R devel - Apr 2008 - apply and monthly time series (PR#11352)

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

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,  <stephen.mcintyre at utoronto.ca>
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
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>