On Sunday 07 October 2007, Gabor Grothendieck wrote:
> zoo's merge can do a multiway intersection. We turn each component
> of aa into the times of a dataless zoo object (assuming the elements of
> each component are unique) and merge them together using all = FALSE
> which will only leave those points at times in all components. Extracting
> the time and stripping off the names gives the result.
>
> library(zoo)
> as.vector(time(do.call(merge, c(lapply(aa, function(x) zoo(,x)), all =
> FALSE))))
Hi Gabor, it's always good to see clever solutions. Never thought about using
time series to perform intersections :)
I can't reproduce your solution though:
Error in rval[[1]] : subscript out of bounds
The elements of each component are indeed unique; the argument all = FALSE
intrigues me: which function should use it, zoo() or c() ?
The result of lapply() is a list; when performing c() over that list, the
argument all = FALSE does nothing but to add another branch... hmm, I should
really use more time series...
Adrian
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