Using ! to assign output of shell command to variable in .ipynb

Is there a way to get this on one line:

> ipython
Python 3.11.1 (main, Jan 24 2023, 17:02:06) [Clang 14.0.0 (clang-1400.0.29.202)]
Type 'copyright', 'credits' or 'license' for more information
IPython 8.8.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]: l = ! ls

In [2]: l[:3]
Out[2]: ['___x.ipynb', 'data', 'e.txt']

In [3]: l = (! ls)[:3]
  Cell In[3], line 1
    l = (! ls)[:3]
         ^
SyntaxError: invalid syntax

?

Ref: https://jakevdp.github.io/PythonDataScienceHandbook/01.05-ipython-and-shell-commands.html

I came to the same conclusion as Azeem, that there is no direct way to do this since iPython only seems to support direct assignment of shell output with =. l = !ls | head -n 3 is the probably the best and most readable way to do this in one line in iPython, but here's a Python alternative that will work anywhere:

import subprocess
l = subprocess.run(['ls'], stdout=subprocess.PIPE).stdout.decode('utf-8').splitlines()[:3]

There doesn't seem to exist a direct way to achieve that in one line under IPython.

According to that blog post:

> anything appearing after ! on a line will be executed not by the Python kernel, but by the system command-line

and, IPython's System shell access:

> Any input line beginning with a ! character is passed verbatim (minus the !, of course) to the underlying operating system.

Except for the variables in curly braces that are expanded before the command is passed to the underlying OS shell.

One possible solution could be to leverage the shell head command for this:

ls | head -n 3

and, in IPython:

files = !ls | head -n 3

Another alternative could be to resort to Python solutions such as os.listdir():

import os
files = [e for e in os.listdir() if not e.startswith('.') and os.path.isfile(e) ][:3]

or, glob.glob():

import glob
files = glob.glob('*')[:3]