# [R] Calculate average of many subsets based on columns in another dataframe

Peter Lomas peter.br.lomas at gmail.com
Wed Feb 10 23:18:12 CET 2016

```Thanks David, Bert,

>From what I'm reading on ?findInterval, It may not be workable because
of overlapping date ranges.  findInterval seems to take a series of
bin breakpoints as its argument. I'm currently exploring data.table

Just on David's point, the extension of this with "groups" would look
as below.  I just don't want to complicate it before I've solved the
simplest issue.

set.seed(345)
date.range <- seq(as.POSIXct("2015-01-01"),as.POSIXct("2015-06-01"),
by="DSTday")
observations <- data.frame(date=date.range, a=runif(152,1,100),
b=runif(152,1,100), c=runif(152,1,100) )
groups <- data.frame(start=sample(date.range[1:50], 20), end =
sample(date.range[51:152], 20), group=sample(letters[1:3], 20,
replace=TRUE), average = NA)

#Potential Solutions (too inefficient)
for(i in 1:NROW(groups)){
groups[i, "average"] <- mean(observations[observations\$date >=
groups[i, "start"] & observations\$date <=groups[i, "end"],
as.character(groups[i, "group"])])
}

Thanks again,
Peter

On Wed, Feb 10, 2016 at 2:26 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
> A strategy:
>
>  1. Convert your dates and intervals to numerics that give the days
> since a time origin. See as.POSIXlt (or ** ct for details and an
> example that does this). Should be fast...
>
> 2. Use the findInterval() function to get the interval into which each
> date falls. This **is** "vectorized" and should be fairly fast.
>
> 3. Use the ave() function using the intervals as your factor that
> splits your data column(s) for which you wish to compute statistics.
> The basic statistics functions like mean, sum, etc. **are**
> vectorized, so this should be fast.
>
> As David said, the *apply functions will probably not be much, if at
> all, faster than an explicit for() loop. Most of the time will be
> spent spent comparing the dates to the intervals to find in which each
> falls, and findInterval is a fast way to do this.
>
> ... I think.
>
> If you try this, let me know (perhaps privately) how/if it works.
>
> Cheers,
> Bert
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, Feb 10, 2016 at 1:08 PM, David Winsemius <dwinsemius at comcast.net> wrote:
>>
>>> On Feb 10, 2016, at 12:18 PM, Peter Lomas <peter.br.lomas at gmail.com> wrote:
>>>
>>> Hello, I have a dataframe with a date range, and another dataframe
>>> with observations by date.  For each date range, I'd like to average
>>> the values within that range from the other dataframe.  I've provided
>>> code below doing what I would like, but using a for loop is too
>>> inefficient for my actual case (takes about an hour).  So I'm looking
>>> for a way to vectorize.
>>>
>>>
>>> set.seed(345)
>>> date.range <- seq(as.POSIXct("2015-01-01"),as.POSIXct("2015-06-01"),
>>> by="DSTday")
>>> observations <- data.frame(date=date.range, values=runif(152,1,100) )
>>> groups <- data.frame(start=sample(date.range[1:50], 20), end =
>>> sample(date.range[51:152], 20), average = NA)
>>>
>>> #Potential Solution (too inefficient)
>>>
>>> for(i in 1:NROW(groups)){
>>> groups[i, "average"] <- mean(observations[observations\$date >=
>>> groups[i, "start"] & observations\$date <=groups[i, "end"], "values"])
>>> }
>>>
>> The 'average' column could be added to groups with this value:
>>
>> mapply( function(start,end){ mean(observations[['values']][
>>                      observations\$date >= start & observations\$date <=end])},
>>         groups\$start, groups\$end)
>>
>>  [1] 50.96831 49.42286 47.27240 49.07534 47.66570 49.30977 48.47503 47.74036
>>  [9] 46.02527 58.76492 48.86580 49.90655 45.79705 48.84071 39.53846 46.44601
>> [17] 47.06631 47.74199 49.16980 46.85131
>>
>> I don't really think this is fully "vectorized" in the usual R-meaning of the word. And I don't expect it to be any faster than the for-loop. Perhaps some of the range functions in the data.table package could accelerate your processing. If you don't get any volunteers in this list, you could repost the question on StackOverflow after a suitable pause that avoids accusations of cross-posting. SO has several skilled users of data.table functions.
>>
>>> As an extension to this, there will end up being multiple value
>>> columns, and each range will also identify which column to average.  I
>>> think if I can figure out the first problem I can try to extend it
>>> myself.
>>
>> Sorry, I didn't understand what was being described in that paragraph.
>>
>> --
>>
>> David Winsemius
>> Alameda, CA, USA
>>
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