What is the equivalent of R matlab-csaps ()

Question

csaps() in matlab makes a cubic spline according to the specific 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 would like to get the same result in R. I tried base::smooth.spline() , but the spar smoothing option is specified differently, which I don't seem to be matlab p (can you?). The closest result I could get was with the smooth.Pspline() function of the pspline package. Here is some code to make things roll 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 

Csaps () help page here

smooth.spline() help can be found here (the code is not listed because I think the connection between spar and p quite hairy, so maybe you shouldn't go this way)

pspline::smooth.Pspline() help here

This quest from another person since 2008 seems to have remained unanswered, making me feel like this guy .

R is crowded with splines, so if the saavy among you can point me to one that does the same thing as matlab csaps() (or the trick along these lines), I would be very grateful.

[EDIT 19-8-2013] spar must be specified as (1-p)/p (not 1/p ), and then the results will be accurate to numerical precision. See answer below.

Answer 1

My colleague found the answer: one converts matlab p to pspline::smooth.Pspline() spar not as 1/p , but as (1-p)/p , and then the results will agree on what degree of numerical accuracy:

 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 

Answer 2

Here is what I found in p. 16 of MATLAB / R link by David Hebeler. [However, I do not use Matlab].

Set the natural cubic spline (S '' (x) = 0 at both ends) to the points (xi, yi) whose coordinates are in the vectors x and y; evaluate at points whose coordinates are in the xx vector, keeping the corresponding ys in yy

Matlab:

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

R

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