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rregressionpredictnlsnon-linear-regression

Predict function in R


I am trying to use to predict function to predict 100 points new points. I have a data.frame with one vector that is 100 doubls long.

I am trying the predict function: predict(model, newdata=mydat)

The function only returns a vector of length four. This could be due to the fact that the model was made only with four points, but I am unsure.

EDIT:

Creation of mydat

mydat <- data.frame(V1 = seq(0, max(myExperimentSummary$V1), length.out = 100))

The model I am using

model
#Nonlinear regression model
#  model: mean ~ (1/(1 + exp(-b * (V1 - c))))
#   data: myExperimentSummary
#      b       c 
#-0.6721  3.2120 
# residual sum-of-squares: 0.04395
# 
#Number of iterations to convergence: 1 
#Achieved convergence tolerance: 5.204e-06

EDIT2: Fixing the typos

EDIT3:

fitcoef = nlsLM(mean~(a/(1+exp(-b*(V5-c)))), data = myExperimentSummary,
                start=c(a=1,b=.1,c=25))

fitmodel = nls(mean~(1/(1+exp(-b*(V1-c)))), data = myExperimentSummary,
               start=coef(fitcoef))

mydat <- data.frame(V1 = seq(0, max(myExperimentSummary$V1), length.out = 100))

predict(fitmodel, mydat)

Solution

  • If your data are still as in your previous question:

    dat <- read.table(text = " V1  N mean  
                              0.1  9 0.9 
                                1  9 0.8 
                               10  9 0.1 
                                5  9 0.2",
                      header = TRUE)
    
    model <- nls(mean ~ -a/(1 + exp(-b * (V1-o))), data = dat,
                 start=list(a=-1.452, b=-0.451, o=1.292))
    

    Then I can not reproduce your problem:

    mydat <- data.frame(V1 = seq(0, max(dat$V1), length.out = 100))
    
    y <- predict(model, mydat)
    
    length(y)
    # [1] 100