Hi Steve, Fixed now. Thanks, Matthew On Thu, 2011-01-06 at 09:48 -0500, Steve Lianoglou wrote: > 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 > > > > >
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