Using R to fit a Sigmoidal Curve

Question

I have read a post ( Sigmoidal Curve Fit in R ). It was labeled duplicated, but I can\'t see anything related with the posts. And the answer given for the posts was not enou

Answer 1

Luckily R offers a selfstarting model for the logistic model. It uses a slight reparametrization, but is really the same model as yours: Asym/(1+exp((xmid-input)/scal))

A selfstarting model can estimate good starting values for you, so you don't have to specify them.

plot(y ~ x)
fit <- nls(y ~ SSlogis(x, Asym, xmid, scal), data = data.frame(x, y))

summary(fit)
#Formula: y ~ SSlogis(x, Asym, xmid, scal)
#
#Parameters:
#      Estimate Std. Error t value Pr(>|t|)
#Asym 1.473e+04  2.309e+04   0.638    0.551
#xmid 4.094e+00  2.739e+00   1.495    0.195
#scal 9.487e-01  5.851e-01   1.622    0.166
#
#Residual standard error: 941.9 on 5 degrees of freedom
#
#Number of iterations to convergence: 0 
#Achieved convergence tolerance: 4.928e-06

lines(seq(0.5, 4, length.out = 100), 
      predict(fit, newdata = data.frame(x = seq(0.5, 4, length.out = 100))))

Of course your data doesn't really support the model. The estimated mid point is just at the right limit of your data range and thus the parameter estimates (in particular for the asymptote) are very uncertain.

Answer 2

I see two issues.

1. the default algorithm of nls is very sensitive to the starting parameter. In your example data I found it useful to use algorithm='port'. Alternatively switching to a "robust" implementation might also help.

2. It helps understanding the role of the parameter in your model.

The simple interpretation for your model is: The sigmoid goes in y from 0 to a. It reaches the "half way" point at x=c. b has the role of a slope, and if negative the model would go from a to 0 instead.

Specifically to the test data posted by you I would estimate the start values as following:

• First thing I notice - your data is not exactly 'close' to zero so maybe it might be useful to add an offset d which is around 1000.
• a is then 5000 or greater
• c is somewhere greater 2 - maybe 3
• b one needs to guess - from x 2 to 3.5 your signal jumps from 1000 to 6000 gives 5000 difference - divided by a - a slope of 1/1.5 = 0.66 or greater... lets round to one.

So ultimately using the formula

fitmodel <- nls(y ~a/(1 + exp(-b * (x-c)) ) + d, start=list(a=5000,b=1,c=3, d=1000))

gives a fit (also works without the d). Trying around I found setting algorithm='port' made the command even less sensitive to the start values.

Answer 3

The code I used to fit your data:

 df <- data.frame(x=c(3.9637878,3.486667,3.0095444,2.5324231,2.0553019,1.5781806,1.1010594,0.6242821),                     
                  y=c(6491.314,6190.092,2664.021,2686.414,724.707,791.243,1809.586,541.243))

library(drc)
fm <- drm(y ~ x, data = df, fct = G.3())

plot(fm)
summary(fm)

The way it looks after fitting: