Fixed subsetting / join bug with keys and factor to character coercion. #2926
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jangorecki
approved these changes
Jun 14, 2018
| io = if (missing(on)) haskey(i) else identical(unname(on), head(key(i), length(on))) | ||
| i = .shallow(i, retain.key = io) | ||
| ans = bmerge(i, x, leftcols, rightcols, io, xo, roll, rollends, nomatch, mult, ops, nqgrp, nqmaxgrp, verbose=verbose) | ||
| i = .shallow(i, retain.key = TRUE) |
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this change of dynamic value to TRUE does not impose any more changes to bmerge and Cbmerge?
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I strongly believe so. Cbmerge just accepts a bool for the io argument telling if i is ordered. If not, it calls forderv. bmerge.R used to accept any existing key on i as valid for the current join without crosscheck. Now, it carefully tests if the key on i is appropriate for the problem at hand. Therefore, the now always retained key on i should make no difference IMHO.
mattdowle
approved these changes
Aug 17, 2018
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Closes #2881.
Unfortunately, my PR #2389, that introduced partial key retainment when assigning to columns, introduced a regression in
bmerge.R.bmerge.Rassumed implicitely that anyset()operation removes a key completely if it assigns to a key column. In contrast, the mentioned PR made it possible that a reduced key still exists when assigning to a key column.I have done the following:
bmerge.Rso that it checks internally if the key oniis correct, instead of using theioargument. This was also mentioned as a TODO inbmerge.Rsourcecode.ioargument), so I changed all calls tobmerge.R.sortedattribute at the end ofdata.table.Rhad to be modified.No existing tests were modified and a new test added.
This regression should occur very rarely, but if it does, it is very bad. Therefore, it might make sense to issue a quick bugfix release to CRAN rather soon?