The first solution with diff works for uniform abscissa only with integer
values.
z <- seq(0, 10, length=100)
all(diff(z) == z[2] - z[1] )
## FALSE
In this case, as you recommended, I could use signif or round or a
tolerance for real numbers. In my particular case, in order to set a
tolerance, I need the scale used and I don't have this information. I
prefer to test the "near uniformity".
I didn't know the function zapsmall. It could be useful!
Thanks Sarah and Bert!!!
Marc
2015-04-06 19:51 GMT+02:00 Bert Gunter <gunter.ber...@gene.com>:
> ... correction: you need to use absolute value for the comparison, of
> course.
>
> all(abs(diff(z) - z[2] + z[1]) < tol)
>
> -- Bert
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> Clifford Stoll
>
>
>
>
> On Mon, Apr 6, 2015 at 10:47 AM, Bert Gunter <bgun...@gene.com> wrote:
> > Perhaps ?diff might be useful here:
> >
> > z <- runif(20)
> > all(diff(z) == z[2] - z[1] )
> > ## FALSE
> >
> > z <- seq_len(10)
> > all(diff(z) == z[2] - z[1] )
> > ##TRUE
> >
> > You can use signif or round as before to allow for "near uniformity"
> > or use ?zapsmall or an explicit comparison with a tolerancec instead
> > of ==, e.g. all(diff(z) - z[2] + z[1] < tol)
> >
> > Cheers,
> > Bert
> >
> > Bert Gunter
> > Genentech Nonclinical Biostatistics
> > (650) 467-7374
> >
> > "Data is not information. Information is not knowledge. And knowledge
> > is certainly not wisdom."
> > Clifford Stoll
> >
> >
> >
> >
> > On Mon, Apr 6, 2015 at 10:11 AM, Marc Lamblin <marcgg.lamb...@gmail.com>
> wrote:
> >> The aim is to control if a given abscissa/grid is uniform or not.
> Abscissa
> >> in generic vector of real ordered numbers.
> >>
> >> Here a reproducibile code:
> >>
> >> # uniform abscissa/grid
> >> abscissa1 <- seq(0, 1, length=100)
> >> # non-uniform abscissa/grid
> >> abscissa2 <- sort(runif(100))
> >>
> >> control1 <- all(signif(abscissa1[1:(length(abscissa1) - 1) + 1] -
> >> abscissa1[1:(length(abscissa1) - 1)]) ==
> signif(rep((range(abscissa1)[2] -
> >> range(abscissa1)[1])/(length(abscissa1) - 1), length(abscissa1) - 1)))
> >> control2 <- all(signif(abscissa2[1:(length(abscissa2) - 1) + 1] -
> >> abscissa2[1:(length(abscissa2) - 1)]) ==
> signif(rep((range(abscissa2)[2] -
> >> range(abscissa2)[1])/(length(abscissa2) - 1), length(abscissa2) - 1)))
> >>
> >> control1
> >> control2
> >>
> >> As expected control1 is TRUE and control2 is FALSE. Actually in this
> code
> >> it is possible also to use
> >> diff inside signif.
> >> Do you mean that the control to perform can be done in this manner
> >>
> >> if (length(unique(diff(vec))) == 1) {
> >> control <- TRUE
> >> } else {
> >> control <- FALSE
> >> }
> >>
> >> I have tried to apply this control on abscissa1 which is uniform but
> >> length(unique(diff(abscissa1))) was greater than one; probably, as you
> >> said, this is due to the fact that in this way I don't take into account
> >> the machine precision.
> >> What I want to understand is if there is a SAFE solution, even if until
> now
> >> this control is working correctly. I have seen in the documentation of
> >> signif that by default the number of digits considered are 6. The
> number of
> >> digits to consider depends on the scale used. It doesn't make sense to
> >> increase the number of digits with respect to default because, in this
> >> case, you are not using an handy scale.
> >> Maybe it could be better directly to ask user if the abscissa passed as
> >> argument is uniform or not.
> >> Thanks a lot for the link!!!
> >>
> >> Marc
> >>
> >>
> >>
> >>
> >> 2015-04-06 16:32 GMT+02:00 Sarah Goslee <sarah.gos...@gmail.com>:
> >>
> >>> Without a reproducible example that includes some sample data (fake is
> >>> fine), the code you used (NOT in HTML format), and some clear idea of
> >>> what output you expect, it's impossible to figure out how to help you.
> >>> Here are some suggestions for creating a good reproducible example:
> >>>
> >>>
> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
> >>>
> >>> Without knowing what you want, it looks like abscissa is a vector, and
> >>> so I'm not sure how this defines a grid, but
> >>> length(unique(diff(vec)))
> >>> might help. Note that this DOES NOT account for machine precision in
> any
> >>> way.
> >>>
> >>> Sarah
> >>>
> >>> On Mon, Apr 6, 2015 at 7:50 AM, Marc Lamblin <marcgg.lamb...@gmail.com
> >
> >>> wrote:
> >>> > I need to control of a given grid is uniform. This control using
> signif
> >>> > until now works:
> >>> >
> >>> > if (all(signif(abscissa[1:(length(abscissa) - 1) + 1] -
> >>> > abscissa[1:(length(abscissa) - 1)]) ==
> signif(rep((range(abscissa)[2] -
> >>> > range(abscissa)[1])/(length(abscissa) - 1),
> length(abscissa) -
> >>> > 1)))) {
> >>> > # other stuff
> >>> > }
> >>> >
> >>> > Does someone have some suggestions to improve this control? Thanks in
> >>> > advance!! :)
> >>> >
> >>> > Marc
> >>> >
> >>> > [[alternative HTML version deleted]]
> >>> >
> >>>
> >>>
> >>> --
> >>> Sarah Goslee
> >>> http://www.functionaldiversity.org
> >>>
> >>
> >> [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> 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.
>
[[alternative HTML version deleted]]
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