Jared -
Actually I didn't realize that nls would accept a formula
until I tried my simple example in reaction to your post :-)
I don't recall nls() ever reporting the cause of the singular
gradient as being bad starting values -- I know I spend a lot
of time in my lectures on non-linear regression emphasizing that
bad starting values are the usual culprit when the dreaded
"singular gradient" message rears its ugly head.
I think your function has a fairly large "flat" region, wherein
changes in some of the parameters don't really effect the residual
sums of squares that much. I think you can visualize it like this:
therss = function(NS,LogKi,BMax)sum((dat$Y - myfunc(NS,LogKi,BMax))^2)
tst = expand.grid(NS=seq(.007,.009,by=.0005),
+ LogKi=seq(-9.5,-8.5,by=.05),
+ BMax =seq(1.8e05,2.8e05,by=.1e05))
tst$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)
If you look at the panels where LogKi is around -8.9 (the reported
maximum), the residual-sums-of-squares surface is pretty flat.
I think you can also see regions where there isn't much change in
the residual sums of squares in this plot:
wireframe(rss~LogKi+BMax|factor(NS),data=tst,as.table=TRUE)
I also ran your data through proc nlp in sas (I know there are a lot of
SAS-bashers on this list, but I worked there many years ago and I know
the quality of their software), and got the following results:
Optimization Results
Parameter Estimates
Gradient
Objective
N Parameter Estimate Function
1 NS 0.006766 -0.121333
2 LogKi -8.966402 -0.000509
3 BMax 237013 1.109368E-11
The message that nlp reported was
NOTE: At least one element of the (projected) gradient is greater than 1e-3.
Finally, I ran the the same model and data using nlfit in matlab, with all
values set to their defaults. It reported the following without warning:
ans =
1.0e+05 *
0.000000086522054 -0.000089870065555 2.371354822440646
which agrees almost exactly with R.
Hope this helps.
- Phil
On Mon, 13 Dec 2010, Jared Blashka wrote:
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 <spec...@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
spec...@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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