Set an exponential curve using nls with a custom data frame in R

Question

I had a lot of questions on SO, but I cannot find a fix for my problem in related answers. I have a table like this like PAO1.data :

 Name P AO Prog1 0.654 59.702 Prog2 0.149 49.595 Prog3 0.505 50.538 Prog4 0.777 59.954 Prog5 0.237 49.611 Prog6 0.756 50.630 Prog7 0.560 118.014 Prog8 0.015 53.779 Prog9 0.789 68.096 Prog10 0.825 79.558 

I tried using nls to match an exponential curve to data.

 df = data.frame(PAO1.data) p = df$P ao = df$AO RMSE <- function(fit, act){ sqrt(mean((fit - act)^2)) } expmodel = nls(formula = p ~ exp(ao), data = df, start = list(ao = 0.01)) fit1 = fitted.values(expmodel) err1 = RMSE(fit1, p) plot(ao, p) lines(ao, predict(expmodel)) print(err1) 

When I try to start it, I get these warning messages when creating expmodel :

 Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 

At the same time, I get this error in lines() :

 Error in xy.coords(x, y) : 'x' and 'y' lengths differ Calls: lines -> lines.default -> plot.xy -> xy.coords 

Another question I read on SO with the error "lengths differenting" actually had different lengths in x and y . However, my x and y (here ao and p ) have exactly the same number of values.

Please note that it is unlikely that the exponential curve will actually be a good fit, but I am testing several different models and I want to know how to use nls correctly, so I can do the same with other models.

Some related curve fitting issues:

This question suggests that the key is raw data. The smallest value for AO in my table is 0.015, and I chose 0.01, which, in my opinion, is pretty close. This question asks about nls, and the answer is asked using a polynomial using lm. I especially need to know how to use nls for many complex models in the future, and this will not work for me. This question looked promising, but I can’t find the problem in my code by looking at this question and the answer - I also have similar statements in my code.

How can I do it?

Edit:

Here are screenshots of the solutions posted in Roland's comments: (Actual data set is larger)

After changing the call to nls to expmodel = nls(formula = p ~ exp(beta * ao), data = df, start = list(beta = 0.01))

After sorting the AO values ​​using lines(sort(ao), predict(expmodel))

Answer 1

 df <- read.table(text = "Name P AO Prog1 0.654 59.702 Prog2 0.149 49.595 Prog3 0.505 50.538 Prog4 0.777 59.954 Prog5 0.237 49.611 Prog6 0.756 50.630 Prog7 0.560 118.014 Prog8 0.015 53.779 Prog9 0.789 68.096 Prog10 0.825 79.558", header = TRUE) #use correct syntax: expmodel = nls(formula = P ~ exp(beta * AO), data = df, start = list(beta = 0.01)) plot(P ~ AO, data = df) #you could use lines after sorting, but this is more convenient: curve(predict(expmodel, newdata = data.frame(AO = x)), from = 49, to = 120, add = TRUE) 

Obviously, this is not a good model for your data. As you know, an exponential function passes through (0,1). You should consider adding an interception.