Or put it differently: How can I keep my ts index? Most of the time I use a time series in a calculation it's not a ts object anymore. What strategy should I follow when writing functions to return a ts object and keep the index information?
E.g.:
#standard Hodrick Prescott Filter
hpfilter <- function(x,lambda=1600){
eye <- diag(length(x))
result <- solve(eye+lambda*crossprod(diff(eye,lag=1,d=2)),x)
### this is what I am talking about :)
### intuitively i´d maybe add something like this
re开发者_开发技巧sult <- ts(result,start=start(x),end=end(x),frequency=frequency(x))
###
return(result)
}
However, I feel that this clumsy and cumbersome. Is there a more elegant way to do it (maybe I should into classes..)?
With time series, subsetting and quite some other functions cause conversion to a matrix or a vector. You don't have to rebuild the time series, you can just transfer the attributes of the original ts
to the result.
hpfilter <- function(x,lambda=1600){
eye <- diag(length(x))
result <-
solve(eye+lambda*crossprod(diff(eye,lag=1,d=2)),x)
attributes(result) <- attributes(x)
return(result)
}
You can use subsetting also to change (but not to append) the values in the time series :
hpfilter <- function(x,lambda=1600){
eye <- diag(length(x))
x[] <-
solve(eye+lambda*crossprod(diff(eye,lag=1,d=2)),x)
return(x)
}
With the coredata
function in the zoo
package you can access the data portion of a ts
or zoo
object. I would change you code to
library(zoo)
#standard Hodrick Prescott Filter
hpfilter <- function(x,lambda=1600){
eye <- diag(length(x))
coredata(x) <- solve(eye + lambda * crossprod(diff(eye, lag=1, d=2)), coredata(x))
return(x)
}
and run
foo <- ts(rnorm(10), frequency = 4, start = c(1959, 2))
bar <- hpfilter(foo)
which yields
> foo
Qtr1 Qtr2 Qtr3 Qtr4
1959 0.8939882 -1.8442215 -0.8959187
1960 -0.2658590 0.5855087 -0.7167737 -1.9318533
1961 0.3489802 -0.6300171 -0.6523006
> bar
Qtr1 Qtr2 Qtr3 Qtr4
1959 -0.3589312 -0.3939791 -0.4282439
1960 -0.4618490 -0.4952099 -0.5286198 -0.5616964
1961 -0.5941750 -0.6266472 -0.6591151
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