[R] Memory usage in prcomp

Martin Maechler
2016-03-22T17:00:27Z

>>>>> Roy Mendelssohn <- NOAA Federal <roy.mendelssohn at
noaa.gov>>
>>>>>     on Tue, 22 Mar 2016 07:42:10 -0700 writes:

    > Hi All:
    > I am running prcomp on a very large array, roughly [500000, 3650].  The
array itself is 16GB.  I am running on a Unix machine and am running ?top? at
the same time and am quite surprised to see that the application memory usage is
76GB.  I have the ?tol? set very high  (.8) so that it should only pull out a
few components.  I am surprised at this memory usage because prcomp uses the SVD
if I am not mistaken, and when I take guesses at the size of the SVD matrices
they shouldn?t be that large.   While I can fit this  in, for a variety of
reasons I would like to reduce the memory footprint.  She questions:

    > 1.  I am running with ?center=FALSE? and ?scale=TRUE?.  Would I save
memory if I scaled the data first myself, saved the result, cleared out the
workspace, read the scaled data back in and did the prcomp call?  Basically are
the intermediate calculations for scaling kept in memory after use.

    > 2. I don?t know how prcomp memory usage compares to a direct call to
?svn? which allows me to explicitly set how many  singular vectors to compute (I
only need like the first five at most).  prcomp is convenient because it does a
lot of the other work for me

For your example, where p := ncol(x)  is 3650  but you only want
the first 5 PCs, it would be *considerably* more efficient to
use svd(..., nv = 5) directly.

So I would take  stats:::prcomp.default  and modify it
correspondingly.

This seems such a useful idea in general that I consider
updating the function in R with a new optional 'rank.'  argument which
you'd set to 5 in your case.

Scrutinizing R's underlying svd() code however, I know see that
there are typicall still two other [n x p] matrices created (on
in R's La.svd(), one in C code) ... which I think should be
unnecessary in this case... but that would really be another
topic (for R-devel , not R-help).

Martin