comparing experimental data with nonlinear model
1
0
Entering edit mode
2.1 years ago

Hello,

I am comparing my experimental data with some models to find how much my data fit with these models. I use R studio, but it gives error and I am not able the causes:

mu<-c(0.010, 0.012, 0.044, 0.055, 0.055, 0.044, 0.045);
S <-c(0.3, 0.6, 1.5, 3.75, 11.25, 18.75, 30)
df<-data.frame(mu,S)
df
m1<-nls(mu~S)

a=(0.11*S)
c=(S^2)/600
d=(873+S+c)
muA<-(a)/(d)
muA
mAn<-nls(muA~S,start=list(df),alg = "plinear")

# error:  
Error in object[[3L]][[1L]] : object of type 'symbol' is not subsettable
anova(m1,mAn)

Error in anovalist.nls(object, ...) : 
'anova' is only defined for sequences of "nls" objects

Andrew inhibition medel

Much appreciated if you help in this issue

F-test R • 887 views
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1
Entering edit mode

You provided a linear model m1<-nls(mu~S) whereas nls can only be applied on nonlinear model

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0
Entering edit mode
2.1 years ago

I have data point ( mu vs S) Specific growth rate vs Substrate concentration. I want to compare y data point with inhibited models such as Aiba, handle, Andrew, Edward models by computing the mean square error of the models. 1-Which code do you recommend? How I can find the best value for a,b, c?

m2<-nls(mu~(a^(-S/c))/(b+S), data=df, start=list(a=0.11,b=783,c=500), algorithm="port")

summary(m2)

but I have this error Error in numericDeriv(form[[3L]], names(ind), env, dir = -1 + 2 * (internal parts < : Missing value or an infinity produced when evaluating the model

2-If I only find the difference between a data point and the model, is it enough for mean square error? mu is the vector of experimental data at different concentrations of S, muB is the vector of Aiba at the same concentration of S: rmse(mu, muB)

Aiba

3- How I present the F-test and R-square of the models. R-square usually uses for the fitness o data points to the linear model.

Thanks

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