'Import functions that use ipython magic
In ipython cells you can execute shell commands as follows:
ipython:
print("asdf")
!echo asdf
However, if you try to import this code from file,
asdf.py:
def asdf():
print("asdf")
!echo asdf
ipython:
from asdf import asdf
it results in an error.
!echo asdf
^
SyntaxError: invalid syntax
The use case for that is repetative massive scripts in google colab utilizing ffmpeg, wget, mount, e.t.c.
While you can use os.system
or subprocess
, they are not as interactive in providing real-time stdout.
Solution 1:[1]
Use case solution
If you want to reuse code that has ipython magic functions in it, simply use %run
instead of the import
.
ipython:
%run utils.ipynb
utils.ipynb:
def asdf():
print("asdf")
!echo asdf
Example above works fine.
Taken from this answer.
On reusing ipython code
Why the original question was an A-B problem of some kind? The reason is that you should not really drag ipython additional functionality to the pure python execution.
- If you have a, let's say, google colab notebook with useful utils,
then just
!wget
it from the public link and%run
it. - If you have a python script to execute, do a regular
import
on.py
files.
Sure, there are ways to import .ipynb
files to python code: Jupyter notebook's Notebook Loader, ipynb package, nbimporter package (but even the nbimporter's author says you should not really do that. Just convert your notebook to .py
using VSCode or another tool).
So, if you use ipython research environment and features, just stick with it. Or switch to pure python environment with proper modules, testing, e.t.c.
Also, if you're struggling with doing wget
:
file_id = "1xE8Db1zvQ7v-z13CO2qc3F6wJmtr3YHu" # can be extracted from public link
file_out_name = "utils.ipynb"
!wget "https://docs.google.com/uc?export=download&id=""$file_id" -O "$file_out_name"
However, on the problem of calling cmd, I see no way of executing shell commands interactively in python with one line. While subprocess
can do that, it takes multiple lines to achieve.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
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Solution 1 |