Thank you! That solution worked! I thought I'd tried something similar to
that, but obviously I didn't. Here's a self-contained example, for
posterity and completeness:
z.df <- data.frame(times=c(Sys.time(), Sys.time() + 3600))
as.POSIXlt(z.df[,1])$hour
And this gives me what I want.
Thank you again!
Matt
On Fri, Sep 24, 2010 at 5:50 PM, Phil Spector <spec...@stat.berkeley.edu>wrote:
> Matthew -
> It's a bit simpler than you think:
>
> as.POSIXlt(pk)$hour
>
> should return what you want. (If not, please provide a reproducible
> example.)
> - Phil Spector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
>
>
> On Fri, 24 Sep 2010, Matthew Pettis wrote:
>
> Hi,
>>
>> I have a list/data.frame 'pk' of POSIXct dates, and I'd like to extract
>> the
>> hour for each row. I know that if I have an individual POSIXct object, I
>> can extract the hour by converting to a new object with:
>>
>> new.lt <- as.POSIXlt(<single POSIXct object>)
>> new.lt$hour
>>
>> But I can't figure out how to apply this for a list of such dates in a
>> vectorized form. I can write a loop, I guess and implement this, but I
>> think I'm missing a way that takes advantage of vectorization. Here is my
>> loop to just print the hour extracts:
>>
>> for (ct in pk) {
>> lt <- as.POSIXlt(ct, origin="1970-01-01")
>> print(lt$hour)
>> }
>>
>> So is there a shorter vectorization idiom that lets me do this? I can't
>> figure out how to use 'lapply' to apply the '$' operator...
>>
>> Thanks,
>> Matt
>>
>> [[alternative HTML version deleted]]
>>
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>>
>>
--
"Pardon him. Theodotus: he is a barbarian, and thinks that the customs of
his tribe and island are the laws of nature." -- George Bernard Shaw,
"Caesar and Cleopatra"
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