Complicated nls formula giving singular gradient message

therss = function(NS,LogKi,BMax)sum((dat$Y - myfunc(NS,LogKi,BMax))^2)
tst = expand.grid(NS=seq(.007,.009,by=.0005),

tst$rss = apply(tst,1,function(x)therss(x[1],x[2],x[3]))
library(lattice)
wireframe(rss~BMax+NS|factor(LogKi),data=tst,as.table=TRUE)

wireframe(rss~LogKi+BMax|factor(NS),data=tst,as.table=TRUE)

Phil,
This is great! I had no idea nls would accept functions in the formula
position. My apologies for not including data to reproduce my issue.?

dat<-data.frame(X=c(-13.0,-11.0,-10.0,-9.5,-9.0,-8.5,-8.0,-7.5,-7.0,-6.5,-6.
0,-5.0,
-13.0,-11.0,-10.0,-9.5,-9.0,-8.5,-8.0,-7.5,-7.0,-6.5,-6.0,-5.0),
Y=c(3146,3321,2773,2415,2183,1091,514,191,109,65,54,50,
3288,3243,2826,2532,2060,896,517,275,164,106,202,53))

With your suggestion, I've changed the formula in nls to the following
function:

myfunc<-function(NS,LogKi,BMax)with(dat,{
KdCPM = KdnM*SpAct*Vol*1000
R<-NS+1
S<-(1+10^(X-LogKi))*KdCPM+Hot
a<-(-1*R)
b<-R*S+NS*Hot+BMax
c<--1*Hot*(S*NS+BMax)
(-1*b+sqrt(b*b-4*a*c))/(2*a)
})

But to get it to compute without errors, I also had to increase the tolerance
level: the step factor keeps being reduced below the min factor.?Looking at
the trace of the nls though, I don't see any changes after the 10th iteration
or so; would increasing the tolerance cause any issue that I'm not thinking
of?

KdnM <- .8687
SpAct <- 4884
Vol <- .125
Hot <- 10191.0
nls(Y~myfunc(NS,LogKi,BMax),data=dat,start=list(NS=.01,LogKi=-7,BMax=10*max(
dat['Y'])),control=nls.control(warnOnly=TRUE,minFactor=1e-5),trace=TRUE)

Also, I've found that if the start value I provide for BMax is too inaccurate
(ex. max(dat['Y']), nls generates the 'singular gradient' error message,
which isn't something I'm used to. Usually nls is kind enough to inform me
that the initial parameter estimates are what caused the problem. Has the
error message changed in a recent update, or is this a different error
message than what I'm thinking about?

Thanks again for all your help,
Jared

On Mon, Dec 13, 2010 at 1:23 PM, Phil Spector <spector at stat.berkeley.edu>
wrote:
      Jared -
      ? nls will happily accept a function on the right hand side
      of the ~ -- you don't have to write out the formula in such
      detail.
      ? What you provided isn't reproducible because you didn't provide
      data, and it's not clear what "Y" in the formula
      represents. ?Let me provide you with an admittedly simpler
      reproducible example.

      ? Suppose we want to estimate the model

      ?response = v * dose / (k + dose)

      where response and dose are variables in a data frame called
      "dat",
      and v and k are the parameters to be estimated.

      Here's the data:

            dat =
            data.frame(dose=c(0.027,0.044,0.073,0.102,0.175,0.257,0.483,0.670),

      + response=c(12.7,16.0,20.4,22.3,26.0,28.2,29.6,31.4))

      Normally we would fit such a simple model as

            nls(response ~ v*dose / (k +
            dose),data=dat,start=list(v=30,k=.05))

      Nonlinear regression model
      ?model: ?response ~ v * dose/(k + dose)
      ? data: ?dat
      ? ? ? v ? ? ? ?k 32.94965 ?0.04568
      ?residual sum-of-squares: 1.091

      Number of iterations to convergence: 4 Achieved convergence
      tolerance: 8.839e-07

      Alternatively, I can write a function like this:

            myfunc = function(v,k)with(dat,v * dose / (k + dose))


      and use the following call to nls:

            nls(response ~
            myfunc(v,k),data=dat,start=list(v=30,k=.05))

      Nonlinear regression model
      ?model: ?response ~ myfunc(v, k)
      ? data: ?dat
      ? ? ? v ? ? ? ?k 32.94965 ?0.04568
      ?residual sum-of-squares: 1.091

      Number of iterations to convergence: 4 Achieved convergence
      tolerance: 8.839e-07

      which gets the identical results.

      Admittedly this function is trivial, but perhaps in your case
      you could use the segments from prism to construct a function
      for the right-hand side of your nls call.

      Hope this helps.
      ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?- Phil Spector
      ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Statistical Computing
      Facility
      ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Department of Statistics
      ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? UC Berkeley
      ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? spector at stat.berkeley.edu






On Mon, 13 Dec 2010, Jared Blashka wrote:

I'm attempting to calculate a regression in R that I normally use
Prism for,
because the formula isn't pretty by any means.

Prism presents the formula (which is in the Prism equation
library as
Heterologous competition with depletion, if anyone is curious) in
these
segments:

KdCPM = KdnM*SpAct*Vol*1000
R=NS+1
S=(1+10^(X-LogKi))*KdCPM+Hot
a=-1*R
b=R*S+NS*Hot+BMax
c = -1*Hot*(S*MS+BMax)
Y = (-1*b+sqrt(b*b-4*a*c))/(2*a)

I'm only trying to solve for NS, LogKi, and BMax. I have
everything else
(KdnM, SpAct, Vol, Hot).

I would use the simple formula at the bottom and then backsolve
for the
terms I'm looking for, but the simple formula at the bottom takes
out the X
term, which is contained within S, which it itself contained in
both b and
c.
So I tried to substitute all the terms back into Y and got the
following

formula<-as.formula("Y ~
(-1*(((NS+1)*((1+10^(X-LogKi))*(KdnM*SpAct*Vol*1000)+Hot))+NS*Hot+BMax)+sqrt
((((NS+1)*((1+10^(X-LogKi))*(KdnM*SpAct*Vol*1000)+Hot))+NS*Hot+BMax)*(((NS+1
)*((1+10^(X-LogKi))*(KdnM*SpAct*Vol*1000)+Hot))+NS*Hot+BMax)-4*(-1*(NS+1))*(
-1*Hot*(((1+10^(X-LogKi))*(KdnM*SpAct*Vol*1000)+Hot)*NS+BMax))))/(2*-1*(NS+1
))")

But trying to use that formula gives me the single gradient
message, which I
wasn't entirely surprised by.
fit<-nls(formula=formula,data=data,start=list(NS=.01,LogKi=-7,BMax=33000))
Error in nls(formula = formula, data = all_no_outliers, start =
list(NS =
0.01, ?:
?singular gradient

I've never worked with a formula this complicated in R. Is it
even possible
to do something like this? Any ideas or points in the right
direction would
be greatly appreciated.

Thanks,
Jared

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