Failure in predicting parameters
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
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Best regards, Luigi