> # using this test data
> set.seed(1)
> x <- 1:20/20
> y <- exp(2 + 3 * x) + rnorm(20)
>
> # if its ok to fit logs so that its linear
> exp(fitted(lm(log(y) ~ x)))
1 2 3 4 5 6 7 8
8.55615 9.94692 11.56376 13.44340 15.62857 18.16894 21.12223 24.55557
9 10 11 12 13 14 15 16
28.54699 33.18720 38.58165 44.85295 52.14363 60.61938 70.47284 81.92793
17 18 19 20
95.24501 110.72673 128.72494 149.64869>
> # or to do it on original scale use linear coefs as starting values
> cc <- coef(lm(log(y) ~ x))
> fitted(nls(y ~ exp(a + b*x), start = list(a = cc[1], b = cc[2])))
[1] 8.592270 9.984536 11.602401 13.482421 15.667073 18.205720
[7] 21.155722 24.583734 28.567211 33.196159 38.575168 44.825776
[13] 52.089214 60.529599 70.337640 81.734946 94.979039 110.369167
[19] 128.253066 149.034820
attr(,"label")
[1] "Fitted values"
On 9/27/06, jessica.gervais at tudor.lu <jessica.gervais at tudor.lu>
wrote:> Hi,
>
> I would like to fit some experimental points by a exponential function.
> I ignore the parameters of this exponential and what I would like is to
> ask R to calculate the best fitting curve an the associated parameters (as
> the linear model function (lm) does for linear models).
> Is it possible ?
> Do anyone have an idea about how to do that ?
>
> Thanks by advance
>
> Jessica Gervais
>
>
> [[alternative HTML version deleted]]
>
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