Nonlinear least square in r

流过昼夜 提交于 2020-01-17 05:31:08

问题


Check please, there is a good fit or not.

I have two vectors here:

x=c(30,110)
y=c(0.000760289, 0.000800320, 0.000830345, 0.000840353, 0.000860370,  
0.000910414, 0.000990490, 0.001090594, 0.001200721, 0.001350912, 
0.001531172, 0.001751533, 0.001961923, 0.002192402, 0.002463031, 
0.002793899, 0.003185067, 0.003636604, 0.004148594, 0.004721127, 
0.005394524, 0.006168989, 0.007014544, 0.007870894, 0.008758242, 
0.009656474, 0.010565620, 0.011485709, 0.012396520, 0.013308162, 
0.014271353, 0.015326859, 0.016525802, 0.017889059, 0.019447890, 
0.021223636, 0.023145810, 0.025174229, 0.027391752, 0.029645106, 
0.032337259, 0.035347424, 0.039375125, 0.043575783, 0.048003973, 
0.052926206, 0.058307309, 0.064581189, 0.071947231, 0.080494476, 
0.089891885, 0.100671526, 0.111971207, 0.124237571, 0.137975539, 
0.153629149, 0.171239194, 0.190712664, 0.212079979, 0.235026373,
0.259457493, 0.282867017, 0.307830359, 0.334773680, 0.364001719, 
0.395742526, 0.425596389, 0.458391314, 0.494558651, 0.534657357, 
0.579354317, 0.616075034, 0.656680256, 0.701804548, 0.752133146, 
0.808558032, 0.872226001, 0.944664487, 1.027837007, 1.124484096, 
1.238426232)

and this My codes:

a=0.000066332
b=0.091394349
k=50.2552812
fit = nls(y ~ (a*b*exp(b*(x+0.5))*k)/((k*b)+(a*(exp(b*(x+0.5))-1))), 
start=list(a=a,b=b, k=k))
plot(x,y,col="red", xlab="Age",ylab=expression(paste(mu)))
lines(x,predict(fit), col="blue")

Is this a good fitting ? could I also use the maximum likelihood estimation for this x and y to estimate a, b, and k ? Could give me some codes for doing MLE in r ? I need for comparing between NLS and MLE. Really appreaciate for the helps. Thank you

来源:https://stackoverflow.com/questions/31846677/nonlinear-least-square-in-r

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