The help page for 'smooth.spline' says the argument 'df' is 'the
desired equivalent number of degrees of freedom (trace of the smoother
matrix).' It also explains that the output of 'smooth.spline' includes
a component 'fit', and two components of 'fit' are 'knot' and 'coef'.
To learn more, you can run the examples, examine 'str(cars.spl)' and the
other objects produced by those examples. You can also read more in the
references cited there.
If you would like further help from this group, please submit another
question, preferably after first reading the posting guide!
"www.R-project.org/posting-guide.html". There is substantial but
anecdotal evidence to suggest that posts more consistent with that guide
tend to get better answers quicker. For example, if the above does NOT
answer your question, I believe you would have gotten a better reply if
you had provided a simple, self-contained example, rather than having me
rely on one from the 'example' section of the 'smooth.spline' help page.
Hope this helps.
Spencer Graves
Steven Shechter wrote:
> Hi,
> If I set df=2 in my smooth.spline function, is that equivalent to running
> a linear regression through my data? It appears that df=# of data points
> gives the interpolating spline and that df = 2 gives the linear
> regression, but I just want to confirm this.
>
> Thank you,
> Steven
>
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