[datatable-help] Bug in subsetting/iterating over a data.table with1 row(?)

[Matthew Dowle]

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 <mdowle at mdowle.plus.com> 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" <mailinglist.honeypot at gmail.com> wrote in message
> > news:AANLkTinEhEXFGKWNze8goCn-67fuJmcE3Zu8YOZAxPYk at mail.gmail.com...
> > 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
> >
> >
> >
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> >
> 
> 
>