Fix Module.is_public() when the module is not a the root

docs/release_notes.rst

@@ -17,6 +17,7 @@ Bug Fixes
 * Detect inner asynchronous functions for D202 (#467)
 * Fix a bug in parsing Google-style argument description.
   The bug caused some argument names to go unreported in D417 (#448).
+* Correctly detect publicity of modules inside directories (#470).
 
 5.0.2 - January 8th, 2020
 ---------------------------

src/pydocstyle/parser.py

@@ -5,6 +5,7 @@
 from itertools import chain, dropwhile
 from re import compile as re
 from io import StringIO
+from pathlib import Path
 
 from .utils import log
 
@@ -117,7 +118,8 @@ def is_public(self):
 
         This helps determine if it requires a docstring.
         """
-        return not self.name.startswith('_') or self.name.startswith('__')
+        module_name = Path(self.name).name
+        return not module_name.startswith('_') or module_name.startswith('__')
 
     def __str__(self):
         return 'at module level'

src/tests/parser_test.py

@@ -1,6 +1,7 @@
 """Parser tests."""
 
 import io
+import os
 import sys
 import pytest
 import textwrap
@@ -562,18 +563,22 @@ def test_matrix_multiplication_with_decorators(code):
     assert inner_function.decorators[0].name == 'a'
 
 
-def test_module_publicity():
+@pytest.mark.parametrize("parent_path", (
+    os.path.join("some", "directory"),
+    ""
+))
+def test_module_publicity(parent_path):
     """Test that a module that has a single leading underscore is private."""
     parser = Parser()
     code = CodeSnippet("")
 
-    module = parser.parse(code, "filepath")
+    module = parser.parse(code, os.path.join(parent_path, "filepath"))
     assert module.is_public
 
-    module = parser.parse(code, "_filepath")
+    module = parser.parse(code, os.path.join(parent_path, "_filepath"))
     assert not module.is_public
 
-    module = parser.parse(code, "__filepath")
+    module = parser.parse(code, os.path.join(parent_path, "__filepath"))
     assert module.is_public