Hi Matthew, On Thu, Jan 6, 2011 at 5:26 AM, Matthew Dowle <[email protected]> wrote: > > How about writing it this way. This way should invoke the incremental binary > search for efficiency too, rather than a repeated binary search for each by. > >> dt2[dt1[, .SD[1], by=list(name, place)],mult="all"] > name place length > [1,] a home 10 > [2,] a home 100 > >> dt2[dt3[, .SD[1], by=list(name, place)],mult="all"] > name place length > [1,] a home 10 > [2,] a home 100 > [3,] b work 20
While doing it this way works for this trivial case, I'm actually doing a fair bit of book keeping/computation in my j expression and returning a list of elements that you can't really get from simple joins and stuff. > But doing it your way should work too, so I'll add as a bug. Thanks. I'm currently working around this by adding a dummy row into my dt1 data.table, which I then remove after the `dogroups` stuff finishes. -steve > Another way to get the first row of each group is a fast self-join via i. > There was a thread on that some time ago when i was changed to be evaluated > within the frame of DT too. Something like : > > dt3[J(unique(name)), mult="first"] # first of each group > > HTH > Matthew > > > "Steve Lianoglou" <[email protected]> wrote in message > news:[email protected]... > Hi, > > I'm calculating some statistics over a large data.table via `dt[, > {somestuff}, by=list(key1,key2)]`. > Sometimes my dt data.table ends up only having one row, which results > in the following error: > > "Didn't allocate enough rows for result of first group." > > Here is a toy/trivial example. > > R> dt1 <- data.table(name='a', place='home', count=1, key='name,place') > R> dt2 <- data.table(name=c('a', 'a', 'a', 'b'), > place=c('home', 'work', 'home', 'work'), > length=c(10,20,100, 20), key='name,place') > > R> dt1[, list(length=dt2[J(.SD$name[1], .SD$place[1]), > mult='all']$length), by=list(name, place)] > Error in `[.data.table`(dt1, , list(length = dt2[J(.SD$name[1], > .SD$place[1]), : > Didn't allocate enough rows for result of first group. > > When my data.table has > 1 row, it works: > > R> dt3 <- data.table(name=c('a', 'b'), place=c('home', 'work'), > count=1:2, key='name,place') > R> dt3[, list(length=dt2[J(.SD$name[1], .SD$place[1]), > mult='all']$length), by=list(name, place)] > name place length > [1,] a home 10 > [2,] a home 100 > [3,] b work 20 > > I believe if the result of my {somestuff} expression only ever > returned one row, this bug wouldn't happen, but .... it doesn't just > do that :-) > > It looks like the fix is where the `byretn` value is calculated in the > `[.data.table` but that code is a somehow inscrutable at first glance > ... can anyone propose a quick fix? > > Thanks, > -steve > > -- > Steve Lianoglou > Graduate Student: Computational Systems Biology > | Memorial Sloan-Kettering Cancer Center > | Weill Medical College of Cornell University > Contact Info: http://cbio.mskcc.org/~lianos/contact > > > > _______________________________________________ > datatable-help mailing list > [email protected] > https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help > -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact _______________________________________________ datatable-help mailing list [email protected] https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help
