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

[Steve Lianoglou]

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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>



-- 
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