, however n[2] == 0.1 reports
> FALSE as well.
> >> >
> >> > The whole problem is solved by switching from all(n == test) to
> all(as.character(n) == as.character(test)). However that is weird, isn’t it?
> >> >
> >> > Does this work as intended? Thanks for any help, advise and
> suggestions in advance.
> >>
> >> I guess this has something to do with how the sequence is built and
> >> the inherent error of floating point arithmetic. In fact, if you
> >> return test minus n, you'll get:
> >>
> >> [1] 0.000000e+00 0.000000e+00 2.220446e-16 0.000000e+00
> >>
> >> and the error gets bigger when you continue the sequence; e.g., this
> >> is for c(1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7):
> >>
> >> [1] 0.000000e+00 0.000000e+00 2.220446e-16 2.220446e-16 4.440892e-16
> >> [6] 4.440892e-16 4.440892e-16 0.000000e+00
> >>
> >> So, independently of this is considered a bug or not, instead of
> >>
> >> length(n) == length(test) && all(n == test)
> >>
> >> I would use the following condition:
> >>
> >> isTRUE(all.equal(n, test))
> >>
> >> Iñaki
> >>
> >> >
> >> > Best regards,
> >> > Felix
> >> >
> >> >
> >> > [[alternative HTML version deleted]]
> >> >
> >> > ______________________________________________
> >> > R-devel@r-project.org mailing list
> >> > https://stat.ethz.ch/mailman/listinfo/r-devel
> >>
> >>
> >>
> >> --
> >> Iñaki Ucar
> >>
> >> ______________________________________________
> >> R-devel@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-devel
> >>
> >>
> >> ______________________________________________
> >> R-devel@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-devel
>
>
>
> --
> Iñaki Ucar
>
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