> peter dalgaard
> on Fri, 25 Mar 2016 09:41:00 +0100 writes:
> As I see it, the display showing the first p << n PCs
> adding up to 100% of the variance is plainly wrong. I
> suspect it comes about via a mental short-circuit: If we
> try to control p using a tolerance, then that amounts to
> saying that the remaining PCs are effectively
> zero-variance, but that is (usually) not the intention at
> all.
> The common case is that the remainder terms have a roughly
> _constant_, small-ish variance and are interpreted as
> noise. Of course the magnitude of the noise is important
> information.
Thank you, Peter, Kasper, Steve.
@Kasper, I've known about *approximate* first few PC methods.
(Are there also exact ones which are more efficient than those
based on LAPACK'S DGESDD ?)
... and so indeed, prcomp() will not be the method of choice for
really large problems. Still, of course, we should try to "do our best".
Given your sentiments, notably Peter's, I now envisage to
do the non-backcompatible change to *summary.prcomp()* and
compute "importances" based on all proportions up to 'p' (= 100%).
What I think would be nice is for the print.summary.prcomp()
method to only show the first 'k' (my notation), i.e., those
which were chosen by 'tol' and/or 'rank.'.
Martin
>> On 25 Mar 2016, at 00:02 , Steve Bronder
>> wrote:
>>
>> I agree with Kasper, this is a 'big' issue. Does your
>> method of taking only n PCs reduce the load on memory?
>>
>> The new addition to the summary looks like a good idea,
>> but Proportion of Variance as you describe it may be
>> confusing to new users. Am I correct in saying Proportion
>> of variance describes the amount of variance with respect
>> to the number of components the user chooses to show? So
>> if I only choose one I will explain 100% of the variance?
>> I think showing 'Total Proportion of Variance' is
>> important if that is the case.
>>
>>
>> Regards,
>>
>> Steve Bronder Website: stevebronder.com Phone:
>> 412-719-1282 Email: sbron...@stevebronder.com
>>
>>
>> On Thu, Mar 24, 2016 at 2:58 PM, Kasper Daniel Hansen <
>> kasperdanielhan...@gmail.com> wrote:
>>
>>> Martin, I fully agree. This becomes an issue when you
>>> have big matrices.
>>>
>>> (Note that there are awesome methods for actually only
>>> computing a small number of PCs (unlike your code which
>>> uses svn which gets all of them); these are available in
>>> various CRAN packages).
>>>
>>> Best, Kasper
>>>
>>> On Thu, Mar 24, 2016 at 1:09 PM, Martin Maechler <
>>> maech...@stat.math.ethz.ch
wrote:
>>>
Following from the R-help thread of March 22 on "Memory
usage in prcomp",
I've started looking into adding an optional 'rank.'
argument to prcomp allowing to more efficiently get
only a few PCs instead of the full p PCs, say when p =
1000 and you know you only want 5 PCs.
(https://stat.ethz.ch/pipermail/r-help/2016-March/437228.html
As it was mentioned, we already have an optional 'tol'
argument which allows *not* to choose all PCs.
When I do that, say
C <- chol(S <- toeplitz(.9 ^ (0:31))) # Cov.matrix and
its root all.equal(S, crossprod(C)) set.seed(17) X <-
matrix(rnorm(32000), 1000, 32) Z <- X %*% C ## ==>
cov(Z) ~= C'C = S all.equal(cov(Z), S, tol = 0.08) pZ
<- prcomp(Z, tol = 0.1) summary(pZ) # only ~14 PCs (out
of 32)
I get for the last line, the summary.prcomp(.) call :
> summary(pZ) # only ~14 PCs (out of 32)
Importance of components: PC1 PC2 PC3 PC4 PC5 PC6 PC7
PC8 Standard deviation 3.6415 2.7178 1.8447 1.3943
1.10207 0.90922
>>> 0.76951
0.67490 Proportion of Variance 0.4352 0.2424 0.1117
0.0638 0.03986 0.02713
>>> 0.01943
0.01495 Cumulative Proportion 0.4352 0.6775 0.7892
0.8530 0.89288 0.92001
>>> 0.93944
0.95439 PC9 PC10 PC11 PC12 PC13 PC14 Standard deviation
0.60833 0.51638 0.49048 0.44452 0.40326 0.3904
Proportion of Variance 0.01214 0.00875 0.00789 0.00648
0.00534 0.0050 Cumulative Proportion 0.96653 0.97528
0.98318 0.98966 0.99500 1.
>
which computes the *proportions* as if there were only
14 PCs in total (but there were 32 originally).
I would think that the summary should or could in
addition show the usual "proportion of variance
explained" like result which does involve all 32
variances or std.dev.s ... which are returned from the
svd() anyway, even in the case when I use my new
'rank.' argument which only ret
