thr3ads.net - R help - [R] Obtain gradient at multiple values for exponential decay model [Apr 2018]

If this information is useful, please help other people find it:
Share via:

David Winsemius

2018-Apr-06 15:15 UTC

[R] Obtain gradient at multiple values for exponential decay model

> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
> 
> 
>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>> 
>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>> From: "David Winsemius" <dwinsemius at comcast.net>
>>> 
>>> 
>>> Not correct. You already have `predict`. It is capale of using the
`newdata` values to do interpolation with the values of the coefficients in the
model. See:
>>> 
>>> ?predict
>>> 
>> 
>> The ? details did not mention interpolation explicity; thanks.
>> 
>>> The original question asked for a derivative (i.e. a
"gradient"), but so far it's not clear that you understand the
mathematical definiton of that term. We also remain unclear whether this is
homework.
>>> 
>> 
>> The motivation of this post was simple differentiation of a tangent
point (dy/dx) manually, then wondering how to re-think in modern-day computing
terms. Hence the original question about asking the appropriate functions/syntax
to read further ("curiosity"), not the answer (indeed,
"homework"). :)
>> 
>> Personal curiosity should be considered "homework".
> 
> Besides symbolic differentiation, there is also the option of numeric
differentiation. Here's an amateurish attempt:
> 
> myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
>                                            exp( predict (graphmodeld,
newdata=data.frame(t=x) )))/
>                                          .0001 }
> myNumDeriv(c(100, 250, 350))
I realized that this would not work in the context of your construction. I had
earlier made a more symbolic version using R formulae:

 graphdata<-read.csv(text='t,c
 0,100
 40,78
 80,59
 120,38
 160,25
 200,21
 240,16
 280,12
 320,10
 360,9
 400,7')
 graphmodeld<-lm(log(c)~t, graphdata)
 graphmodelp<-exp(predict(graphmodeld))
 plot(c~t, graphdata)
 lines(graphdata[,1],graphmodelp)
 myNumDeriv(c(100, 250, 350), graphmodeld )
#----------------------------------------------
          1           2           3 
-0.31464102 -0.11310753 -0.05718414 

> 
> 
> 
> David Winsemius
> Alameda, CA, USA
> 
> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
> 
> ______________________________________________
> 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.
David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.' 
-Gehm's Corollary to Clarke's Third Law

David Winsemius

2018-Apr-06 18:25 UTC

head link

[R] Obtain gradient at multiple values for exponential decay model

> On Apr 6, 2018, at 8:15 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
> 
>> 
>> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
>> 
>> 
>>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>>> 
>>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>>> From: "David Winsemius" <dwinsemius at
comcast.net>
>>>> 
>>>> 
>>>> Not correct. You already have `predict`. It is capale of using
the `newdata` values to do interpolation with the values of the coefficients in
the model. See:
>>>> 
>>>> ?predict
>>>> 
>>> 
>>> The ? details did not mention interpolation explicity; thanks.
>>> 
>>>> The original question asked for a derivative (i.e. a
"gradient"), but so far it's not clear that you understand the
mathematical definiton of that term. We also remain unclear whether this is
homework.
>>>> 
>>> 
>>> The motivation of this post was simple differentiation of a tangent
point (dy/dx) manually, then wondering how to re-think in modern-day computing
terms. Hence the original question about asking the appropriate functions/syntax
to read further ("curiosity"), not the answer (indeed,
"homework"). :)
>>> 
>>> Personal curiosity should be considered "homework".
>> 
>> Besides symbolic differentiation, there is also the option of numeric
differentiation. Here's an amateurish attempt:
>> 
>> myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
>>                                           exp( predict (graphmodeld,
newdata=data.frame(t=x) )))/
>>                                         .0001 }
>> myNumDeriv(c(100, 250, 350))
> 
> I realized that this would not work in the context of your construction. I
had earlier made a more symbolic version using R formulae:
> 
> graphdata<-read.csv(text='t,c
> 0,100
> 40,78
> 80,59
> 120,38
> 160,25
> 200,21
> 240,16
> 280,12
> 320,10
> 360,9
> 400,7')
> graphmodeld<-lm(log(c)~t, graphdata)
> graphmodelp<-exp(predict(graphmodeld))
> plot(c~t, graphdata)
> lines(graphdata[,1],graphmodelp)
Again I attempt to correct my incomplete code with this definition that had be
modified to take a model object as its second argument:

myNumDeriv <- function(x, mod) (exp( predict (mod,
newdata=data.frame(t=x+.0001))) - exp( predict (mod, newdata=data.frame(t=x)
)))/.0001

> myNumDeriv(c(100, 250, 350), graphmodeld )
> #----------------------------------------------
>          1           2           3 
> -0.31464102 -0.11310753 -0.05718414 
> 
> 
>> 
>> 
>> 
>> David Winsemius
>> Alameda, CA, USA
>> 
>> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
>> 
>> ______________________________________________
>> 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.
> 
> David Winsemius
> Alameda, CA, USA
> 
> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
> 
> ______________________________________________
> 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.
David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.' 
-Gehm's Corollary to Clarke's Third Law

Jeff Newmiller

2018-Apr-07 06:19 UTC

head link

[R] Obtain gradient at multiple values for exponential decay model

I have never found the R symbolic differentiation helpful because my 
functions are typically quite complicated, but was prompted by Steve 
Ellison's suggestion to try it out in this case:

################# reprex (see reprex package)
graphdta <- read.csv( text "t,c
0,100
40,78
80,59
120,38
160,25
200,21
240,16
280,12
320,10
360,9
400,7
", header = TRUE )

nd <- c( 100, 250, 300 )
graphmodeld <- lm( log(c) ~ t, data = graphdta )
graphmodelplin <- predict( graphmodeld
                          , newdata = data.frame( t = nd )
                          )
graphmodelp <- exp(graphmodelplin)
graphmodelp
#>        1        2        3
#> 46.13085 16.58317 11.79125
# derivative of exp( a + b*t ) is b * exp( a + b*t )
graphmodeldpdt <- coef( graphmodeld )[ 2 ] * graphmodelp
graphmodeldpdt
#>           1           2           3
#> -0.31464113 -0.11310757 -0.08042364

# Ellison suggestion - fancy, only works for simple functions
dc <- deriv( expression( exp( a + b * t ) )
            , namevec = "t"
            )
dcf <- function( t ) {
   cgm <- coef( graphmodeld )
   a <- cgm[ 1 ]
   b <- cgm[ 2 ]
   eval(dc)
}
result <- dcf( nd )
result
#> [1] 46.13085 16.58317 11.79125
#> attr(,"gradient")
#>                t
#> [1,] -0.31464113
#> [2,] -0.11310757
#> [3,] -0.08042364
attr( result, "gradient" )[ , 1 ]
#> [1] -0.31464113 -0.11310757 -0.08042364
#################

On Fri, 6 Apr 2018, David Winsemius wrote:
>
>> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
>>
>>
>>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>>>
>>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>>> From: "David Winsemius" <dwinsemius at
comcast.net>
>>>>
>>>>
>>>> Not correct. You already have `predict`. It is capale of using
the `newdata` values to do interpolation with the values of the coefficients in
the model. See:
>>>>
>>>> ?predict
>>>>
>>>
>>> The ? details did not mention interpolation explicity; thanks.
>>>
>>>> The original question asked for a derivative (i.e. a
"gradient"), but so far it's not clear that you understand the
mathematical definiton of that term. We also remain unclear whether this is
homework.
>>>>
>>>
>>> The motivation of this post was simple differentiation of a tangent
point (dy/dx) manually, then wondering how to re-think in modern-day computing
terms. Hence the original question about asking the appropriate functions/syntax
to read further ("curiosity"), not the answer (indeed,
"homework"). :)
>>>
>>> Personal curiosity should be considered "homework".
>>
>> Besides symbolic differentiation, there is also the option of numeric
differentiation. Here's an amateurish attempt:
>>
>> myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
>>                                            exp( predict (graphmodeld,
newdata=data.frame(t=x) )))/
>>                                          .0001 }
>> myNumDeriv(c(100, 250, 350))
>
> I realized that this would not work in the context of your construction. I
had earlier made a more symbolic version using R formulae:
>
> graphdata<-read.csv(text='t,c
> 0,100
> 40,78
> 80,59
> 120,38
> 160,25
> 200,21
> 240,16
> 280,12
> 320,10
> 360,9
> 400,7')
> graphmodeld<-lm(log(c)~t, graphdata)
> graphmodelp<-exp(predict(graphmodeld))
> plot(c~t, graphdata)
> lines(graphdata[,1],graphmodelp)
> myNumDeriv(c(100, 250, 350), graphmodeld )
> #----------------------------------------------
>          1           2           3
> -0.31464102 -0.11310753 -0.05718414
>
>
>>
>>
>>
>> David Winsemius
>> Alameda, CA, USA
>>
>> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
>>
>> ______________________________________________
>> 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.
>
> David Winsemius
> Alameda, CA, USA
>
> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
>
> ______________________________________________
> 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.
>
---------------------------------------------------------------------------
Jeff Newmiller                        The     .....       .....  Go Live...
DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live
Go...
                                       Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k

J C Nash

2018-Apr-07 12:37 UTC

head link

[R] Obtain gradient at multiple values for exponential decay model

R users may want to note that there are some extensions in packages for symbolic
derivatives.

In particular, Duncan Murdoch added some "all in R" tools in the
package nlsr that I maintain. This
is a substitute for the nls() function that uses a fairly unsatisfactory forward
difference derivative
approximation. Moreover, the solver in nlsr is a variant of the Marquardt
stabilized Gauss-Newton
approach, rather than the straight Gauss-Newton that often gives a
"singular gradient" error. Nothing
is free, of course, and the Marquardt approach often uses more iterations when
the problem is
uncomplicated. On the other hand, it is rather like a pit bull in trying to find
a solution.

There is also the Deriv package, also "all in R". Both it and nlsr
allow extensions to the derivatives
table, though I think most users will need to do some homework to become
comfortable with doing
that.

There is also a Google Summer of Code proposal this year to wrap the Julia
Automatic Differentiation
tools to R. This would allow derivatives of coded functions to be computed
avoiding approximations.
I believe it will be at least partly successful in achieving its goals.

Unfortunately (and I would rather hope that someone can say eventually that I am
wrong), I believe
no tool is universally applicable.

JN

On 2018-04-07 02:19 AM, Jeff Newmiller wrote:> I have never found the R symbolic differentiation helpful because my
functions are typically quite complicated, but was
> prompted by Steve Ellison's suggestion to try it out in this case:
> 
> ################# reprex (see reprex package)
> graphdta <- read.csv( text > "t,c
> 0,100
> 40,78
> 80,59
> 120,38
> 160,25
> 200,21
> 240,16
> 280,12
> 320,10
> 360,9
> 400,7
> ", header = TRUE )
> 
> nd <- c( 100, 250, 300 )
> graphmodeld <- lm( log(c) ~ t, data = graphdta )
> graphmodelplin <- predict( graphmodeld
> ???????????????????????? , newdata = data.frame( t = nd )
> ???????????????????????? )
> graphmodelp <- exp(graphmodelplin)
> graphmodelp
> #>??????? 1??????? 2??????? 3
> #> 46.13085 16.58317 11.79125
> # derivative of exp( a + b*t ) is b * exp( a + b*t )
> graphmodeldpdt <- coef( graphmodeld )[ 2 ] * graphmodelp
> graphmodeldpdt
> #>?????????? 1?????????? 2?????????? 3
> #> -0.31464113 -0.11310757 -0.08042364
> 
> # Ellison suggestion - fancy, only works for simple functions
> dc <- deriv( expression( exp( a + b * t ) )
> ?????????? , namevec = "t"
> ?????????? )
> dcf <- function( t ) {
> ? cgm <- coef( graphmodeld )
> ? a <- cgm[ 1 ]
> ? b <- cgm[ 2 ]
> ? eval(dc)
> }
> result <- dcf( nd )
> result
> #> [1] 46.13085 16.58317 11.79125
> #> attr(,"gradient")
> #>??????????????? t
> #> [1,] -0.31464113
> #> [2,] -0.11310757
> #> [3,] -0.08042364
> attr( result, "gradient" )[ , 1 ]
> #> [1] -0.31464113 -0.11310757 -0.08042364
> #################
> 
> On Fri, 6 Apr 2018, David Winsemius wrote:
> 
>>
>>> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
>>>
>>>
>>>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com>
wrote:
>>>>
>>>>> Sent: Friday, April 06, 2018 at 5:55 AM
>>>>> From: "David Winsemius" <dwinsemius at
comcast.net>
>>>>>
>>>>>
>>>>> Not correct. You already have `predict`. It is capale of
using the `newdata` values to do interpolation with the
>>>>> values of the coefficients in the model. See:
>>>>>
>>>>> ?predict
>>>>>
>>>>
>>>> The ? details did not mention interpolation explicity; thanks.
>>>>
>>>>> The original question asked for a derivative (i.e. a
"gradient"), but so far it's not clear that you understand the
>>>>> mathematical definiton of that term. We also remain unclear
whether this is homework.
>>>>>
>>>>
>>>> The motivation of this post was simple differentiation of a
tangent point (dy/dx) manually, then wondering how to
>>>> re-think in modern-day computing terms. Hence the original
question about asking the appropriate functions/syntax to
>>>> read further ("curiosity"), not the answer (indeed,
"homework"). :)
>>>>
>>>> Personal curiosity should be considered "homework".
>>>
>>> Besides symbolic differentiation, there is also the option of
numeric differentiation. Here's an amateurish attempt:
>>>
>>> myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
>>> ?????????????????????????????????????????? exp( predict
(graphmodeld, newdata=data.frame(t=x) )))/
>>> ???????????????????????????????????????? .0001 }
>>> myNumDeriv(c(100, 250, 350))
>>
>> I realized that this would not work in the context of your
construction. I had earlier made a more symbolic version
>> using R formulae:
>>
>> graphdata<-read.csv(text='t,c
>> 0,100
>> 40,78
>> 80,59
>> 120,38
>> 160,25
>> 200,21
>> 240,16
>> 280,12
>> 320,10
>> 360,9
>> 400,7')
>> graphmodeld<-lm(log(c)~t, graphdata)
>> graphmodelp<-exp(predict(graphmodeld))
>> plot(c~t, graphdata)
>> lines(graphdata[,1],graphmodelp)
>> myNumDeriv(c(100, 250, 350), graphmodeld )
>> #----------------------------------------------
>> ???????? 1?????????? 2?????????? 3
>> -0.31464102 -0.11310753 -0.05718414
>>
>>
>>>
>>>
>>>
>>> David Winsemius
>>> Alameda, CA, USA
>>>
>>> 'Any technology distinguishable from magic is insufficiently
advanced.'?? -Gehm's Corollary to Clarke's Third Law
>>>
>>> ______________________________________________
>>> 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.
>>
>> David Winsemius
>> Alameda, CA, USA
>>
>> 'Any technology distinguishable from magic is insufficiently
advanced.'?? -Gehm's Corollary to Clarke's Third Law
>>
>> ______________________________________________
>> 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.
>>
> 
> ---------------------------------------------------------------------------
> Jeff Newmiller??????????????????????? The???? .....?????? .....? Go Live...
> DCN:<jdnewmil at dcn.davis.ca.us>??????? Basics: ##.#.?????? ##.#.?
Live Go...
> ????????????????????????????????????? Live:?? OO#.. Dead: OO#..? Playing
> Research Engineer (Solar/Batteries??????????? O.O#.?????? #.O#.? with
> /Software/Embedded Controllers)?????????????? .OO#.?????? .OO#.? rocks...1k
> 
> ______________________________________________
> 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.

Possibly Parallel Threads

Search for more reasonably related threads

R help - Apr 2018 - Obtain gradient at multiple values for exponential decay model

[R] Obtain gradient at multiple values for exponential decay model

[R] Obtain gradient at multiple values for exponential decay model

[R] Obtain gradient at multiple values for exponential decay model

[R] Obtain gradient at multiple values for exponential decay model

Possibly Parallel Threads