https://github.com/python/cpython/commit/7f1e66ae0e7c6bfa26cc128291d8705abd1a3c93
commit: 7f1e66ae0e7c6bfa26cc128291d8705abd1a3c93
branch: main
author: Raymond Hettinger <rhettin...@users.noreply.github.com>
committer: rhettinger <rhettin...@users.noreply.github.com>
date: 2025-07-11T12:36:17-07:00
summary:

Minor edit: Improve comment readability and ordering (gh-136557)

files:
M Lib/collections/__init__.py

diff --git a/Lib/collections/__init__.py b/Lib/collections/__init__.py
index d2ddc1cd9ec2ea..b8653f40a942f0 100644
--- a/Lib/collections/__init__.py
+++ b/Lib/collections/__init__.py
@@ -776,23 +776,26 @@ def __repr__(self):
     # When the multiplicities are all zero or one, multiset operations
     # are guaranteed to be equivalent to the corresponding operations
     # for regular sets.
+    #
     #     Given counter multisets such as:
     #         cp = Counter(a=1, b=0, c=1)
     #         cq = Counter(c=1, d=0, e=1)
+    #
     #     The corresponding regular sets would be:
     #         sp = {'a', 'c'}
     #         sq = {'c', 'e'}
+    #
     #     All of the following relations would hold:
-    #         set(cp + cq) == sp | sq
-    #         set(cp - cq) == sp - sq
-    #         set(cp | cq) == sp | sq
-    #         set(cp & cq) == sp & sq
     #         (cp == cq) == (sp == sq)
     #         (cp != cq) == (sp != sq)
     #         (cp <= cq) == (sp <= sq)
     #         (cp < cq) == (sp < sq)
     #         (cp >= cq) == (sp >= sq)
     #         (cp > cq) == (sp > sq)
+    #         set(cp + cq) == sp | sq
+    #         set(cp - cq) == sp - sq
+    #         set(cp | cq) == sp | sq
+    #         set(cp & cq) == sp & sq
 
     def __eq__(self, other):
         'True if all counts agree. Missing counts are treated as zero.'

_______________________________________________
Python-checkins mailing list -- python-checkins@python.org
To unsubscribe send an email to python-checkins-le...@python.org
https://mail.python.org/mailman3//lists/python-checkins.python.org
Member address: arch...@mail-archive.com

Reply via email to