How can I manually generate a .pyc file from a .py file
For some reason, I can not depend on Python's "import" statement to generate .pyc file automatically
Is there a way to implement a function as f开发者_运维问答ollowing?
def py_to_pyc(py_filepath, pyc_filepath):
...
You can use compileall
in the terminal. The following command will go recursively into sub directories and make pyc files for all the python files it finds. The compileall module is part of the python standard library, so you don't need to install anything extra to use it. This works exactly the same way for python2 and python3.
python -m compileall .
You can compile individual files(s) from the command line with:
python -m compileall <file_1>.py <file_n>.py
It's been a while since I last used Python, but I believe you can use py_compile
:
import py_compile
py_compile.compile("file.py")
I found several ways to compile python scripts into bytecode
Using
py_compile
in terminal:python -m py_compile File1.py File2.py File3.py ...
-m
specifies the module(s) name to be compiled.Or, for interactive compilation of files
python -m py_compile - File1.py File2.py File3.py . . .
Using
py_compile.compile
:import py_compile py_compile.compile('YourFileName.py')
Using
py_compile.main()
:It compiles several files at a time.
import py_compile py_compile.main(['File1.py','File2.py','File3.py'])
The list can grow as long as you wish. Alternatively, you can obviously pass a list of files in main or even file names in command line args.
Or, if you pass
['-']
in main then it can compile files interactively.Using
compileall.compile_dir()
:import compileall compileall.compile_dir(direname)
It compiles every single Python file present in the supplied directory.
Using
compileall.compile_file()
:import compileall compileall.compile_file('YourFileName.py')
Take a look at the links below:
https://docs.python.org/3/library/py_compile.html
https://docs.python.org/3/library/compileall.html
I would use compileall. It works nicely both from scripts and from the command line. It's a bit higher level module/tool than the already mentioned py_compile that it also uses internally.
Normally the following command compilies a python project:
python -m compileall <project-name>
In Python2 it compiles all .py
files to .pyc
files in a project which contains packages as well as modules.
Whereas in Python3 it compiles all .py
files to __pycache__
folders in a project which contains packages as well as modules.
With browning from this post:
You can enforce the same layout of
.pyc
files in the folders as in Python2 by using:
python3 -m compileall -b <pythonic-project-name>
The option
-b
triggers the output of.pyc
files to their legacy-locations (i.e. the same as in Python2).
To match the original question requirements (source path and destination path) the code should be like that:
import py_compile
py_compile.compile(py_filepath, pyc_filepath)
If the input code has errors then the py_compile.PyCompileError exception is raised.
- create a new python file in the directory of the file.
- type
import (the name of the file without the extension)
- run the file
- open the directory, then find the pycache folder
- inside should be your .pyc file
There are two ways to do this
- Command line
- Using python program
If you are using command line, use python -m compileall <argument>
to compile python code to python binary code.
Ex: python -m compileall -x ./*
Or, You can use this code to compile your library into byte-code:
import compileall
import os
lib_path = "your_lib_path"
build_path = "your-dest_path"
compileall.compile_dir(lib_path, force=True, legacy=True)
def moveToNewLocation(cu_path):
for file in os.listdir(cu_path):
if os.path.isdir(os.path.join(cu_path, file)):
compile(os.path.join(cu_path, file))
elif file.endswith(".pyc"):
dest = os.path.join(build_path, cu_path ,file)
os.makedirs(os.path.dirname(dest), exist_ok=True)
os.rename(os.path.join(cu_path, file), dest)
moveToNewLocation(lib_path)
look at ☞ docs.python.org for detailed documentation
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