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Fix IPython detection routine #237

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Merged
merged 1 commit into from
Jan 12, 2022
Merged

Fix IPython detection routine #237

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2 changes: 1 addition & 1 deletion monai/deploy/core/application.py
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Original file line number Diff line number Diff line change
Expand Up @@ -95,7 +95,7 @@ def __init__(
self.path = get_class_file_path(self.__class__)

# Set the runtime environment
if str(self.path).startswith("<ipython-"):
if str(self.path) == "ipython":
self.in_ipython = True
else:
self.in_ipython = False
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10 changes: 7 additions & 3 deletions monai/deploy/utils/importutil.py
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Original file line number Diff line number Diff line change
Expand Up @@ -99,7 +99,11 @@ def get_application(path: Union[str, Path]) -> Optional["Application"]:
def get_class_file_path(cls: Type) -> Path:
"""Get the file path of a class.

If file path is not available, it tries to get the internal class name used in IPython(starting with <ipython-...>).
If the file path is not available, it tries to see each frame information
in the stack to check whether the file name ends with "interactiveshell.py"
and the function name is "run_code".
If so, it returns Path("ipython") to notify that the class is defined
inside IPython.

Args:
cls (Type): A class to get file path from.
Expand All @@ -114,8 +118,8 @@ def get_class_file_path(cls: Type) -> Path:
# If in IPython shell, use inspect.stack() to get the caller's file path
stack = inspect.stack()
for frame in stack:
if frame.filename.startswith("<ipython-"):
return Path(frame.filename)
if frame.filename.endswith("interactiveshell.py") and frame.function == "run_code":
return Path("ipython")
# If not in IPython shell, re-raise the error
raise

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