Are you aware that there are many different kinds of splines?
With "spline" and "splinefun", you can use method = "fmm" (Forsyth,
Malcolm and Moler), "natural", or "periodic". I'm not familiar with
"fmm", but it seems to be adequately explained by the "Manual spline
evaluation" you quoted from the documentation.
Natural splines are perhaps the simplest: I(x-x0)*(x-x0)^j, where
x0 is a knot, and I(z) = 1 if z>0 and 0 otherwise.
However, computations using natural splines are numerically
unstable. The standard solution to this problem is to use B-splines,
which are 0 outside a finite interval.
Let's look at your example:
n <- 9
x <- 1:n
y <- rnorm(n)
plot(x, y, main = paste("spline[fun](.) through", n, "points"))
spl <- smooth.spline(x,y)
lines(spl)
The 'smooth.spline' function uses B-splines. To see what they
look like, let's do the following:
library(fda)
Bspl.basis <- create.bspline.basis(unique(spl$fit$knot))
# Check to make sure:
all.equal(knots(Bspl.basis, interior=FALSE), spl$fit$knot)
# TRUE
# What do B-splines look like?
plot(Bspl.basis)
abline(v=knots(Bspl.basis), lty='dotted', col='red')
# 7 interior knots, 2 end knots replicated 4 times each, for a spline
of order 4, degree 3 (cubic splines)
# total of 15 knots
# Each spline uses 5 consecutive knots, which means there will be 11
basis functions.
# NOTE: 'smooth.spline' rescaled the interval [1, 9] to [0, 1].
# Evaluate the 11 B-splines at 'x'
Bspl.basis.x <- eval.basis((x-1)/8, Bspl.basis)
round(Bspl.basis.x, 4)
# Now the manual computation:
y.spl <- Bspl.basis.x %*% spl$fit$coef
# Plot to confirm:
plot(x, y, main = paste("spline[fun](.) through", n, "points"))
spl.xy <- spline(x, y)
lines(spl.xy)
points(x, y.spl, pch=2, col='red')
Hope this helps.
Spencer
[EMAIL PROTECTED] wrote:
Fair enough. FOr a spline interpolation I can do the following:
n <- 9
x <- 1:n
y <- rnorm(n)
plot(x, y, main = paste("spline[fun](.) through", n, "points"))
lines(spline(x, y))
Then look at the coefficients generated as:
f <- splinefun(x, y)
ls(envir = environment(f))
[1] "ties" "ux" "z"
splinecoef <- get("z", envir = environment(f))
slinecoef
$method
[1] 3
$n
[1] 9
$x
[1] 1 2 3 4 5 6 7 8 9
$y
[1] 0.93571604 0.44240485 0.45451903 -0.96207396 -1.13246522 -0.60032698
[7] -1.77506105 -0.09171419 -0.23262573
$b
[1] -1.53673409 0.22775629 -0.81788209 -1.16966436 0.73558677 -0.68744178
[7] 0.08639287 1.86770869 -2.92992167
$c
[1] 1.3657783 0.3987121 -1.4443504 1.0925682 0.8126830 -2.2357115 3.0095462
[8] -1.2282303 -3.5694000
$d
[1] -0.32235542 -0.61435416 0.84563953 -0.09329507 -1.01613149 1.74841922
[7] -1.41259217 -0.78038989 -0.78038989
WHen I look at ?spline there is even an example of "manually" using these
coefficeients:
## Manual spline evaluation --- demo the coefficients :
.x <- get("ux", envir = environment(f))
u <- seq(3,6, by = 0.25)
(ii <- findInterval(u, .x))
dx <- u - .x[ii]
f.u <- with(splinecoef,
y[ii] + dx*(b[ii] + dx*(c[ii] + dx* d[ii])))
stopifnot(all.equal(f(u), f.u))
For the smooth.spline as
spl <- smooth.spline(x,y)
I can also look at the coefficients:
spl$fit
$knot
[1] 0.000 0.000 0.000 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
[13] 1.000 1.000 1.000
$nk
[1] 11
$min
[1] 1
$range
[1] 8
$coef
[1] 0.90345898 0.73823276 0.40777431 -0.08046715 -0.54625461 -0.85205147
[7] -0.96233408 -0.91373830 -0.66529714 -0.47674774 -0.38246971
attr(,"class")
[1] "smooth.spline.fit"
But there isn't an example on how to "manual" use these coefficients. This is what I was
asking about. Once I hae the coefficients how do I "manually" interpolate using the
coefficients given and x.
Thank you.
Kevin
---- Spencer Graves <[EMAIL PROTECTED]> wrote:
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
I do NOT know how to do what you want, but with a self-contained
example, I suspect many people on this list -- probably including me --
could easily solve the problem. Without such an example, there is a
high probability that any answer might (a) not respond to your need, and
(b) take more time to develop, just because we don't know enough of what
you are asking.
Spencer
[EMAIL PROTECTED] wrote:
Like I indicated. I understand the coefficients in a B-spline context. If I use
the the 'spline' or 'splinefun' I can get the coefficients and they are grouped
as 'a', 'b', 'c', and 'd' coefficients. But the coefficients for smooth.spline
is just an array. I basically want to take these coefficients and outside of
'R' use them to form an interpolation. In other words I want 'R' to do the hard
work and then export the results so they can be used else where.
Thank you.
Kevin
Spencer Graves wrote:
I believe that a short answer to your question is that the
"smooth" is a linear combination of B-spline basis functions, and the
coefficients are the weights assigned to the different B-splines in
that basis.
Before offering a much longer answer, I would want to know what
problem you are trying to solve and why you want to know. For a brief
description of B-splines, see
"http://en.wikipedia.org/wiki/B-spline";. For a slightly longer
commentary on them I suggest the "scripts\ch01.R" in the DierckxSpline
package: That script computes and displays some B-splines using
"splineDesign", "spline.des" in the 'splines' package plus comparable
functions in the 'fda' package. For more info on this, I found the
first chapter of Paul Dierckx (1993) Curve and Surface Fitting with
Splines (Oxford U. Pr.). Beyond that, I've learned a lot from the
'fda' package and the two companion volumes by Ramsay and Silverman
(2006) Functional Data Analysis, 2nd ed. and (2002) Applied Functional
Data Analysis (both Springer).
If you'd like more help from this listserve, PLEASE do read the
posting guide http://www.R-project.org/posting-guide.html and provide
commented, minimal, self-contained, reproducible code.
Hope this helps. Spencer Graves
[EMAIL PROTECTED] wrote:
I like what smooth.spline does but I am unclear on the output. I can
see from the documentation that there are fit.coef but I am unclear
what those coeficients are applied to.With spline I understand the
"noraml" coefficients applied to a cubic polynomial. But these
coefficients I am not sure how to interpret. If I had a description
of the algorithm maybe I could figure it out but as it is I have this
question. Any help?
Kevin
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and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
