but wondered if anyone had suggestions for something simpler.
Thanks!Peter
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Peter Tittmann
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Hi,
Attachments:- nlfit.png
On 10-12-01 04:08 PM, Peter Tittmann wrote:
>
>
> Hi,
>
> Thanks very much for taking the time on this. I realize that this is
> sort of a pure R question but I'll ask anyway:
>
> can you tell me what exactly the %*% operator does to be able to
> multiply the vcov() and v0 matrices since the vcov is a 2x2 and the v0
> is a 2x20? Its sort of a weird question but Im trying to replicate the
> behavior in python.
>
> Thanks again!
what's the problem? %*% is standard matrix multiplication --
multiplying an (m,n) by an (n,p) matrix should Just Work ...
(hope I'm not missing something obvious myself)
Ben Bolker
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Dennis and Ben,
On Wednesday, December 1, 2010 at 2:41 PM, Dennis Murphy wrote:
Hi:
Ben's got it right - it's simply a matrix multiplication operator. In the example, the multiplication produces a 20 x 20 matrix, the only relevant part of which is the diagonal.
Essentially, the formula works on an individual basis (Bates and Watts, 1988, section 2.3.2, pp. 58-59). The term under the square root is a quadratic form v'Av, where v is a row of the derivative matrix evaluated at x and the LSEs and A is the covariance matrix of the estimated coefficients. All I did was to try to 'vectorize' the operation by computing the covariance matrix of all the y's and then just pull out the diagonals (the variances).
Ref: Bates and Watts (1988). Nonlinear Regression Analysis and Its Applications. NY: Wiley.
HTH,
Dennis
`ggplot2.theme_bw()+ggplot2.scale_x_continuous('height (m)')+ggplot2.scale_y_continuous('DBH (cm)’)`
Peter
On Aug 30, 2016, at 5:45 AM, Mrunmayee Sahoo <mrunmay...@gmail.com> wrote:hithe code helps me a lot. can you please tell me how to add legend for the lines and dots here.thank you in advance
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#Input data (measured variables)
d <- matrix(c(70,10,64,20,58,40,52,79), ncol=2)
colnames(d) <- c('x', 'y')
trial.table <- as.table(d)
#model
m <- nls(y ~ A - 20*log10(x) - B*x, data = dd, start = list(A = 1, B = 1))
From the model, we can easily calculate the partial derivatives as following df/dA = 1 and df/dB = x.
And as I understood (and I am sorry if I lost something), the SE calculations depends on the model, and we should use the partial derivatives to calculate v0.
But how do I do that?
Best wishes,
Carlos.
PS: This is the first time I write in this forum, so I sorry if the message is somehow unclear or out of the format.