Thanks for your example Peter.
It does seem like one means of getting the job done and if I'm unable to
figure out a means with zoo::rollapply I may follow your example.
Hopefully someone else can chime in.
Thanks,
Nate
-----------------------------------------------
Nathan A. Miller
Romberg Tiburon Center
San Francisco State University
3150 Paradise Dr
Tiburon, CA 94920
www.nate-miller.org
On Wed, Mar 13, 2013 at 4:34 PM, Peter Ehlers <ehl...@ucalgary.ca> wrote:
> On 2013-03-12 17:10, Nathan Miller wrote:
>
>> Hello,
>>
>> I have a challenge!
>>
>> I have a large dataset with three columns, "date","temp", "location".
>> "date" is in the format %m/%d/%y %H:%M, with a "temp" recorded every 10
>> minutes. These temperatures of surface temperatures and so fluctuate
>> during
>> the day, heating up and then cooling down, so the data is a series of
>> peaks
>> and troughs. I would like to develop a function that would go through a
>> dataset consisting of many sequential dates and determine for each day the
>> maximum hourly slope of temp~date for each site (the fastest hourly rate
>> of
>> heating). The output would be the date, the maximum hourly slope for that
>> date, and the location. It would also be great if I could extract when
>> during the day the maximum hourly slope occurred.
>>
>> I have been playing around with using the package lubridate to identify
>> each hour of the day using something like this to create a separate column
>> grouping the data into hours
>>
>> library(lubridate)
>> data$date2 <- floor_date(data$date, "hour")
>>
>> I was then imagining something like this though this code doesn't work as
>> written.
>>
>> ddply(data, .(location, date2), function(d)
>> max(rollapply(slope(d$temp~d$**date, data=d)))
>>
>> Essentially what I'm imagining is calculating the slope (though I'd have
>> to
>> write a quick slope function) of the date/temp relationship, use rollapply
>> to apply this function across the dataset, and determine the maximum
>> slope,
>> grouped by location and hour (using date2). Hmm... and per day!
>>
>> This seems complicated. Can others think of a simpler, more elegant means
>> of extracting this type of data? I struggled to put together a working
>> example with a set of data, but if this doesn't make sense let me know and
>> I'll see what I can do.
>>
>>
>> Thanks,
>> Nate
>>
>
> First, let's ignore location; if you can do it for one location,
> you can surely do it for others.
>
> Second, let's ignore date; if you can do it for one date, you
> can surely do it for others.
>
> That leaves us with the question of what you want to do for one
> given date. If you want the maximum slope for any 60-minute interval
> on that date (which I take your question to mean), then rollapply
> should do the job. But I'm not very familiar with zoo, so here's a
> crude approach:
>
> d <- data.frame(time = 1:72, temp = rnorm(72))
> slope <- rep(NA, 72)
> for(i in 6:72) {
> slope[i] <- coef(lm(temp ~ time, data = d, subset = (i-5):i))[2]
> }
> maxslope <- max(slope, na.rm = TRUE)
> idx <- which.max(slope)
>
> Obviously, this can be extended to cover more than a 24-hour period.
>
> Now, let's wait for Gabor to show us the trivial way with zoo::rollapply.
>
> Peter Ehlers
>
>
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