What is the R equivalent of matlab's csaps()

问题

csaps() in matlab does a cubic spline according to a particular definition of the smoothing parameter p. Here is some matlab code and its result:

     % x variable
    age = 75:99  

    % y variable
    diffs = [-39   -2 -167  -21  -13   32  -37 -132 -143  -91  -93  -88  -62 -112  -95  -28  -90  -40  -27  -23  -28  -11   -8   -6    1]

    % 0.0005 is the parameter p, and the later specification of 
    % age are the desired x for prediction
    csaps(age,diffs,0.0005,age)
    % result (column headers removed):
     -63.4604  -64.0474  -64.6171  -65.1397  -65.6111  -66.0165  -66.3114  
     -66.4123  -66.2229  -65.6726  -64.7244  -63.3582  -61.5676  -59.3568  
     -56.7364  -53.7382  -50.4086  -46.7922  -42.9439  -38.9183  -34.7629  
     -30.5180  -26.2186  -21.8912  -17.5532

I'd like to get the same result in R. I've tried base::smooth.spline(), but the smoothing parameter spar is specified in a different way that I can't seem to relate to matlab's p (can you?). The closest result I've been able to get has been with the smooth.Pspline() function of the pspline package. Here is some code to get things rolling in R:

age <- 75:99
diffs <- c(-39L, -2L, -167L, -21L, -13L, 32L, -37L, -132L, -143L, -91L, 
-93L, -88L, -62L, -112L, -95L, -28L, -90L, -40L, -27L, -23L, 
-28L, -11L, -8L, -6L, 1L)
predict(pspline::smooth.Pspline(
                           x = age,
                           y = diffs, 
                           norder = 2, 
                           method = 1,
                           spar = 1 / 0.0005     # p given in MP and matlab as 0.0005
                         ),age)
# which gives something close, but not exactly the same:
 [1] -63.46487 -64.05103 -64.61978 -65.14158 -65.61214 -66.01662 -66.31079
 [8] -66.41092 -66.22081 -65.67009 -64.72153 -63.35514 -61.56447 -59.35372
[15] -56.73367 -53.73584 -50.40680 -46.79098 -42.94333 -38.91850 -34.76393
[22] -30.51985 -26.22131 -21.89474 -17.55757

The csaps() help page is here

smooth.spline() help can be found here (code not given because I think maybe the relation between spar and p is pretty hairy, so maybe not worth going down this path)

pspline::smooth.Pspline() help is here

This other person's quest from 2008 appears to have gone unanswered, making me feel like this guy.

R is chock full of spline doings, so if the saavy amongst ye can point me to the one that does the same thing as matlab's csaps() (or a trick along those lines) I'd be most appreciative.

[EDIT 19-8-2013] spar needs to be specified as (1-p)/p (rather than 1/p) and then results will agree to as far as numerical precision can take you. See answer below.

回答1:

My colleague found the answer: One converts matlab's p to pspline::smooth.Pspline()'s spar not as 1/p, but as (1-p)/p, and then results will agree out to whatever the degree of numerical precision is:

c(predict(pspline::smooth.Pspline(
                          x = age,
                          y = diffs, 
                          norder = 2, 
                          method = 1,
                          spar = (1-0.0005) / 0.0005     # p given in MP and matlab as 
                         ),age))
 [1] -63.46035 -64.04741 -64.61705 -65.13972 -65.61114 -66.01646 -66.31144
 [8] -66.41232 -66.22285 -65.67263 -64.72443 -63.35823 -61.56761 -59.35675
[15] -56.73643 -53.73821 -50.40864 -46.79221 -42.94387 -38.91828 -34.76291
[22] -30.51801 -26.21863 -21.89122 -17.55320

回答2:

Here is what I found in p. 16 of MATLAB/R Reference by David Hiebeler. [I don't use Matlab,however].

Fit natural cubic spline(S′′(x) = 0 at both endpoints) to points (xi, yi)whose coordinates are in vectors x and y; evaluate at points whose x coordinates are in vector xx, storing corresponding y’s in yy

Matlab:

pp=csape(x,y,’variational’);
yy=ppval(pp,xx) but note that
csape is in Matlab’s Spline
Toolbox

tmp=spline(x,y,method=’natural’,
xout=xx); yy=tmp$y

来源：https://stackoverflow.com/questions/18283718/what-is-the-r-equivalent-of-matlabs-csaps

标签

matlab

splines