As can data.table (i.e. do 'having' in one statement) :
> DT = data.table(DF)
> DT[,list(n=length(NAME),mean(SCORE)),by="NAME"][n==3]
NAME n V2
[1,] James 3 64.00000
[2,] Tom 3 78.66667
>
but data.table isn't restricted to SQL functions (such as avg), any R
functions can be used, sometimes for their side effects (such as plotting)
rather than just returning data.
Further data.table has a thing called 'join inherited scoping'. Say we
knew the specific groups, we can go directly to them (without even looking
at the rest of the data in the table) in very short and convenient syntax,
which also happens to run quickly on large data sets (but can be useful just
for the syntax alone) :
> setkey(DT,NAME)
> DT[c("James","Tom"),mean(SCORE),mult="all"]
NAME V1
[1,] James 64.00000
[2,] Tom 78.66667
>
Notice there is no "group by" or even a "by" in the above. It inherits the
scope from the join because mult="all" means that "James" matches to
multiple rows, as does "Tom", creating two groups. It does it by binary
search to the beginning of each group, binary search to the end of the
group, and runs the R expression inside the scope of that group.
An example of join inherited scoping for the side effects only :
> pdf("out.pdf")
> DT[c("James","Tom"),plot(SCORE),mult="all"]
NULL data table
> dev.off()
# out.pdf now contains 2 plots
which you couldn't do in SQL because SQL has no plotting (or any of R's
other packages).
It aims to do this quickly. Where 'quickly' means 1) shorter code is
quicker to write, read, debug and maintain and also 2) quicker to compute,
and its 1 that often dominates 2.
Finally, consider the following two statements which are both equivalent :
> sqldf("select NAME, avg(SCORE) from DF group by NAME having count(*) = 3")
NAME avg(SCORE)
1 James 64.00000
2 Tom 78.66667
> DT[ J(DT[,length(NAME),by="NAME"][V1==3,NAME]), mean(SCORE), mult="all"]
NAME avg(SCORE)
1 James 64.00000
2 Tom 78.66667
Now ok I hear you groaning (!) that the 2nd looks (on first glance) ugly,
but bear with me ... in the SQL solution do you know for sure that
avg(SCORE) isn't computed wastefully for the all the groups that don't have
count(*)=3 ? It might well do the 'group by' first for all the groups, then
do the 'having' afterwards as a 'where' on the result. It might depend on
the particular SQL database being used (mySQL, sqllite, etc) or the
installation parameters, any indexes etc. Some investigation would be
required (taking time) if someone doesn't already know. In the data.table
however, the syntax explictly makes it clear than mean(SCORE) is only
computed for the particular groups. For certain, always. Maybe this
particular example is not a good one, but I'm trying to demonstrate an
overall syntax which is scalable (i.e. this syntax can do more complicated
things that SQL can't, or can't do well). Notice that the method earlier
on i.e. "DT[,list(n=length(NAME),mean(SCORE)),by="NAME"][n==3]" is simpler
but wasteful as it does compute mean(SCORE) for all the groups. But the
syntax explicity conveys what is being done, and the user has the choice.
"Gabor Grothendieck" <ggrothendi...@gmail.com> wrote in message
news:971536df1001051122l58389037p4e16288aedfde...@mail.gmail.com...
Here is the solution using sqldf which can do it in one statement:
> # read in data
> Lines <- "OBS NAME SCORE
+ 1 Tom 92
+ 2 Tom 88
+ 3 Tom 56
+ 4 James 85
+ 5 James 75
+ 6 James 32
+ 7 Dawn 56
+ 8 Dawn 91
+ 9 Clara 95
+ 10 Clara 84"
>
> DF <- read.table(textConnection(Lines), header = TRUE)
>
> # run
> library(sqldf)
> sqldf("select NAME, avg(SCORE) from DF group by NAME having count(*) = 3")
NAME avg(SCORE)
1 James 64.00000
2 Tom 78.66667
On Tue, Jan 5, 2010 at 2:03 PM, Gabor Grothendieck
<ggrothendi...@gmail.com> wrote:
> Have a look at this post and the rest of that thread:
>
> https://stat.ethz.ch/pipermail/r-help/2010-January/223420.html
>
> On Tue, Jan 5, 2010 at 1:29 PM, Geoffrey Smith <g...@asu.edu> wrote:
>> Hello, does anyone know how to take the mean for a subset of
>> observations?
>> For example, suppose my data looks like this:
>>
>> OBS NAME SCORE
>> 1 Tom 92
>> 2 Tom 88
>> 3 Tom 56
>> 4 James 85
>> 5 James 75
>> 6 James 32
>> 7 Dawn 56
>> 8 Dawn 91
>> 9 Clara 95
>> 10 Clara 84
>>
>> Is there a way to get the mean of the SCORE variable by NAME but only
>> when
>> the number of observations is equal to 3? In other words, is there a way
>> to
>> get the mean of the SCORE variable for Tom and James, but not for Dawn
>> and
>> Clara? Thank you.
>>
>> --
>> Geoffrey Smith
>> Visiting Assistant Professor
>> Department of Finance
>> W. P. Carey School of Business
>> Arizona State University
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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