The colMeans comes closest,
for a single series the assume you have 100 years of monthly data.
The mean you want to scale by is the mean for a restricted period in the
center
of the series.. say 1950-1960
for this period you have the average jan (1950-1960) average feb, ect.
your final series would be
jan 1900 - average jan(1950-60)
feb 1990 - average feb
....
jan 2000 - average jan(1950-60)
Which gives you a scaling that is not relative to the mean of the whole, but
relative to a base period which is selctable.
BTW switching to zoo has greatly simplified the code.
On Wed, Aug 11, 2010 at 11:21 AM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> On Wed, Aug 11, 2010 at 12:22 PM, steven mosher <mosherste...@gmail.com>
> wrote:
> > Given a long zoo matrix, the goal is to "sweep" out a statistic from the
> > entire length of the
> > sequences.
> >
> >
>
> longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+")))
> > cnames<-c(12345,23456,34567,45678,56789,67890)
> > colnames(longzoomatrix)<-cnames
> > longzoomatrix[1:24,]
> > 12345 23456 34567 45678 56789
> > 67890
> > Jan 1900 -0.17123165 1.02087086 0.79514870 -0.54519494 -0.13025459
> > -0.009980402
> > Feb 1900 1.21729926 -0.74541038 -0.08138406 -2.01180775 0.19256998
> > 0.551965871
> > Mar 1900 1.13222481 -1.25315703 0.01013473 0.08366155 -0.84246010
> > -1.405959298
> > Apr 1900 -0.02352559 -1.25001473 -1.53570550 -0.17945324 0.33368133
> > 2.045125104
> > May 1900 2.08204920 1.28091067 -0.80888146 0.31796730 0.83248551
> > 1.439049603
> > Jun 1900 0.62209570 -0.66189249 -0.57923119 -0.04346112 -2.71353384
> > -0.346826902
> > Jul 1900 -1.39758918 -0.54525469 -0.05230070 -0.36725079 1.28281798
> > 1.391174712
> > Aug 1900 0.12594069 0.09303970 0.69916411 -1.01902352 -0.82720898
> > -0.208113626
> > Sep 1900 -0.34310543 0.41718435 0.79455765 1.13234707 0.14652667
> > -0.551426097
> > Oct 1900 1.70634123 -1.20073104 -1.08771551 -0.01715296 0.24931996
> > -0.753481196
> > Nov 1900 0.15224070 -0.05108370 -0.97410069 0.51130170 0.13880814
> > -2.160811186
> > Dec 1900 0.34726817 0.61830719 0.84429979 -0.26253635 0.95243068
> > -0.533562966
> > Jan 1901 0.28647563 -0.40650198 -1.19640622 0.70267162 0.18867804
> > 0.098855045
> > Feb 1901 1.27269836 0.31797472 -1.13038040 1.33654480 0.08885501
> > -0.134690872
> > Mar 1901 -1.36934330 -0.17244539 0.81705554 -0.09113888 0.90241413
> > 0.473939164
> > Apr 1901 -0.89768498 0.82497595 0.15684387 2.25294476 -1.72886103
> > -0.104769411
> > May 1901 -0.27898445 -1.24348285 1.36203180 0.02422083 -1.33745980
> > 1.098856752
> > Jun 1901 -0.67968801 0.42082064 0.47056133 -0.12981223 0.19445803
> > -0.284638114
> > Jul 1901 0.03791761 -0.22118130 1.96044737 -1.18280989 0.90075205
> > 0.055720535
> > Aug 1901 1.12904079 0.57177055 0.64300572 -0.16284983 0.07951656
> > -0.159396821
> > Sep 1901 -1.43513934 0.03036697 1.09039400 0.99201776 0.98744827
> > -0.057234838
> > Oct 1901 0.73828382 0.53967835 2.16608282 -0.82929778 -1.99666687
> > 0.352778450
> > Nov 1901 0.06561583 -1.20126258 0.67427027 0.15493106 0.08867697
> > 1.223073528
> > Dec 1901 -1.23347027 -1.09699304 0.59398031 -0.22269292 -0.21569543
> > 1.389667825
> >
> > The statistic to be swept out is itself a zoo series with matching column
> > names.
> > There are twelve valies for each column representing an monthly average
> for
> > that
> > series.
> >
> > The average is to be subtracted
> >
> > sweepzoo<-zoo(matrix(rnorm(72),ncol=6), frequency=12)
> > colnames(sweepzoo)<-cnames
> > sweepzoo
> > 12345 23456 34567 45678 56789 67890
> > 1(1) -2.5569706 -0.4375741 -0.1803866 -0.6303760 -0.08995198 2.7293244
> > 2(1) 1.4154202 0.2559212 0.2104513 0.7439446 0.84897905 -0.4144865
> > 3(1) -1.3709275 1.0472759 1.5975148 0.3190503 1.10430959 -1.8285194
> > 4(1) -1.1436430 2.2071763 -0.2637954 -0.4915366 -0.03925020 1.3311624
> > 5(1) -0.8003656 1.6421541 -1.4603128 0.4493069 0.28194066 -0.4728086
> > 6(1) 0.9236015 0.3780122 -1.3848196 0.4263684 0.99584590 -1.4536475
> > 7(1) 0.8810281 0.0381152 0.3810457 -0.6884233 -0.11018089 0.4221188
> > 8(1) 0.3819421 -0.8431364 1.9876901 0.7072257 0.45524929 2.7013515
> > 9(1) -1.1247988 1.3083178 -0.3438442 0.3300832 0.67013503 1.2912443
> > 10(1) -0.3643043 1.0756782 -1.2026318 0.4477054 0.54486700 -0.3369889
> > 11(1) 0.8294049 1.8170357 0.5691249 1.9213791 -0.29295754 -0.2617228
> > 12(1) -1.0085265 -0.7556545 -1.4033321 -0.4646647 -0.14984913 -0.4848657
> >
> > A brute force way to do this is to repeat the 12 values for each column
> so
> > that
> > the number of rows in the "sweepzoo" is equal to the nmber of rows on
> the
> > long zoo, object and then just subtract them. longzoomatrix-sweepzoo
> >
> > As a function sweep() wont work because it expects a vector whose
> > dimensions
> > matches the dimension of the MARGIN.
> >
> > Is there a elegant way to do this short of creating a "sweep zoo" that
> > matches
> > the row dimension of longzoo? ( would be a nice addition to sweep)
> >
>
> If the question is how to demean each column then try this:
>
> > z <- zoo(cbind(a = 1:3, b = 4:6))
> > scale(z, scale = FALSE)
> a b
> 1 -1 -1
> 2 0 0
> 3 1 1
> attr(,"scaled:center")
> a b
> 2 5
>
> > # or
> > sweep(z, 2, colMeans(z))
> a b
> 1 -1 -1
> 2 0 0
> 3 1 1
>
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