fix(chain): sort matches by match_index before groupby

rebulk/chain.py

@@ -216,7 +216,10 @@ def _fix_matches_offset(chain_part_matches: Iterable[Match], input_string: str,
     @staticmethod
     def _group_by_match_index(matches: Iterable[Match]) -> dict[int, list[Match]]:
         grouped_matches_dict: dict[int, list[Match]] = {}
-        for match_index, match in itertools.groupby(matches, lambda m: m.match_index):
+        # groupby only groups consecutive equal keys, so sort by match_index first
+        # to avoid splitting (and overwriting) groups when matches are unordered.
+        sorted_matches = sorted(matches, key=lambda m: m.match_index)
+        for match_index, match in itertools.groupby(sorted_matches, lambda m: m.match_index):
             grouped_matches_dict[match_index] = list(match)
         return grouped_matches_dict
 

rebulk/test/test_chain.py

@@ -8,11 +8,12 @@
 from rebulk.pattern import FunctionalPattern, RePattern, StringPattern
 
 from ..chain import Chain
+from ..match import Match
 from ..rebulk import Rebulk
 from ..validators import chars_surround
 
 if TYPE_CHECKING:
-    from ..match import Match, Matches
+    from ..match import Matches
 
 
 def test_chain_close() -> None:
@@ -489,3 +490,22 @@ def chain_breaker(matches: Matches) -> bool:
     matches[0].value = 1
     matches[1].value = 2
     matches[2].value = 3
+
+
+def test_group_by_match_index_sorts_unordered_input() -> None:
+    # groupby only groups consecutive equal keys, so _group_by_match_index must
+    # sort by match_index first; otherwise unordered matches split (and overwrite)
+    # groups sharing an index.
+    matches: list[Match] = []
+    for start, match_index in ((0, 2), (5, 0), (10, 1), (15, 2)):
+        match = Match(start, start + 1, name="test")
+        match.match_index = match_index
+        matches.append(match)
+
+    grouped = Chain._group_by_match_index(matches)
+
+    assert sorted(grouped) == [0, 1, 2]
+    assert [m.start for m in grouped[0]] == [5]
+    assert [m.start for m in grouped[1]] == [10]
+    # both match_index=2 matches are kept, in their original relative order
+    assert [m.start for m in grouped[2]] == [0, 15]