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.lu...@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.lu...@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$B) - y
> > > estim <- nls.lm(par = list(m = halfFluo, s = slopes, M = MaxFluo, B =
> > > high[1]),
> > + fn = rutledge_param, x = 1:45, y = high)
> > Error in dimnames(x) <- dn :
> > length of 'dimnames' [2] not equal to array extent
> > ```
> > Probably because 'slopes' is a vector instead of a scalar. Since the
> > slope is changing, I don't think is right to use a scalar, but I tried
> > and I got:
> > ```
> > > estim <- nls.lm(par = list(m = halfFluo, s = 1, M = MaxFluo, B = high[1]),
> > + fn = rutledge_param, x = 1:45, y = high)
> > > estim
> > Nonlinear regression via the Levenberg-Marquardt algorithm
> > parameter estimates: 6010.94, 1, 12021.88, 4700.49288888889
> > residual sum-of-squares: 1.14e+09
> > reason terminated: Relative error in the sum of squares is at most `ftol'.
> > ```
> > The values reported are the same I used at the beginning apart from
> > the last (the background parameter) which is 4700 instead of zero. If
> > I plug it, I get an L shaped plot that is worse than that at the
> > beginning:
> > ```
> > after = init = rutledge(halfFluo, 1, MaxFluo, 4700.49288888889, high)
> > points(1:45, after, type="l", col="blue")
> > ```
> > What did I get wrong here?
> > Thanks
> >
> > On Sun, Mar 14, 2021 at 8:05 PM Bill Dunlap <williamwdun...@gmail.com>
> > wrote:
> > >
> > > > rutledge_param <- function(p, x, y) ((p$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.lu...@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@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.
> >
> >
> >
> > --
> > Best regards,
> > Luigi
>
>
>
> --
> Best regards,
> Luigi
--
Best regards,
Luigi
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