Dear All,
I have a problem I haver been struggling with for a while: I need to
carry out a non-linear fit (and this is the
easy part).
I have a set of discrete values {x1,x2...xN} and the corresponding
{y1, y2...yN}. The difficulty is that I would like the linear fit to
preserve the sum of the values y1+y2+...yN.
I give an example below (for which there may even be an analytical
solution, but that is not the point here)
############################################################################
library(minpack.lm)
set.seed(124)
z <- rexp(3000,3)
zf <- z[z<= 0.5 | z>=0.9]
myhist <- hist(zf, plot=FALSE)
df <- data.frame(x=myhist$mids, y=myhist$density)
myfit <- nlsLM(y~(A*exp(-lambda*x))
,data=df, start=list(A=1,lambda=1))
> sum(myhist$density)
[1] 5> sum(predict(myfit))
[1] 4.931496
############################################################################
I would like sum(predict(myfit)) to be exactly 5 from the start,
without renormalising a posteriori the fit.
Any suggestion is appreciated.
Cheers
Lorenzo
I recommend posting this on a mathematics discussion forum like Stack Exchange and (re-)reading the Posting Guide for this mailing list. I think you are going to need to re-write your model function to algebraically combine your original model along with the constraint, and then use the original model alone for prediction... but I haven't tried it so might be quite far off the mark. On June 20, 2018 8:50:48 AM PDT, Lorenzo Isella <lorenzo.isella at gmail.com> wrote:>Dear All, >I have a problem I haver been struggling with for a while: I need to >carry out a non-linear fit (and this is the >easy part). >I have a set of discrete values {x1,x2...xN} and the corresponding >{y1, y2...yN}. The difficulty is that I would like the linear fit to >preserve the sum of the values y1+y2+...yN. >I give an example below (for which there may even be an analytical >solution, but that is not the point here) > >############################################################################ >library(minpack.lm) > > > >set.seed(124) > >z <- rexp(3000,3) > > >zf <- z[z<= 0.5 | z>=0.9] > >myhist <- hist(zf, plot=FALSE) > > > >df <- data.frame(x=myhist$mids, y=myhist$density) > > > >myfit <- nlsLM(y~(A*exp(-lambda*x)) > ,data=df, start=list(A=1,lambda=1)) > > > >> sum(myhist$density) >[1] 5 >> sum(predict(myfit)) >[1] 4.931496 > >############################################################################ >I would like sum(predict(myfit)) to be exactly 5 from the start, >without renormalising a posteriori the fit. > >Any suggestion is appreciated. >Cheers > >Lorenzo > >______________________________________________ >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.-- Sent from my phone. Please excuse my brevity.
> On Jun 20, 2018, at 8:50 AM, Lorenzo Isella <lorenzo.isella at gmail.com> wrote: > > Dear All, > I have a problem I haver been struggling with for a while: I need to > carry out a non-linear fit (and this is the > easy part). > I have a set of discrete values {x1,x2...xN} and the corresponding > {y1, y2...yN}. The difficulty is that I would like the linear fit to > preserve the sum of the values y1+y2+...yN. > I give an example below (for which there may even be an analytical > solution, but that is not the point here) > > ############################################################################ > library(minpack.lm) > > > > set.seed(124) > > z <- rexp(3000,3) > > > zf <- z[z<= 0.5 | z>=0.9] > > myhist <- hist(zf, plot=FALSE) > > > df <- data.frame(x=myhist$mids, y=myhist$density) > > > > myfit <- nlsLM(y~(A*exp(-lambda*x)) > ,data=df, start=list(A=1,lambda=1)) > > > >> sum(myhist$density) > [1] 5 >> sum(predict(myfit)) > [1] 4.931496 > > ############################################################################ > I would like sum(predict(myfit)) to be exactly 5 from the start, > without renormalising a posteriori the fit.Wouldn't that happen if you minimized that absolute deviations from the fit rather than minimizing the sums of squares??> > Any suggestion is appreciated. > Cheers > > Lorenzo >David Winsemius Alameda, CA, USA 'Any technology distinguishable from magic is insufficiently advanced.' -Gehm's Corollary to Clarke's Third Law