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Github user willb commented on the issue:
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Thanks for the feedback @srowen! I think 18e6bfe addresses everything from
a code perspective, but it missed removing the comment about assuming that
distinct words have distinct vector
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Github user willb commented on the issue:
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@srowen Yes. But if we're querying for words similar to a given vector,
then there's no word to filter out (and, indeed, no way to know which word we
might want to filter out if multiple words might
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Can we assume in general that distinct words will not have identical (viz.,
by `.equals`) vector representations? I ask because the behavior in the
current documentation (in that pyspark test I
Github user srowen commented on the issue:
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I don't think you can assume that, though it will almost always be true.
But I don't think the goal is to filter vectors of other words that happen to
be identical. The point was just to filter the
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Thanks, @hhbyyh. This is what I get for running `sbt "testOnly ..."` I'll
push a fix.
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Github user hhbyyh commented on the issue:
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Hi @willb Good catch. This is a valid issue.
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