See ?apply
If X is not an array but has a dimension attribute, apply attempts to coerce
it to an array via as.matrix if it is two-dimensional (e.g., data
frames) or via as.array.
So for example, try this noting that time is 1, 2, 3, ...
apply(monthly, 2, time)
Try
lm(monthly ~ time(monthly))
On Wed, Apr 30, 2008 at 9:00 AM, <[EMAIL PROTECTED]> wrote:
> Full_Name: Stephen McIntyre
> Version: 2.7
> OS: Windows XP
> Submission from: (NULL) (99.231.2.44)
>
>
> When I use the apply function to calculate a trend for a matrix of monthly
> time
> series, it yields a different answer than when the trend is calculated one at
> a
> time (by a factor of 12) rather than the identical answer as it should. Here's
> an example:
>
> download.file("http://www.climateaudit.org/data/models/monthly.tab","temp.dat",mode="wb";)
> load("temp.dat")
> trend= function(x) lm(x~c(time(x)))$coef[2]
> b= apply(monthly,2,trend)
> a= c(trend(monthly[,1]),trend(monthly[,2]),trend(monthly[,3]))
> a/b #12 12 12
>
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