Merge pull request #66 from chaimleib/dev

faster interval overlaps #56
chaimleib · Dec 11, 2017 · 1ff87c1 · 1ff87c1
2 parents 2d1b463 + ce4b52a
commit 1ff87c1
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Showing 3 changed files with 8 additions and 6 deletions.
diff --git a/CHANGELOG.md b/CHANGELOG.md
@@ -3,6 +3,7 @@
 ## Version 3.0.0
 - Dropped support for Python 2.6, 3.2, and 3.3
 - Add support for Python 3.5 and 3.6
+- Faster `Interval` overlap checking (@tuxzz, #56)
 - Updated README:
     - new restructuring methods from 2.1.0
     - example of `from_tuples()` added

diff --git a/README.rst b/README.rst
@@ -372,6 +372,7 @@ Version 3.0.0
 
 -  Dropped support for Python 2.6, 3.2, and 3.3
 -  Add support for Python 3.5 and 3.6
+-  Faster ``Interval`` overlap checking (@tuxzz, #56)
 -  Updated README:
 
    -  new restructuring methods from 2.1.0

diff --git a/intervaltree/interval.py b/intervaltree/interval.py
@@ -39,12 +39,12 @@ def overlaps(self, begin, end=None):
         :rtype: bool
         """
         if end is not None:
-            return (
-                (begin <= self.begin < end) or
-                (begin < self.end <= end) or
-                (self.begin <= begin < self.end) or
-                (self.begin < end <= self.end)
-            )
+            # An overlap means that some C exists that is inside both ranges:
+            #   begin <= C < end
+            # and 
+            #   self.begin <= C < self.end
+            # See https://stackoverflow.com/questions/3269434/whats-the-most-efficient-way-to-test-two-integer-ranges-for-overlap/3269471#3269471
+            return begin < self.end and end > self.begin
         try:
             return self.overlaps(begin.begin, begin.end)
         except: