Optimize opening a file when it already exists on disk

Author: brettcannon

importlib_resources/__init__.py

@@ -46,20 +46,24 @@ def open(package: Package, file_name: FileName) -> BinaryIO:
     """Return a file-like object opened for binary-reading of the resource."""
     file_name = _normalize_path(file_name)
     package = _get_package(package)
-    package_path = os.path.dirname(os.path.abspath(package.__spec__.origin))
-    full_path = os.path.join(package_path, file_name)
-    # Just assume the loader is a resource loader; all the relevant
-    # importlib.machinery loaders are and an AttributeError for get_data() will
-    # make it clear what is needed from the loader.
-    loader = typing.cast(importlib.abc.ResourceLoader, package.__spec__.loader)
-    try:
-        data = loader.get_data(full_path)
-    except OSError:
-        package_name = package.__spec__.name
-        message = f'{file_name!r} resource not found in {package_name!r}'
-        raise FileNotFoundError(message)
+    package_path = pathlib.Path(package.__spec__.origin).resolve().parent
+    full_path = package_path / file_name
+    if full_path.exists():
+        return full_path.open('rb')
     else:
-        return io.BytesIO(data)
+        # Just assume the loader is a resource loader; all the relevant
+        # importlib.machinery loaders are and an AttributeError for get_data()
+        # will make it clear what is needed from the loader.
+        loader = typing.cast(importlib.abc.ResourceLoader,
+                             package.__spec__.loader)
+        try:
+            data = loader.get_data(str(full_path))
+        except OSError:
+            package_name = package.__spec__.name
+            message = f'{file_name!r} resource not found in {package_name!r}'
+            raise FileNotFoundError(message)
+        else:
+            return io.BytesIO(data)
 
 
 def read(package: Package, file_name: FileName, encoding: str = 'utf-8',