"Fixed" tracing bug if module name and current working directory are identically named

[danielryckman]

For Python projects named reponame/packagename where reponame equals packagename (i.e. torchtitan/torchtitan), .split() will improperly split at the first instance in the string. The path after the root module will then still have module name on it. For example, it splits "/home/daniel/torchtitan/torchtitan/config/manager.py" into ['/home/daniel', 'torchtitan/config/manager.py'] instead of ['/home/daniel/torchtitan', 'config/manager.py']. Thus, after prepending the module name, we get torchtitan/torchtitan/config/manager.py, which doesn't exist.

[Essoz]

✅ The fix solves the immediate problem, but I share your concern about potential edge cases.

Potential issue:
This still relies on string matching for the module separator pattern f"/{root_module}/". If a root module has the same name as a submodule or class within it (e.g., a package structure like mylib.submodule.mylib), the rsplit could still grab the wrong segment.

Better approach: Instead of parsing the file path string directly, we:

Determine the absolute path of the root module (using importlib.util.find_spec() or pathlib)
Use pathlib.Path.relative_to() to remove the root module path from the file path
Convert the resulting relative path to a module path
This avoids string-based parsing altogether and is much more robust to naming collisions.

Example Code (you can probably use a coding agent to help you implement this, should be a pretty lightweight change).

from pathlib import Path
from importlib.util import find_spec

def get_module_path_from_file_path(file_path: str, root_module: str) -> str | None:
    """
    Get the module path from a file path using importlib and pathlib.
    """
    try:
        # Find the root module's location
        spec = find_spec(root_module)
        if spec is None or spec.origin is None:
            return None
        
        root_module_path = Path(spec.origin).parent
        file_path_obj = Path(file_path).resolve()
        
        # Get the relative path from root module to file
        try:
            relative_path = file_path_obj.relative_to(root_module_path)
        except ValueError:
            # file_path is not under the root module
            return None
        
        # Convert to module path
        module_name = str(relative_path.with_suffix(''))
        module_path = f"{root_module}.{module_name.replace('/', '.')}"
        return module_path
    except Exception:
        return None

This approach is import-aware and doesn't rely on fragile string parsing.

[Essoz]

See comment.