[R] How to find inverse of glm model?

[Gregg Powell]

Hello Luigi,

Great follow-up — Looks like you’re on the right track using nls() for nonlinear regression. You're fitting a logistic-like sigmoidal model (as in Rutledge’s paper), I think both errors you’re encountering are signs of bad starting values or a maybe model formulation issue.

Breakdown of Your Model Attempt
Your model formula:
> Fluo ~ (MaxFluo / (1 + exp(-(Cycle - (HalfFluoCycle / Slope)))) + 1)

This expression has perhaps two issues:

1.   Parameter placement: In the paper, the term inside exp() should look like -(Cycle - HalfFluoCycle)/Slope — not Cycle - (HalfFluoCycle / Slope).

2.   +1 additive constant: You called it backgroundSignal, but it’s hard-coded as +1. That constant should be a parameter (e.g., Bkg), otherwise the model fit has no way to adjust it.

Alternative Model Form:
Here’s a likely corrected version of the model based on Rutledge’s sigmoidal fit:

> mod1 <- nls(Fluo ~ MaxFluo / (1 + exp(-(Cycle - HalfFluoCycle)/Slope)) > + Bkg,
>            data = df,
>            start = list(MaxFluo = max(df$Fluo),
>                         HalfFluoCycle = 20,
>                         Slope = 2,
>                         Bkg = min(df$Fluo)))

Notes on nls() Convergence
nls() is very sensitive to starting values. Always inspect your plot and roughly estimate parameters:

MaxFluo: Use max(Fluo)

Bkg: Use min(Fluo)

HalfFluoCycle: Approximate where the curve inflects

Slope: Try values like 1, 2, 5 depending on steepness

You can also add trace=TRUE to watch convergence progress:

> mod1 <- nls(..., trace=TRUE)

About SSmicmen
>From what I've gathered - this is meant for fitting the Michaelis-Menten model (enzyme kinetics), not sigmoidal growth. That's why mod2 fails — wrong self-starting model function.


Visualizing the Fit
After fitting:

> plot(Fluo ~ Cycle, data = df)
> lines(df$Cycle, predict(mod1), col = "red", lwd = 2)

You might also want to compute the inverse (e.g., solving for Cycle given a fluorescence value) — that would involve solving the nonlinear equation manually or using uniroot().

All the best,
gregg




On Wednesday, April 23rd, 2025 at 8:23 AM, Luigi Marongiu <marongiu.luigi using gmail.com> wrote:

> 

> 

> Further on this, I have found a formula from a paper from Rutledge
> (DOI: 10.1093/nar/gnh177), which I rendered as
> `MaxFluo / (1+ exp(-(Cycle-(HalfFluoCycle/Slope)))) + backgroundSignal`
> I then see that one can use `nls` to fit non-linear regression, so I tried:
> `df <- data.frame(Cycle=1:35, Fluo=c(15908.54, 16246.92, 16722.43, 17104.29, 16794.93, 17031.44, 17299.05, 17185.49, 17362.28, 17127.43, 17368.96, 17513.19, 17593.95, 17626.37, 18308.29, 18768.12, 19955.26, 22493.85, 27088.12, 36539.44, 53694.18, 84663.18, 138835.64, 223331.89, 336409.69, 457729.88, 561233.12, 637536.31, 688985.88, 723158.56, 746575.62, 766274.75, 776087.75, 785144.81, 791573.81) ) plot(Fluo~Cycle, df) mod1 = nls(Fluo~(MaxFluo / (1+ exp(-(Cycle-(HalfFluoCycle/Slope)))) + 1), data=df, start=list(MaxFluo=max(df$Fluo), HalfFluoCycle=15, Slope=0.1)) mod2 = nls(Fluo~SSmicmen(Fluo, Cycle), data=df)`
> but I got errors in both cases:
> ```
> 

> > mod1
> 

> Error in nlsModel(formula, mf, start, wts, scaleOffset = scOff,
> nDcentral = nDcntr) :
> singular gradient matrix at initial parameter estimates
> 

> > mod2
> 

> Error in qr.qty(QR.rhs, .swts * ddot(attr(rhs, "gradient"), lin)) :
> NA/NaN/Inf in foreign function call (arg 5)
> ```
> How can I properly set this regression model?
> Thank you
> 

> 

> On Wed, Apr 16, 2025 at 7:08 AM Luigi Marongiu marongiu.luigi using gmail.com wrote:
> 

> > Thank you. This topic is more complicated than anticipated. Best regards, Luigi
> > 

> > On Tue, Apr 15, 2025 at 11:09 PM Andrew Robinson apro using unimelb.edu.au wrote:
> > 

> > > A statistical (off-topic!) point to consider: when the GLM was fitted, you conditioned on x and let y be the random variable. Therefore the model supports predictions of y conditional on x. You’re now seeking to make predictions of x conditional on y. That’s not the same thing, even in OLS.
> > > 

> > > It might not matter for your application but it’s probably worth thinking about. Simulation experiments might shed some light on that.
> > > 

> > > Cheers, Andrew
> > > 

> > > --
> > > Andrew Robinson
> > > Chief Executive Officer, CEBRA and Professor of Biosecurity,
> > > School/s of BioSciences and Mathematics & Statistics
> > > University of Melbourne, VIC 3010 Australia
> > > Tel: (+61) 0403 138 955
> > > Email: apro using unimelb.edu.au
> > > Website: https://researchers.ms.unimelb.edu.au/~apro@unimelb/
> > > 

> > > I acknowledge the Traditional Owners of the land I inhabit, and pay my respects to their Elders.
> > > 

> > > On 16 Apr 2025 at 1:01 AM +1000, Gregg Powell via R-help r-help using r-project.org, wrote:
> > > 

> > > Take a look at this Luigi...
> > > 

> > > # The model is: logit(p) = β₀ + β₁*Cycles
> > > # Where p is the probability (or normalized value in your case)
> > > 

> > > # The inverse function would be: Cycles = (logit⁻¹(p) - β₀)/β₁
> > > # Where logit⁻¹ is the inverse logit function (also called the expit >function)
> > > 

> > > # Extract coefficients from your model
> > > intercept <- coef(b_model)[1]
> > > slope <- coef(b_model)[2]
> > > 

> > > # Define the inverse function
> > > inverse_predict <- function(p) {
> > > # p is the probability or normalized value you want to find the >cycles for
> > > # Inverse logit: log(p/(1-p)) which is the logit function
> > > logit_p <- log(p/(1-p))
> > > 

> > > # Solve for Cycles: (logit(p) - intercept)/slope
> > > cycles <- (logit_p - intercept)/slope
> > > 

> > > return(cycles)
> > > }
> > > 

> > > # Example: What cycle would give a normalized value of 0.5?
> > > inverse_predict(0.5)
> > > 

> > > This function takes a probability (normalized value between 0 and 1) and returns the cycle value that would produce this probability according to your model.
> > > Also:
> > > This works because GLM with binomial family uses the logit link function by default
> > > The inverse function will return values outside your original data range if needed
> > > This won't work for p=0 or p=1 exactly (you'd get -Inf or Inf), so you might want to add checks
> > > 

> > > All the best,
> > > Gregg
> > > 

> > > On Tuesday, April 15th, 2025 at 5:57 AM, Luigi Marongiu marongiu.luigi using gmail.com wrote:
> > > 

> > > I have fitted a glm model to some data; how can I find the inverse
> > > function of this model? Since I don't know the y=f(x) implemented by
> > > glm (this works under the hood), I can't define a f⁻¹(y).
> > > Is there an R function that can find the inverse of a glm model?
> > > Thank you.
> > > 

> > > The working example is:
> > > `V = c(120.64, 66.14, 34.87, 27.11, 8.87, -5.8, 4.52, -7.16, -17.39, -14.29, -20.26, -14.99, -21.05, -20.64, -8.03, -21.56, -1.28, 15.01, 75.26, 191.76, 455.09, 985.96, 1825.59, 2908.08, 3993.18, 5059.94, 6071.93, 6986.32, 7796.01, 8502.25, 9111.46, 9638.01, 10077.19, 10452.02, 10751.81, 11017.49, 11240.37, 11427.47, 11570.07, 11684.96, 11781.77, 11863.35, 11927.44, 11980.81, 12021.88, 12058.35, 12100.63, 12133.57, 12148.89, 12137.09) df = data.frame(Cycles = 1:35, Values = V[1:35]) M = max(df$Values) df$Norm = df$Values/M df$Norm[df$Norm<0] = 0 b_model = glm(Norm ~ Cycles, data=df, family=binomial) plot(Norm ~ Cycles, df, main="Normalized view", xlab=expression(bold("Time")), ylab=expression(bold("Signal (normalized)")), type="p", col="cyan") lines(b_model$fitted.values ~ df$Cycles, col="blue", lwd=2)`
> > > 

> > > ______________________________________________
> > > R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide https://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > 

> > --
> > Best regards,
> > Luigi
> 

> 

> 

> 

> --
> Best regards,
> Luigi
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