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.goslee at 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.lamblin at
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]]
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.lamblin at 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.goslee at 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.lamblin at 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 at 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.
... 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 <bgunter at 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.lamblin at 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.goslee at 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.lamblin at 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 at 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.