Re: [R] quantile from quantile table calculation without original data
Hi Petr,
In principle, I like David's approach the best.
However, I note that there's a bug in the squared step.
Furthemore, the variance of the sample quantiles should increase as
they move away from the modal region.
I've built on David's approach, but changed it to a two stage
optimization algorithm.
The parameter estimates from the first stage are used to compute density values.
Then the second stage is weighted, using the scaled density values.
I tried to create an iteratively reweighted algorithm.
However, it didn't converge.
(But that doesn't necessarily mean it can't be done).
The following code returns the value: 1.648416e-05
qfit.lnorm <- function (p, q, lower.tail=TRUE, ...,
par0 = c (-0.5, 0.5) )
{ n <- length (p)
qsample <- q
objf <- function (par)
{ qmodel <- qlnorm (p, par [1], par [2], lower.tail)
sum ( (qmodel - qsample)^2) / n
}
objf.w <- function (wpar, w)
{ qmodel <- qlnorm (p, wpar [1], wpar [2], lower.tail)
sum (w * (qmodel - qsample)^2)
}
wpar0 <- optim (par0, objf)$par
w <- dlnorm (p, wpar0 [1], wpar0 [2], lower.tail)
optim (wpar0, objf.w,, w=w)
}
par <- qfit.lnorm (temp$percent, temp$size, FALSE)$par
plnorm (0.1, par [1], par [2])
On Tue, Mar 9, 2021 at 2:52 AM PIKAL Petr wrote:
>
> Hallo David, Abby and Bert
>
> Thank you for your solutions. In the meantime I found package
> rriskDistributions, which was able to calculate values for lognormal
> distribution from quantiles.
>
> Abby
> > 1-psolution
> [1] 9.980823e-06
>
> David
> > plnorm(0.1, -.7020649, .4678656)
> [1] 0.0003120744
>
> rriskDistributions
> > plnorm(0.1, -.6937355, .3881209)
> [1] 1.697379e-05
>
> Bert suggested to ask for original data before quantile calculation what is
> probably the best but also the most problematic solution. Actually, maybe
> original data are unavailable as it is the result from particle size
> measurement, where the software always twist the original data and spits only
> descriptive results.
>
> All your results are quite consistent with the available values as they are
> close to 1, so for me, each approach works.
>
> Thank you again.
>
> Best regards.
> Petr
>
> > -Original Message-
> > From: David Winsemius
> > Sent: Sunday, March 7, 2021 1:33 AM
> > To: Abby Spurdle ; PIKAL Petr
> >
> > Cc: r-help@r-project.org
> > Subject: Re: [R] quantile from quantile table calculation without original
> > data
> >
> >
> > On 3/6/21 1:02 AM, Abby Spurdle wrote:
> > > I came up with a solution.
> > > But not necessarily the best solution.
> > >
> > > I used a spline to approximate the quantile function.
> > > Then use that to generate a large sample.
> > > (I don't see any need for the sample to be random, as such).
> > > Then compute the sample mean and sd, on a log scale.
> > > Finally, plug everything into the plnorm function:
> > >
> > > p <- seq (0.01, 0.99,, 1e6)
> > > Fht <- splinefun (temp$percent, temp$size) x <- log (Fht (p) )
> > > psolution <- plnorm (0.1, mean (x), sd (x), FALSE) psolution
> > >
> > > The value of the solution is very close to one.
> > > Which is not a surprise.
> > >
> > > Here's a plot of everything:
> > >
> > > u <- seq (0.01, 1.65,, 200)
> > > v <- plnorm (u, mean (x), sd (x), FALSE) plot (u, v, type="l", ylim =
> > > c (0, 1) ) points (temp$size, temp$percent, pch=16) points (0.1,
> > > psolution, pch=16, col="blue")
> >
> > Here's another approach, which uses minimization of the squared error to
> > get the parameters for a lognormal distribution.
> >
> > temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069, 0.3781,
> > 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1, 0.25, 0.5, 0.75, 0.9,
> > 0.95,
> > 0.99)), .Names = c("size", "percent"
> > ), row.names = c(NA, -9L), class = "data.frame")
> >
> > obj <- function(x) {sum( qlnorm(1-temp$percent, x[[1]], x[[2]])-temp$size
> > )^2}
> >
> > # Note the inversion of the poorly named and flipped "percent" column,
> >
> > optim( list(a=-0.65, b=0.42), obj)
> >
> > #
> >
> > $par
> > a b
> > -0.7020649 0.4678656
> >
> > $value
> > [1] 3.110316e-12
> >
> > $counts
> > function gradient
> >51 NA
> >
> > $convergence
> > [1] 0
> >
> > $message
> > NULL
> >
Re: [R] quantile from quantile table calculation without original data
Hallo David, Abby and Bert
Thank you for your solutions. In the meantime I found package
rriskDistributions, which was able to calculate values for lognormal
distribution from quantiles.
Abby
> 1-psolution
[1] 9.980823e-06
David
> plnorm(0.1, -.7020649, .4678656)
[1] 0.0003120744
rriskDistributions
> plnorm(0.1, -.6937355, .3881209)
[1] 1.697379e-05
Bert suggested to ask for original data before quantile calculation what is
probably the best but also the most problematic solution. Actually, maybe
original data are unavailable as it is the result from particle size
measurement, where the software always twist the original data and spits only
descriptive results.
All your results are quite consistent with the available values as they are
close to 1, so for me, each approach works.
Thank you again.
Best regards.
Petr
> -Original Message-
> From: David Winsemius
> Sent: Sunday, March 7, 2021 1:33 AM
> To: Abby Spurdle ; PIKAL Petr
>
> Cc: r-help@r-project.org
> Subject: Re: [R] quantile from quantile table calculation without original
> data
>
>
> On 3/6/21 1:02 AM, Abby Spurdle wrote:
> > I came up with a solution.
> > But not necessarily the best solution.
> >
> > I used a spline to approximate the quantile function.
> > Then use that to generate a large sample.
> > (I don't see any need for the sample to be random, as such).
> > Then compute the sample mean and sd, on a log scale.
> > Finally, plug everything into the plnorm function:
> >
> > p <- seq (0.01, 0.99,, 1e6)
> > Fht <- splinefun (temp$percent, temp$size) x <- log (Fht (p) )
> > psolution <- plnorm (0.1, mean (x), sd (x), FALSE) psolution
> >
> > The value of the solution is very close to one.
> > Which is not a surprise.
> >
> > Here's a plot of everything:
> >
> > u <- seq (0.01, 1.65,, 200)
> > v <- plnorm (u, mean (x), sd (x), FALSE) plot (u, v, type="l", ylim =
> > c (0, 1) ) points (temp$size, temp$percent, pch=16) points (0.1,
> > psolution, pch=16, col="blue")
>
> Here's another approach, which uses minimization of the squared error to
> get the parameters for a lognormal distribution.
>
> temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069, 0.3781,
> 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1, 0.25, 0.5, 0.75, 0.9,
> 0.95,
> 0.99)), .Names = c("size", "percent"
> ), row.names = c(NA, -9L), class = "data.frame")
>
> obj <- function(x) {sum( qlnorm(1-temp$percent, x[[1]], x[[2]])-temp$size
> )^2}
>
> # Note the inversion of the poorly named and flipped "percent" column,
>
> optim( list(a=-0.65, b=0.42), obj)
>
> #
>
> $par
> a b
> -0.7020649 0.4678656
>
> $value
> [1] 3.110316e-12
>
> $counts
> function gradient
>51 NA
>
> $convergence
> [1] 0
>
> $message
> NULL
>
>
> I'm not sure how principled this might be. There's no consideration in this
> approach for expected sampling error at the right tail where the magnitudes
> of the observed values will create much larger contributions to the sum of
> squares.
>
> --
>
> David.
>
> >
> >
> > On Sat, Mar 6, 2021 at 8:09 PM Abby Spurdle
> wrote:
> >> I'm sorry.
> >> I misread your example, this morning.
> >> (I didn't read the code after the line that calls plot).
> >>
> >> After looking at this problem again, interpolation doesn't apply, and
> >> extrapolation would be a last resort.
> >> If you can assume your data comes from a particular type of
> >> distribution, such as a lognormal distribution, then a better
> >> approach would be to find the most likely parameters.
> >>
> >> i.e.
> >> This falls within the broader scope of maximum likelihood.
> >> (Except that you're dealing with a table of quantile-probability
> >> pairs, rather than raw observational data).
> >>
> >> I suspect that there's a relatively easy way of finding the parameters.
> >>
> >> I'll think about it...
> >> But someone else may come back with an answer first...
> >>
> >>
> >> On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle
> wrote:
> >>> I note three problems with your data:
> >>> (1) The name "percent" is misleading, perhaps you want "probability"?
> >>> (2) There are straight (or near-straight) regions, each of which, is
> >>> equally (or near-equally) spaced, which is not what I would expect
> >>> in p
Re: [R] quantile from quantile table calculation without original data
I am aware of that... I have my own functions for this purpose that use splinefun. But if you are trying to also do other aspects of probability distribution calculations, it looked like using fBasics would be easier than re-inventing the wheel. I could be wrong, though, since I haven't used fBasics myself. On March 8, 2021 12:41:40 AM PST, Martin Maechler wrote: >> Jeff Newmiller >> on Fri, 05 Mar 2021 10:09:41 -0800 writes: > >> Your example could probably be resolved with approx. If >> you want a more robust solution, it looks like the fBasics >> package can do spline interpolation. > >base R's spline package does spline interpolation !! > >Martin -- Sent from my phone. Please excuse my brevity. __ 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.
Re: [R] quantile from quantile table calculation without original data
> Jeff Newmiller > on Fri, 05 Mar 2021 10:09:41 -0800 writes: > Your example could probably be resolved with approx. If > you want a more robust solution, it looks like the fBasics > package can do spline interpolation. base R's spline package does spline interpolation !! Martin __ 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.
Re: [R] quantile from quantile table calculation without original data
On 3/6/21 1:02 AM, Abby Spurdle wrote:
I came up with a solution.
But not necessarily the best solution.
I used a spline to approximate the quantile function.
Then use that to generate a large sample.
(I don't see any need for the sample to be random, as such).
Then compute the sample mean and sd, on a log scale.
Finally, plug everything into the plnorm function:
p <- seq (0.01, 0.99,, 1e6)
Fht <- splinefun (temp$percent, temp$size)
x <- log (Fht (p) )
psolution <- plnorm (0.1, mean (x), sd (x), FALSE)
psolution
The value of the solution is very close to one.
Which is not a surprise.
Here's a plot of everything:
u <- seq (0.01, 1.65,, 200)
v <- plnorm (u, mean (x), sd (x), FALSE)
plot (u, v, type="l", ylim = c (0, 1) )
points (temp$size, temp$percent, pch=16)
points (0.1, psolution, pch=16, col="blue")
Here's another approach, which uses minimization of the squared error to
get the parameters for a lognormal distribution.
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")
obj <- function(x) {sum( qlnorm(1-temp$percent, x[[1]],
x[[2]])-temp$size )^2}
# Note the inversion of the poorly named and flipped "percent" column,
optim( list(a=-0.65, b=0.42), obj)
#
$par
a b
-0.7020649 0.4678656
$value
[1] 3.110316e-12
$counts
function gradient
51 NA
$convergence
[1] 0
$message
NULL
I'm not sure how principled this might be. There's no consideration in
this approach for expected sampling error at the right tail where the
magnitudes of the observed values will create much larger contributions
to the sum of squares.
--
David.
On Sat, Mar 6, 2021 at 8:09 PM Abby Spurdle wrote:
I'm sorry.
I misread your example, this morning.
(I didn't read the code after the line that calls plot).
After looking at this problem again, interpolation doesn't apply, and
extrapolation would be a last resort.
If you can assume your data comes from a particular type of
distribution, such as a lognormal distribution, then a better approach
would be to find the most likely parameters.
i.e.
This falls within the broader scope of maximum likelihood.
(Except that you're dealing with a table of quantile-probability
pairs, rather than raw observational data).
I suspect that there's a relatively easy way of finding the parameters.
I'll think about it...
But someone else may come back with an answer first...
On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle wrote:
I note three problems with your data:
(1) The name "percent" is misleading, perhaps you want "probability"?
(2) There are straight (or near-straight) regions, each of which, is
equally (or near-equally) spaced, which is not what I would expect in
problems involving "quantiles".
(3) Your plot (approximating the distribution function) is
back-the-front (as per what is customary).
On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr wrote:
Dear all
I have table of quantiles, probably from lognormal distribution
dput(temp)
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")
and I need to calculate quantile for size 0.1
plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
ss <- approxfun(temp$size, temp$percent)
points((0:100)/50, ss((0:100)/50))
abline(v=.1)
If I had original data it would be quite easy with ecdf/quantile function but
without it I am lost what function I could use for such task.
Please, give me some hint where to look.
Best regards
Petr
Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních
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[[alternative HTML version deleted]]
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Re: [R] quantile from quantile table calculation without original data
I came up with a solution.
But not necessarily the best solution.
I used a spline to approximate the quantile function.
Then use that to generate a large sample.
(I don't see any need for the sample to be random, as such).
Then compute the sample mean and sd, on a log scale.
Finally, plug everything into the plnorm function:
p <- seq (0.01, 0.99,, 1e6)
Fht <- splinefun (temp$percent, temp$size)
x <- log (Fht (p) )
psolution <- plnorm (0.1, mean (x), sd (x), FALSE)
psolution
The value of the solution is very close to one.
Which is not a surprise.
Here's a plot of everything:
u <- seq (0.01, 1.65,, 200)
v <- plnorm (u, mean (x), sd (x), FALSE)
plot (u, v, type="l", ylim = c (0, 1) )
points (temp$size, temp$percent, pch=16)
points (0.1, psolution, pch=16, col="blue")
On Sat, Mar 6, 2021 at 8:09 PM Abby Spurdle wrote:
>
> I'm sorry.
> I misread your example, this morning.
> (I didn't read the code after the line that calls plot).
>
> After looking at this problem again, interpolation doesn't apply, and
> extrapolation would be a last resort.
> If you can assume your data comes from a particular type of
> distribution, such as a lognormal distribution, then a better approach
> would be to find the most likely parameters.
>
> i.e.
> This falls within the broader scope of maximum likelihood.
> (Except that you're dealing with a table of quantile-probability
> pairs, rather than raw observational data).
>
> I suspect that there's a relatively easy way of finding the parameters.
>
> I'll think about it...
> But someone else may come back with an answer first...
>
>
> On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle wrote:
> >
> > I note three problems with your data:
> > (1) The name "percent" is misleading, perhaps you want "probability"?
> > (2) There are straight (or near-straight) regions, each of which, is
> > equally (or near-equally) spaced, which is not what I would expect in
> > problems involving "quantiles".
> > (3) Your plot (approximating the distribution function) is
> > back-the-front (as per what is customary).
> >
> >
> > On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr wrote:
> > >
> > > Dear all
> > >
> > > I have table of quantiles, probably from lognormal distribution
> > >
> > > dput(temp)
> > > temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
> > > 0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
> > > 0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
> > > ), row.names = c(NA, -9L), class = "data.frame")
> > >
> > > and I need to calculate quantile for size 0.1
> > >
> > > plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
> > > ss <- approxfun(temp$size, temp$percent)
> > > points((0:100)/50, ss((0:100)/50))
> > > abline(v=.1)
> > >
> > > If I had original data it would be quite easy with ecdf/quantile function
> > > but without it I am lost what function I could use for such task.
> > >
> > > Please, give me some hint where to look.
> > >
> > >
> > > Best regards
> > >
> > > Petr
> > > Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních
> > > partnerů PRECHEZA a.s. jsou zveřejněny na:
> > > https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information
> > > about processing and protection of business partner's personal data are
> > > available on website:
> > > https://www.precheza.cz/en/personal-data-protection-principles/
> > > Důvěrnost: Tento e-mail a jakékoliv k němu připojené dokumenty jsou
> > > důvěrné a podléhají tomuto právně závaznému prohlá±ení o vyloučení
> > > odpovědnosti: https://www.precheza.cz/01-dovetek/ | This email and any
> > > documents attached to it may be confidential and are subject to the
> > > legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/
> > >
> > >
> > > [[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.
__
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.
Re: [R] quantile from quantile table calculation without original data
I'm sorry.
I misread your example, this morning.
(I didn't read the code after the line that calls plot).
After looking at this problem again, interpolation doesn't apply, and
extrapolation would be a last resort.
If you can assume your data comes from a particular type of
distribution, such as a lognormal distribution, then a better approach
would be to find the most likely parameters.
i.e.
This falls within the broader scope of maximum likelihood.
(Except that you're dealing with a table of quantile-probability
pairs, rather than raw observational data).
I suspect that there's a relatively easy way of finding the parameters.
I'll think about it...
But someone else may come back with an answer first...
On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle wrote:
>
> I note three problems with your data:
> (1) The name "percent" is misleading, perhaps you want "probability"?
> (2) There are straight (or near-straight) regions, each of which, is
> equally (or near-equally) spaced, which is not what I would expect in
> problems involving "quantiles".
> (3) Your plot (approximating the distribution function) is
> back-the-front (as per what is customary).
>
>
> On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr wrote:
> >
> > Dear all
> >
> > I have table of quantiles, probably from lognormal distribution
> >
> > dput(temp)
> > temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
> > 0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
> > 0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
> > ), row.names = c(NA, -9L), class = "data.frame")
> >
> > and I need to calculate quantile for size 0.1
> >
> > plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
> > ss <- approxfun(temp$size, temp$percent)
> > points((0:100)/50, ss((0:100)/50))
> > abline(v=.1)
> >
> > If I had original data it would be quite easy with ecdf/quantile function
> > but without it I am lost what function I could use for such task.
> >
> > Please, give me some hint where to look.
> >
> >
> > Best regards
> >
> > Petr
> > Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních
> > partnerů PRECHEZA a.s. jsou zveřejněny na:
> > https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about
> > processing and protection of business partner's personal data are available
> > on website: https://www.precheza.cz/en/personal-data-protection-principles/
> > Důvěrnost: Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné
> > a podléhají tomuto právně závaznému prohlá±ení o vyloučení odpovědnosti:
> > https://www.precheza.cz/01-dovetek/ | This email and any documents attached
> > to it may be confidential and are subject to the legally binding
> > disclaimer: https://www.precheza.cz/en/01-disclaimer/
> >
> >
> > [[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.
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Re: [R] quantile from quantile table calculation without original data
I note three problems with your data:
(1) The name "percent" is misleading, perhaps you want "probability"?
(2) There are straight (or near-straight) regions, each of which, is
equally (or near-equally) spaced, which is not what I would expect in
problems involving "quantiles".
(3) Your plot (approximating the distribution function) is
back-the-front (as per what is customary).
On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr wrote:
>
> Dear all
>
> I have table of quantiles, probably from lognormal distribution
>
> dput(temp)
> temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
> 0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
> 0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
> ), row.names = c(NA, -9L), class = "data.frame")
>
> and I need to calculate quantile for size 0.1
>
> plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
> ss <- approxfun(temp$size, temp$percent)
> points((0:100)/50, ss((0:100)/50))
> abline(v=.1)
>
> If I had original data it would be quite easy with ecdf/quantile function but
> without it I am lost what function I could use for such task.
>
> Please, give me some hint where to look.
>
>
> Best regards
>
> Petr
> Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních
> partnerů PRECHEZA a.s. jsou zveřejněny na:
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>
> [[alternative HTML version deleted]]
>
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Re: [R] quantile from quantile table calculation without original data
On 3/5/21 1:14 AM, PIKAL Petr wrote:
Dear all
I have table of quantiles, probably from lognormal distribution
dput(temp)
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")
and I need to calculate quantile for size 0.1
plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
ss <- approxfun(temp$size, temp$percent)
points((0:100)/50, ss((0:100)/50))
abline(v=.1)
If I had original data it would be quite easy with ecdf/quantile function but
without it I am lost what function I could use for such task.
The quantiles are in reverse order so tryoing to match the data to
quantiles from candidate parameters requires subtracting them from unity:
> temp$size
[1] 1.6000 0.9466 0.8062 0.6477 0.5069 0.3781 0.3047 0.2681 0.1907
> qlnorm(1-temp$percent, -.5)
[1] 6.21116124 3.14198142 2.18485959 1.19063854 0.60653066 0.30897659
0.16837670 0.11708517 0.05922877
> qlnorm(1-temp$percent, -.9)
[1] 4.16346589 2.10613313 1.46455518 0.79810888 0.40656966 0.20711321
0.11286628 0.07848454 0.03970223
> qlnorm(1-temp$percent, -2)
[1] 1.38589740 0.70107082 0.48750807 0.26566737 0.13533528 0.06894200
0.03756992 0.02612523 0.01321572
> qlnorm(1-temp$percent, -1.6)
[1] 2.06751597 1.04587476 0.72727658 0.39632914 0.20189652 0.10284937
0.05604773 0.03897427 0.01971554
> qlnorm(1-temp$percent, -1.6, .5)
[1] 0.64608380 0.45951983 0.38319004 0.28287360 0.20189652 0.14410042
0.10637595 0.08870608 0.06309120
> qlnorm(1-temp$percent, -1, .5)
[1] 1.1772414 0.8372997 0.6982178 0.5154293 0.3678794 0.2625681
0.1938296 0.1616330 0.1149597
> qlnorm(1-temp$percent, -1, .4)
[1] 0.9328967 0.7103066 0.6142340 0.4818106 0.3678794 0.2808889
0.2203318 0.1905308 0.1450700
> qlnorm(1-temp$percent, -0.5, .4)
[1] 1.5380866 1.1710976 1.0127006 0.7943715 0.6065307 0.4631076
0.3632657 0.3141322 0.2391799
> qlnorm(1-temp$percent, -0.55, .4)
[1] 1.4630732 1.1139825 0.9633106 0.7556295 0.5769498 0.4405216
0.3455491 0.2988118 0.2275150
> qlnorm(1-temp$percent, -0.55, .35)
[1] 1.3024170 1.0260318 0.9035201 0.7305712 0.5769498 0.4556313
0.3684158 0.3244257 0.2555795
> qlnorm(1-temp$percent, -0.55, .45)
[1] 1.6435467 1.2094723 1.0270578 0.7815473 0.5769498 0.4259129
0.3241016 0.2752201 0.2025322
> qlnorm(1-temp$percent, -0.53, .45)
[1] 1.6767486 1.2339052 1.0478057 0.7973356 0.5886050 0.4345169
0.3306489 0.2807799 0.2066236
> qlnorm(1-temp$percent, -0.57, .45)
[1] 1.6110023 1.1855231 1.0067207 0.7660716 0.5655254 0.4174793
0.3176840 0.2697704 0.1985218
Seems like it might be an acceptable fit. modulo the underlying data
gathering situation which really should be considered.
You can fiddle with that result. My statistical hat (not of PhD level
certification) says that the middle quantiles in this sequence probably
have the lowest sampling error for a lognormal, but I'm rather unsure
about that. A counter-argument might be that since there is a hard lower
bound of 0 for the 0-th quantile that you should be more worried about
matching the 0.1907 value to the 0.01 order statistic, since 99% of the
data is know to be above it. Seems like efforts at matching the 0.50
quantile to 0.5069 for the logmean parameter and matching the 0.01
quantile 0.1907 for estimation of the variance estimate might be
preferred to worrying too much about the 1.6 value which would be in the
right tail (and far away from your region of extrapolation.)
Further trial and error:
> qlnorm(1-temp$percent, -0.58, .47)
[1] 1.6709353 1.2129813 1.0225804 0.7687497 0.5598984 0.4077870
0.3065638 0.2584427 0.1876112
> qlnorm(1-temp$percent, -0.65, .47)
[1] 1.5579697 1.1309763 0.9534476 0.7167775 0.5220458 0.3802181
0.2858382 0.2409704 0.1749275
> qlnorm(1-temp$percent, -0.65, .5)
[1] 1.6705851 1.1881849 0.9908182 0.7314290 0.5220458 0.3726018
0.2750573 0.2293682 0.1631355
> qlnorm(1-temp$percent, -0.65, .4)
[1] 1.3238434 1.0079731 0.8716395 0.6837218 0.5220458 0.3986004
0.3126657 0.2703761 0.2058641
> qlnorm(1-temp$percent, -0.68, .4)
[1] 1.2847179 0.9781830 0.8458786 0.6635148 0.5066170 0.3868200
0.3034251 0.2623852 0.1997799
> qlnorm(1-temp$percent, -0.65, .39)
[1] 1.2934016 0.9915290 0.8605402 0.6791257 0.5220458 0.4012980
0.3166985 0.2748601 0.2107093
>
> qlnorm(1-temp$percent, -0.65, .42)
[1] 1.3868932 1.0416839 0.8942693 0.6930076 0.5220458 0.3932595
0.3047536 0.2616262 0.1965053
> qlnorm(1-temp$percent, -0.68, .42)
[1] 1.3459043 1.0108975 0.8678396 0.6725261 0.5066170 0.3816369
0.2957468 0.2538940 0.1906976
(I did make an effort at searching for quantile matching as a method for
distribution fitting, but came up empty.)
--
David.
Please, give me some hint where to look.
Best regards
Petr
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Re: [R] quantile from quantile table calculation without original data
Your example could probably be resolved with approx. If you want a more robust
solution, it looks like the fBasics package can do spline interpolation. You
may want to spline on the log of your size variable and use exp on the output
if you want to avoid negative results.
On March 5, 2021 1:14:22 AM PST, PIKAL Petr wrote:
>Dear all
>
>I have table of quantiles, probably from lognormal distribution
>
> dput(temp)
>temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
>0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
>0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
>), row.names = c(NA, -9L), class = "data.frame")
>
>and I need to calculate quantile for size 0.1
>
>plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
>ss <- approxfun(temp$size, temp$percent)
>points((0:100)/50, ss((0:100)/50))
>abline(v=.1)
>
>If I had original data it would be quite easy with ecdf/quantile
>function but without it I am lost what function I could use for such
>task.
>
>Please, give me some hint where to look.
>
>
>Best regards
>
>Petr
>Osobn� �daje: Informace o zpracov�n� a ochran� osobn�ch �daj�
>obchodn�ch partner� PRECHEZA a.s. jsou zve�ejn�ny na:
>https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information
>about processing and protection of business partner's personal data are
>available on website:
>https://www.precheza.cz/en/personal-data-protection-principles/
>D�v�rnost: Tento e-mail a jak�koliv k n�mu p�ipojen� dokumenty jsou
>d�v�rn� a podl�haj� tomuto pr�vn� z�vazn�mu prohl�en� o vylou�en�
>odpov�dnosti: https://www.precheza.cz/01-dovetek/ | This email and any
>documents attached to it may be confidential and are subject to the
>legally binding disclaimer: https://www.precheza.cz/en/01-disclaimer/
>
>
> [[alternative HTML version deleted]]
--
Sent from my phone. Please excuse my brevity.
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[R] quantile from quantile table calculation without original data
Dear all
I have table of quantiles, probably from lognormal distribution
dput(temp)
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")
and I need to calculate quantile for size 0.1
plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
ss <- approxfun(temp$size, temp$percent)
points((0:100)/50, ss((0:100)/50))
abline(v=.1)
If I had original data it would be quite easy with ecdf/quantile function but
without it I am lost what function I could use for such task.
Please, give me some hint where to look.
Best regards
Petr
Osobn� �daje: Informace o zpracov�n� a ochran� osobn�ch �daj� obchodn�ch
partner� PRECHEZA a.s. jsou zve�ejn�ny na:
https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about
processing and protection of business partner's personal data are available on
website: https://www.precheza.cz/en/personal-data-protection-principles/
D�v�rnost: Tento e-mail a jak�koliv k n�mu p�ipojen� dokumenty jsou d�v�rn� a
podl�haj� tomuto pr�vn� z�vazn�mu prohl�en� o vylou�en� odpov�dnosti:
https://www.precheza.cz/01-dovetek/ | This email and any documents attached to
it may be confidential and are subject to the legally binding disclaimer:
https://www.precheza.cz/en/01-disclaimer/
[[alternative HTML version deleted]]
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