[Python] How to access a file in zipapp

I've got a simple Python zipapp built with the following command:

python -m pkg/ -c -o test -p '/usr/bin/python3' -m 'test:main' zipapp


I would like to access the binary file from the script

$ cat pkg/test.py
def main():
with open('test.bin', 'rb') as f:
print(f.name)


Directory structure

$ tree pkg/
pkg/
├── test.bin
└── test.py

0 directories, 2 files


But it looks like the script is referring to a file from the current directory:

$ ./test
Traceback (most recent call last):
File "/usr/lib64/python3.7/runpy.py", line 193, in _run_module_as_main
"__main__", mod_spec)
File "/usr/lib64/python3.7/runpy.py", line 85, in _run_code
exec(code, run_globals)
File "./test/__main__.py", line 3, in <module>
File "./test/test.py", line 2, in main
FileNotFoundError: [Errno 2] No such file or directory: 'test.bin'


It's a quite large binary file so I would like to avoid creating a variable. Is there a way to access this file from the script itself?

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