I want to perform cubic spline interpolation :
x <- c(1,1.5, 2, 4,5,6,7,8,9,10,12,15,20, 25,30)
y <- c(-0.402, -0.303, -0.198, 0.211,0.4133,0.606,0.7835,0.9404,1.0733,1.189,1.383,1.594,1.812,1.893,1.922)
smooth.spline (x,y)
you can notice that in the x vector , the x-value of 3 needs to be interpolated. How can I view interpolated values ( or insert an x values and get the interpolated y value), and view the spline functions?
You can use predict to apply the fitted spline function:
> x <- c(1,1.5, 2, 4,5,6,7,8,9,10,12,15,20, 25,30)
> y <- c(-0.402, -0.303, -0.198, 0.211,0.4133,0.606,0.7835,0.9404,1.0733,1.189,1.383,1.594,1.812,1.893,1.922)
> predict(smooth.spline (x,y), 3)
$x
[1] 3
$y
[1] 0.006938673
Some more information on the spline functions you can get using str function:
> str(smooth.spline(x,y))
List of 19
$ x : num [1:15] 1 1.5 2 4 5 6 7 8 9 10 ...
$ y : num [1:15] -0.403 -0.301 -0.199 0.212 0.413 ...
$ w : num [1:15] 1 1 1 1 1 1 1 1 1 1 ...
$ yin : num [1:15] -0.402 -0.303 -0.198 0.211 0.413 ...
$ tol : num 9e-06
$ data :List of 3
..$ x: num [1:15] 1 1.5 2 4 5 6 7 8 9 10 ...
..$ y: num [1:15] -0.402 -0.303 -0.198 0.211 0.413 ...
..$ w: num 1
$ no.weights: logi TRUE
$ lev : num [1:15] 0.766 0.392 0.592 0.708 0.572 ...
$ cv.crit : num 6.3e-06
$ pen.crit : num 7.17e-06
$ crit : num 6.3e-06
$ df : num 10.9
$ spar : num 0.336
$ ratio : num 5.29e-06
$ lambda : num 5.51e-06
$ iparms : Named int [1:5] 1 0 13 0 NA
..- attr(*, "names")= chr [1:5] "icrit" "ispar" "iter" "" ...
$ auxM : NULL
$ fit :List of 5
..$ knot : num [1:21] 0 0 0 0 0.0172 ...
..$ nk : int 17
..$ min : num 1
..$ range: num 29
..$ coef : num [1:17] -0.403 -0.369 -0.301 -0.096 0.145 ...
..- attr(*, "class")= chr "smooth.spline.fit"
$ call : language smooth.spline(x = x, y = y)
- attr(*, "class")= chr "smooth.spline"