评估（即预测）R 外的平滑样条

Question

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 predict in 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:

1. the .Fortran() function calls a Fortran subroutine bvalus whose source is quite simple. However, bvalus in turn calls bvalue which is much more complicated , and calls interv whose source I cannot find. Bad news: bvalue is way too complicated for me to understand (I'm definitely not a Fortran expert). Good news: the external code which must reproduce predict.smooth.spline.fit is also a Fortran code. If worse comes to worst, I could just ask my coworker to include the source from bvalus and bvalue in his code. However, even in this admittedly not so nice scenario, I would still miss the source code for interv (I hope it doesn't call something else!!!).

2. I don't understand what it's being done here (note I'm only interested in the deriv == 0 case):

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?

Answer 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 .)

Answer 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.