GSee and List, With commit 1336, this now works:
require(data.table) L = list(as.data.table(BOD), as.data.table(BOD)) lapply(L, "[", Time == 3L) Please write back if you come across any issues. Arun From: G See <[email protected]> Reply: G See <[email protected]>> Date: November 11, 2013 at 3:40:46 PM To: Gabor Grothendieck <[email protected]>> Cc: [email protected] <[email protected]>> Subject: Re: [datatable-help] lapply without anonymous function heh, after all my efforts to avoid subset(), it can be useful after all. :) Bug report filed, per Eduard's suggestion. On Mon, Nov 11, 2013 at 8:06 AM, Gabor Grothendieck <[email protected]> wrote: > On Sun, Nov 10, 2013 at 2:39 PM, G See <[email protected]> wrote: >> Hi, >> >> I have a list of data.tables and I am trying to extract a subset from >> each of them. I can achieve what I want with this: >> >>> L <- list(data.table(BOD), data.table(BOD)) >>> lapply(L, function(x) x[Time==3L]) >> [[1]] >> Time demand >> 1: 3 19 >> >> [[2]] >> Time demand >> 1: 3 19 >> >> However, I'd rather not type have to create an anonymous function. I >> tried the below, but `[.data.frame` is being dispatched. >> >>> lapply(L, "[", Time==3L) >> Error in `[.data.frame`(x, i) : object 'Time' not found >> >> Even if I am explicit, `[.data.table` does not get dispatched: >> >>> lapply(L, data.table:::`[.data.table`, Time==3L) >> Error in `[.data.frame`(x, i) : object 'Time' not found >> >> I'm guessing this is due to where evaluation takes place. Is there an >> alternate syntax I should use? >> > > subset works: > >> lapply(L, subset, Time == 3L) > [[1]] > Time demand > 1: 3 19 > > [[2]] > Time demand > 1: 3 19 > > > > -- > Statistics & Software Consulting > GKX Group, GKX Associates Inc. > tel: 1-877-GKX-GROUP > email: ggrothendieck at gmail.com _______________________________________________ datatable-help mailing list [email protected] https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help
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