问题
I fitted a smoothing spline to data in R with
library(splines)
Model <- smooth.spline(x, y, df =6)
I would like to take the fitted spline and evaluate it for arbitrary new data in an external code (not in R). In other words, do what the predict.smooth.spline function does. I had a look at the Model object:
> str(Total_work_model)
List of 15
$ x : num [1:14] 0.0127 0.0186 0.0275 0.0343 0.0455 ...
$ y : num [1:14] 3174 3049 2887 2862 2975 ...
$ w : num [1:14] 1 1 1 1 1 1 1 1 1 1 ...
$ yin : num [1:14] 3173 3075 2857 2844 2984 ...
$ data :List of 3
..$ x: num [1:14] 0.0343 0.0455 0.0576 0.0697 0.0798 ...
..$ y: num [1:14] 2844 2984 3048 2805 2490 ...
..$ w: num [1:14] 1 1 1 1 1 1 1 1 1 1 ...
$ lev : num [1:14] 0.819 0.515 0.542 0.568 0.683 ...
$ cv.crit : num 6494075
$ pen.crit: num 3260
$ crit : num 3
$ df : num 8
$ spar : num 0.353
$ lambda : num 8.26e-05
$ iparms : Named int [1:3] 3 0 10
..- attr(*, "names")= chr [1:3] "icrit" "ispar" "iter"
$ fit :List of 5
..$ knot : num [1:20] 0 0 0 0 0.056 ...
..$ nk : int 16
..$ min : num 0.0127
..$ range: num 0.104
..$ coef : num [1:16] 3174 3132 3027 2871 2842 ...
..- attr(*, "class")= chr "smooth.spline.fit"
$ call : language smooth.spline(x = Phi00, y = Total, df = 8)
- attr(*, "class")= chr "smooth.spline"
I think the Model$fit$knot and Model$fit$coef vectors contain the full description of the fit. Note that the knots are 20, while x and y have 14 elements each: I always thought a smoothing spline would have as many knots as fitting points. However, since the first three and the last three knots are identical, 20-6 = 14 which makes sense. The problem is that I don't know how to use Model$fit$knot and Model$fit$coef to make predictions outside R. I tried to have a look at predict.smooth.spline, but surprisingly that's what I get
> library(splines)
> predict.smooth.spline
Error: object 'predict.smooth.spline' not found
EDIT: since apparently some users misunderstood the question, I know how to use predictin R, to get new values of my smoothing spline. The problem is that I want to make those predictions in an external code. Thus I wanted to have a look at the code for the function predict.smooth.spline, so that I could try to reproduce the algorithm outside R. Usually in R you can read the code of a function just by entering its name (without arguments and without parentheses) at the R prompt. But when I try to do that with predict.smooth.spline, I get the above error.
EDIT2: thanks to the great help from @r2evans, I found the source for the predict method of smooth.spline. I (think I) understand most of it:
> stats:::predict.smooth.spline.fit
function (object, x, deriv = 0, ...)
{
if (missing(x))
x <- seq.int(from = object$min, to = object$min + object$range,
length.out = length(object$coef) - 4L)
xs <- (x - object$min)/object$range
extrap.left <- xs < 0
extrap.right <- xs > 1
interp <- !(extrap <- extrap.left | extrap.right)
n <- sum(interp)
y <- xs
if (any(interp))
y[interp] <- .Fortran(C_bvalus, n = as.integer(n), knot = as.double(object$knot),
coef = as.double(object$coef), nk = as.integer(object$nk),
x = as.double(xs[interp]), s = double(n), order = as.integer(deriv))$s
if (any(extrap)) {
xrange <- c(object$min, object$min + object$range)
if (deriv == 0) {
end.object <- Recall(object, xrange)$y
end.slopes <- Recall(object, xrange, 1)$y * object$range
if (any(extrap.left))
y[extrap.left] <- end.object[1L] + end.slopes[1L] *
(xs[extrap.left] - 0)
if (any(extrap.right))
y[extrap.right] <- end.object[2L] + end.slopes[2L] *
(xs[extrap.right] - 1)
}
else if (deriv == 1) {
end.slopes <- Recall(object, xrange, 1)$y * object$range
y[extrap.left] <- end.slopes[1L]
y[extrap.right] <- end.slopes[2L]
}
else y[extrap] <- 0
}
if (deriv > 0)
y <- y/(object$range^deriv)
list(x = x, y = y)
}
However, I have two difficulties:
the
.Fortran()function calls a Fortran subroutinebvaluswhose source is quite simple. However,bvalusin turn callsbvaluewhich is much more complicated, and callsintervwhose source I cannot find. Bad news:bvalueis way too complicated for me to understand (I'm definitely not a Fortran expert). Good news: the external code which must reproducepredict.smooth.spline.fitis also a Fortran code. If worse comes to worst, I could just ask my coworker to include the source frombvalusandbvaluein his code. However, even in this admittedly not so nice scenario, I would still miss the source code forinterv(I hope it doesn't call something else!!!).I don't understand what it's being done here (note I'm only interested in the
deriv == 0case):
k
if (any(extrap)) {
xrange <- c(object$min, object$min + object$range)
if (deriv == 0) {
end.object <- Recall(object, xrange)$y
end.slopes <- Recall(object, xrange, 1)$y * object$range
if (any(extrap.left))
y[extrap.left] <- end.object[1L] + end.slopes[1L] *
(xs[extrap.left] - 0)
if (any(extrap.right))
y[extrap.right] <- end.object[2L] + end.slopes[2L] *
(xs[extrap.right] - 1)
}
Some sort of recursive code? Any help here?
回答1:
smooth.spline is not in the splines package, it's in stats. Additionally, it's not exported, so you have to use the triple-colon method to see it: stats:::predict.smooth.spline. It then points you to predict.smooth.spline.fit, which can be found in a similar manner. (Since it optionally uses .Fortran(), you may have to infer what is going on ... unless you dive into the source.)
回答2:
Exporting a smoothing spline as piecewise polynomials is one way to reconstruct the spline outside R. My package SplinesUtils: https://github.com/ZheyuanLi/SplinesUtils can do this. You can get it by
devtools::install_github("ZheyuanLi/SplinesUtils")
The function to be used here is SmoothSplinesAsPiecePoly. I am just copying in examples for this function in its documentation.
library(SplinesUtils)
## a toy dataset
set.seed(0)
x <- 1:100 + runif(100, -0.1, 0.1)
y <- poly(x, 9) %*% rnorm(9)
y <- y + rnorm(length(y), 0, 0.2 * sd(y))
## fit a smoothing spline
sm <- smooth.spline(x, y)
## coerce "smooth.spline" object to "PiecePoly" object
oo <- SmoothSplineAsPiecePoly(sm)
## print the "PiecePoly"
oo
#61 piecewise polynomials of degree 3 are constructed!
#Use 'summary' to export all of them.
#The first 6 are printed below.
#-0.626 - 0.17 * (x - 1.08) - 0 * (x - 1.08) ^ 2 - 0.0094 * (x - 1.08) ^ 3
#-0.768 - 0.148 * (x - 1.95) - 0.0246 * (x - 1.95) ^ 2 - 0.00569 * (x - 1.95) ^ 3
#-0.919 + 0.0259 * (x - 4.01) + 0.0598 * (x - 4.01) ^ 2 + 0.0086 * (x - 4.01) ^ 3
#-0.834 + 0.124 * (x - 5.08) + 0.0323 * (x - 5.08) ^ 2 + 0.00466 * (x - 5.08) ^ 3
#-0.494 + 0.197 * (x - 7.08) + 0.00433 * (x - 7.08) ^ 2 + 0.0027 * (x - 7.08) ^ 3
#-0.113 + 0.183 * (x - 9.03) + 0.0115 * (x - 9.03) ^ 2 + 0.00377 * (x - 9.03) ^ 3
The knots of the spline are
oo$knots
# [1] 1.079339 1.953102 4.014571 5.081642 7.079678 9.032160
# [7] 10.025823 11.941195 12.935311 14.976821 16.999540 18.043524
#[13] 19.976007 22.086941 22.942429 24.925111 25.953444 27.902678
#[19] 30.073938 30.968070 33.019913 34.937244 36.065475 38.058848
#[25] 38.921589 40.982255 43.029412 44.056587 46.005944 47.904666
#[31] 48.995446 51.038546 51.995524 53.987619 55.914136 56.919893
#[37] 59.003727 60.981366 62.082575 63.991813 64.966479 66.951603
#[43] 69.053262 69.916849 71.967815 73.969337 74.966755 77.078440
#[49] 78.072868 80.055464 81.986932 83.042503 84.965070 86.940538
#[55] 88.042224 89.949098 90.928661 92.911787 95.075254 96.055783
#[61] 97.991055 100.062174
The piecewise polynomial coefficients are
CoefMat <- oo$PiecePoly$coef
## for the first 5 pieces
CoefMat[, 1:5]
# [,1] [,2] [,3] [,4] [,5]
#[1,] -0.626225456 -0.768321912 -0.919380838 -0.83408278 -0.494257767
#[2,] -0.169805245 -0.148267472 0.025888868 0.12418698 0.197353171
#[3,] 0.000000000 0.024649465 0.059832206 0.03228737 0.004331680
#[4,] 0.009403574 0.005688943 -0.008604501 -0.00466386 -0.002702257
The package also has other functionality. See Identify all local extrema of a fitted smoothing spline via R function 'smooth.spline' for an example.
来源：https://stackoverflow.com/questions/35647572/evaluate-i-e-predict-a-smoothing-spline-outside-r