Convert namespace package names into paths

Namespace packages, which have a name like "namespace.package" need to
be converted to paths in order to be found by flake8 - as the code
is in "namespace/package/".

src/flake8/main/setuptools_command.py

@@ -54,7 +54,7 @@ class Flake8(setuptools.Command):
         empty_directory_exists = "" in directories
         packages = self.distribution.packages or []
         for package in packages:
-            package_directory = package
+            package_directory = os.path.join(*package.split("."))
             if package in directories:
                 package_directory = directories[package]
             elif empty_directory_exists:
@@ -71,7 +71,7 @@ class Flake8(setuptools.Command):
             if package_directory.startswith(seen_package_directories):
                 continue
 
-            seen_package_directories += (package_directory + ".",)
+            seen_package_directories += (os.path.join(package_directory, ""),)
             yield package_directory
 
     def module_files(self):

tests/unit/test_setuptools_command.py

@@ -14,6 +14,7 @@ def distribution():
             'foo',
             'foo.bar',
             'foo_biz',
+            'namespace.bar',
         ],
     })
 
@@ -30,4 +31,5 @@ def test_package_files_removes_submodules(command):
     assert sorted(package_files) == [
         'foo',
         'foo_biz',
+        'namespace/bar',
     ]