Select the longest matching key in matchMap #179
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Fixes sirthias#136 By checking keys from longest to shortest, we ensure that we always select the longest matching key. This fix requires sorting the keys each time the rule is invoked. If sirthias#115 gets implemented, this could be improved to make use of it instead.
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Sorry, I wasn't thinking straight in writing that final comment. The map value isn't known until runtime, so of course the optimisation desired in #115 would be unhelpful. |
sirthias
reviewed
Mar 6, 2017
| @@ -449,7 +449,7 @@ abstract class Parser(initialValueStackSize: Int = 16, | |||
| * THIS IS NOT PUBLIC API and might become hidden in future. Use only if you know what you are doing! | |||
| */ | |||
| def __matchMap(m: Map[String, Any]): Boolean = { | |||
| val keys = m.keysIterator | |||
| val keys = m.keys.toSeq.sortBy(-_.length).iterator | |||
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I think we can actually make this more efficient.
We don't need to sort the whole map, only the prefix up until the match.
Check out this SO answer for an idea of how this could be done.
Only sort the keys until we find the one we're interested in, as suggested in sirthias#179 (review)
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Thank you, Ben, and sorry for taking so long to get this merged! |
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Fixes #136
By checking keys from longest to shortest, we ensure that we always
select the longest matching key.
This fix requires sorting the keys each time the rule is invoked. If
#115 gets implemented, this could be improved to make use of it instead.