R help - Dec 2010 - lm() on a matrix of zoo series

I have a matrix of zoo series. each series is in a column.
 x <- as.yearmon(2000 + seq(0, 23)/12)
# 24 months of data, lets make 20 sets of random data
 testData <- matrix(rnorm(480),ncol=20)
# make a zoo object and columns will hold the 20 series
TestZoo  <- zoo(testData,order.by=x)
# now run lm for just one series.
 m <- lm(TestZoo[,1]~time(TestZoo))$coeff[2]
 m
time(TestZoo)
    0.3443124
 m2 <- lm(TestZoo[,2]~time(TestZoo))$coeff[2]
 m2
time(TestZoo)
   -0.1192866

I've been struggling trying to use apply ( or something equally suitable) to
get a vector of "m" for this entire matrix
without resorting to a loop.

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On Tue, Dec 21, 2010 at 3:02 PM, steven mosher <moshersteven at gmail.com>
wrote:
> I have a matrix of zoo series. each series is in a column.
> ?x <- as.yearmon(2000 + seq(0, 23)/12)
> # 24 months of data, lets make 20 sets of random data
> ?testData <- matrix(rnorm(480),ncol=20)
> # make a zoo object and columns will hold the 20 series
> TestZoo ?<- zoo(testData,order.by=x)
> # now run lm for just one series.
> ?m <- lm(TestZoo[,1]~time(TestZoo))$coeff[2]
> ?m
> time(TestZoo)
> ? ?0.3443124
> ?m2 <- lm(TestZoo[,2]~time(TestZoo))$coeff[2]
> ?m2
> time(TestZoo)
> ? -0.1192866
>
> I've been struggling trying to use apply ( or something equally
suitable) to
> get a vector of "m" for this entire matrix
> without resorting to a loop.
>
Try this:

   lm(TestZoo ~ time(TestZoo))


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