One would normally want the original order that so that one can stack a list, operate on the result and then unstack it back with the unstacked result having the same ordering as the original.
LL <- list(z = 1:3, a = list()) # since we can't do s <- stack(LL,. drop = FALSE) do this instead: s <- transform(stack(LL), ind = factor(as.character(ind), levels = names(LL))) unstack(s) On Mon, Jun 27, 2016 at 2:55 PM, Michael Lawrence <lawrence.mich...@gene.com> wrote: > I'll add the drop argument but I'm wondering about the order of the > levels. Should we set the levels to unique(names(x)) or sort them, > too? > > On Mon, Jun 27, 2016 at 10:39 AM, Gabor Grothendieck > <ggrothendi...@gmail.com> wrote: >> stack() seems to drop empty levels. Perhaps there could be a >> drop=FALSE argument if one wanted all the original levels. In the >> example below, we may wish to retain level "b" in s$ind even though >> component LL$b has length 0. >> >>> LL <- list(a = 1:3, b = list()) >>> s <- stack(LL) >>> str(s) >> 'data.frame': 3 obs. of 2 variables: >> $ values: int 1 2 3 >> $ ind : Factor w/ 1 level "a": 1 1 1 >> >> >> -- >> Statistics & Software Consulting >> GKX Group, GKX Associates Inc. >> tel: 1-877-GKX-GROUP >> email: ggrothendieck at gmail.com >> >> ______________________________________________ >> R-devel@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-devel >> -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
