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)

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