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Fix: Correct LCSMatch logic for repeated tokens #3

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Fix: Correct LCSMatch logic for repeated tokens #3

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4 changes: 0 additions & 4 deletions spellpy/spell.py
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Original file line number Diff line number Diff line change
Expand Up @@ -122,12 +122,8 @@ def LCSMatch(self, LCSMap, seq):

maxLen = -1
maxLCSObject = None
set_seq = set(seq)
size_seq = len(seq)
for LCSObject in LCSMap:
set_template = set(LCSObject.logTemplate)
if len(set_seq & set_template) < 0.5 * size_seq:
continue
lcs = self.LCS(seq, LCSObject.logTemplate)
if len(lcs) > maxLen or (len(lcs) == maxLen and len(LCSObject.logTemplate) < len(maxLCSObject.logTemplate)):
maxLen = len(lcs)
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13 changes: 13 additions & 0 deletions tests/test_spellpy.py
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Original file line number Diff line number Diff line change
Expand Up @@ -94,6 +94,19 @@ def test_getTemplate(self):
new_template = self.parser.getTemplate(lcs, seq)
self.assertListEqual(new_template, expected_template)

def test_LCSMatch_with_repeated_tokens(self):
# This test case simulates the scenario described in the issue.
# 'seq' has many repeated tokens, which makes len(set(seq)) small.
# The old implementation would fail this test.
logmessageL = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J']
seq = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'K', 'K', 'K', 'K', 'K', 'K', 'K', 'K', 'K']
logID = 0
newCluster = LCSObject(logTemplate=logmessageL, logIDL=[logID])

retLogClust = self.parser.LCSMatch([newCluster], seq)
self.assertIsNotNone(retLogClust)
self.assertListEqual(retLogClust.logTemplate, newCluster.logTemplate)


def helper(rootNode):
if rootNode.childD == dict():
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