[R] Failure in predicting parameters

Luigi Marongiu m@rong|u@|u|g| @end|ng |rom gm@||@com
Fri Mar 19 07:28:59 CET 2021

```Thank you, I'll try it!

On Thu, Mar 18, 2021 at 9:46 PM Rui Barradas <ruipbarradas using sapo.pt> wrote:
>
> Hello,
>
> Maybe a bit late but there is a contributed package [1] for quantitative
> PCR fitting non-linear models with the Levenberg-Marquardt algorithm.
>
> estim and vector R below are your model and your fitted values vector.
> The RMSE of this fit is smaller than your model's.
>
>
> Isn't this simpler?
>
>
> library(qpcR)
>
> df1 <- data.frame(Cycles = seq_along(high), high)
>
> fit <- pcrfit(
>    data = df1,
>    cyc = 1,
>    fluo = 2
> )
> summary(fit)
>
> coef(estim)
> coef(fit)
>
>
> sqrt(sum(resid(estim)^2))
> #[1] 1724.768
> sqrt(sum(resid(fit)^2))
> #[1] 1178.318
>
>
> highpred <- predict(fit, newdata = df1)
>
> plot(1:45, high, type = "l", col = "red")
> points(1:45, R, col = "blue")
> points(1:45, highpred\$Prediction, col = "cyan", pch = 3)
>
>
> [1] https://CRAN.R-project.org/package=qpcR
>
> Hope this helps,
>
>
> Às 06:51 de 18/03/21, Luigi Marongiu escreveu:
> > It worked. I re-written the equation as:
> > ```
> > rutledge_param <- function(p, x, y) ( (p\$M / ( 1 + exp(-(x-p\$m)/p\$s))
> > ) + p\$B ) - y
> > ```
> > and used Desmos to estimate the slope, so:
> > ```
> > estim <- nls.lm(par = list(m = halfCycle, s = 2.77, M = MaxFluo, B = high[1]),
> >              fn = rutledge_param, x = 1:45, y = high)
> > summary(estim)
> > R <- rutledge(list(half_fluorescence = 27.1102, slope = 2.7680,
> >                     max_fluorescence = 11839.7745, back_fluorescence =
> > -138.8615) , 1:45)
> > points(1:45, R, type="l", col="red")
> > ```
> >
> > Thanks
> >
> > On Tue, Mar 16, 2021 at 8:29 AM Luigi Marongiu <marongiu.luigi using gmail.com> wrote:
> >>
> >> Just an update:
> >> I tried with desmos and the fitting looks good. Desmos calculated the
> >> parameters as:
> >> Fmax = 11839.8
> >> Chalf = 27.1102 (with matches with my estimate of 27 cycles)
> >> k = 2.76798
> >> Fb = -138.864
> >> I forced R to accept the right parameters using a single named list
> >> and re-written the formula (it was a bit unclear in the paper):
> >> ```
> >> rutledge <- function(p, x) {
> >>    m = p\$half_fluorescence
> >>    s = p\$slope
> >>    M = p\$max_fluorescence
> >>    B = p\$back_fluorescence
> >>    y = (M / (1+exp( -((x-m)/s) )) ) + B
> >>    return(y)
> >> }
> >> ```
> >> but when I apply it I get a funny graph:
> >> ```
> >> desmos <- rutledge(list(half_fluorescence = 27.1102, slope = 2.76798,
> >>                          max_fluorescence = 11839.8, back_fluorescence
> >> = -138.864) , high)
> >> ```
> >>
> >> On Mon, Mar 15, 2021 at 7:39 AM Luigi Marongiu <marongiu.luigi using gmail.com> wrote:
> >>>
> >>> Hello,
> >>> the negative data comes from the machine. Probably I should use raw
> >>> data directly, although in the paper this requirement is not reported.
> >>> The p\$x was a typo. Now I corrected it and I got this error:
> >>> ```
> >>>
> >>>> rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(x-p\$m)/p\$s))) + p\$B) - y
> >>>> estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]),
> >>> +             fn = rutledge_param, x = 1:45, y = high)
> >>> Error in dimnames(x) <- dn :
> >>>    length of 'dimnames' [2] not equal to array extent
> >>> ```
> >>> Probably because 'slopes' is a vector instead of a scalar. Since the
> >>> slope is changing, I don't think is right to use a scalar, but I tried
> >>> and I got:
> >>> ```
> >>>> estim <- nls.lm(par = list(m = halfFluo, s = 1, M = MaxFluo, B = high[1]),
> >>> +             fn = rutledge_param, x = 1:45, y = high)
> >>>> estim
> >>> Nonlinear regression via the Levenberg-Marquardt algorithm
> >>> parameter estimates: 6010.94, 1, 12021.88, 4700.49288888889
> >>> residual sum-of-squares: 1.14e+09
> >>> reason terminated: Relative error in the sum of squares is at most `ftol'.
> >>> ```
> >>> The values reported are the same I used at the beginning apart from
> >>> the last (the background parameter) which is 4700 instead of zero. If
> >>> I plug it, I get an L shaped plot that is worse than that at the
> >>> beginning:
> >>> ```
> >>> after = init = rutledge(halfFluo, 1, MaxFluo, 4700.49288888889, high)
> >>> points(1:45, after, type="l", col="blue")
> >>> ```
> >>> What did I get wrong here?
> >>> Thanks
> >>>
> >>> On Sun, Mar 14, 2021 at 8:05 PM Bill Dunlap <williamwdunlap using gmail.com> wrote:
> >>>>
> >>>>> rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y
> >>>>
> >>>> Did you mean that p\$x to be just x?  As is, this returns numeric(0)
> >>>> for the p that nls.lm gives it because p\$x is NULL and NULL-aNumber is
> >>>> numeric().
> >>>>
> >>>> -Bill
> >>>>
> >>>> On Sun, Mar 14, 2021 at 9:46 AM Luigi Marongiu <marongiu.luigi using gmail.com> wrote:
> >>>>>
> >>>>> Hello,
> >>>>> I would like to use the Rutledge equation
> >>>>> (https://pubmed.ncbi.nlm.nih.gov/15601990/) to model PCR data. The
> >>>>> equation is:
> >>>>> Fc = Fmax / (1+exp(-(C-Chalf)/k)) + Fb
> >>>>> I defined the equation and another that subtracts the values from the
> >>>>> expectations. I used minpack.lm to get the parameters, but I got an
> >>>>> error:
> >>>>> ```
> >>>>>
> >>>>>> library("minpack.lm")
> >>>>>> h <- 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)
> >>>>>> high <- h[1:45]
> >>>>>> MaxFluo <- max(high)
> >>>>>> halfFluo <- MaxFluo/2
> >>>>>> halfCycle = 27
> >>>>>> find_slope <- function(X, Y) {
> >>>>> +   Slope <- c(0)
> >>>>> +   for (i in 2:length(X)) {
> >>>>> +     delta_x <- X[i] - X[i-1]
> >>>>> +     delta_y <- Y[i] - Y[i-1]
> >>>>> +     Slope[i] <- delta_y/delta_x
> >>>>> +   }
> >>>>> +   return(Slope)
> >>>>> + }
> >>>>>> slopes <- find_slope(1:45, high)
> >>>>>>
> >>>>>> rutledge <- function(m, s, M, B, x) {
> >>>>> +   divisor = 1 + exp(-1* ((x-m)/s) )
> >>>>> +   y = (M/divisor) + B
> >>>>> +   return(y)
> >>>>> + }
> >>>>>> rutledge_param <- function(p, x, y) ((p\$M / (1 + exp(-1*(p\$x-p\$m)/p\$s))) + p\$B) - y
> >>>>>>
> >>>>>>
> >>>>>> init = rutledge(halfFluo, slopes, MaxFluo, 0, high)
> >>>>>> points(1:45, init, type="l", col="red")
> >>>>>> estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B = high[1]),
> >>>>> +                 fn = rutledge_param, x = 1:45, y = high)
> >>>>> Error in nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B =
> >>>>> high[1]),  :
> >>>>>    evaluation of fn function returns non-sensible value!
> >>>>> ```
> >>>>>
> >>>>> Where could the error be?
> >>>>>
> >>>>>
> >>>>> --
> >>>>> Best regards,
> >>>>> Luigi
> >>>>>
> >>>>> ______________________________________________
> >>>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >>>>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>>>> and provide commented, minimal, self-contained, reproducible code.
> >>>
> >>>
> >>>
> >>> --
> >>> Best regards,
> >>> Luigi
> >>
> >>
> >>
> >> --
> >> Best regards,
> >> Luigi
> >
> >
> >

--
Best regards,
Luigi

```