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Function is_known()
not working as expected (missing word when a rule is active)
#125
Comments
Hi @adbar , The And you are right. The We have 2 options: Option 2: Remove the I like the latter, as it simplifies things. What do you think? |
It turns out I actually use RulesThe rules were first about extending coverage, there are not meant to save space, there are other means for that, see #72 (using binary strings instead of strings in particular). So I'd be in favor of these changes:
is_known()Then we could keep the function but make it simpler:
@juanjoDiaz Would you be interested in implementing part of these changes? And by the way to you want/need write access to the repository? I already added you as co-author in the text since the sum of your contribution is highly significant. Maybe you also want to work on the documentation which I recently (finally) made available. |
Makes sense. However, the more I think about it, the more I think that we should remove the
We can definitely do that.
This sounds good.
Let's do this.
Sure. I can do this once we align on the way forward.
Thanks! I appreciate you giving me credit.
Sure thing! |
Let's work on your PR now (I just had one comment) and break the other points apart in new issue threads (changes in dictionary pickler and tests for binary string). |
Summary
The case happens when a word can be decomposed without being in the dictionary (here because of a rule).
from simplemma import lemmatize, is_known
Legacy
Current state
Trying to debug
Problem
The word "Bergungen" is in the original input list and it can be lemmatized so it is "known". It is already present in the previous package versions. The result should be comparable to "Berge", i.e. something is found.
Also, the greedy option doesn't do anything (although there are rules that can be applied here).
Suggestions:
is_known()
on top of the dictionary.greedy
argument foris_known()
?@juanjoDiaz What do you think?
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