Failure in predicting parameters

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,

Rui Barradas

?s 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 at 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 at 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 at 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 at 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

--
Best regards,
Luigi

--
Best regards,
Luigi