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  1. POnonlinear mixed model using R?
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    <p>I have a data set like this example data, which includes three sites. I want to model the trend of "cumSR" along "timeInt". I did it for each site using the following model ("cumSR~b/(cc+(timeInt)^(-z))"). Now I want to use the same model to get only one curve for these three sites. I think that I need to use nonlinear mixed model, right? I have tried "nlme" function in "nlme" package and "nlmer" function in "lme4" function, but haven't figured out. Could anyone help on this? Thanks so much.</p> <pre><code> dat = data.frame(cumSR=c(17, 25, 34, 43, 46, 49, 51, 52, 55, 57, 59, 60, 18, 26, 34, 38, 41, 43, 45, 49, 51, 54, 57, 58, 59, 23, 29, 33, 37, 38, 40, 44, 46, 47), timeInt=c(2.5, 7.5, 12.5, 17.5, 22.5, 27.5, 32.5, 42.5, 47.5, 52.5, 62.5, 67.5, 2.5, 7.5, 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5, 62.5, 2.5, 7.5, 12.5, 17.5, 22.5, 37.5, 42.5, 47.5, 57.5), site=c("A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B","B", "B", "B", "B", "B", "B", "B", "C","C","C","C","C","C","C","C","C")) par(mfcol=c(1,3)) dat1 &lt;- dat[which(dat$site=="A"),] SAR.nls.logist &lt;- nls(cumSR~b/(cc+(timeInt)^(-z)), data=dat1,start = list(b=10,cc=.1,z=.5)) plot(dat1$timeInt,dat1$cumSR, type="b",col=1) curve(coef(SAR.nls.logist)[1]/(coef(SAR.nls.logist)[2]+(x)^(-coef(SAR.nls.logist)[3])), col = "pink", type="l",add = TRUE,lwd=2) dat1 &lt;- dat[which(dat$site=="B"),] SAR.nls.logist &lt;- nls(cumSR~b/(cc+(timeInt)^(-z)), data=dat1,start = list(b=10,cc=.1,z=.5)) plot(dat1$timeInt,dat1$cumSR, type="b",col=1) curve(coef(SAR.nls.logist)[1]/(coef(SAR.nls.logist)[2]+(x)^(-coef(SAR.nls.logist)[3])), col = "pink", type="l",add = TRUE,lwd=2) dat1 &lt;- dat[which(dat$site=="C"),] SAR.nls.logist &lt;- nls(cumSR~b/(cc+(timeInt)^(-z)), data=dat1,start = list(b=10,cc=.1,z=.5)) plot(dat1$timeInt,dat1$cumSR, type="b",col=1) curve(coef(SAR.nls.logist)[1]/(coef(SAR.nls.logist)[2]+(x)^(-coef(SAR.nls.logist)[3])), col = "pink", type="l",add = TRUE,lwd=2) </code></pre>
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