I want to source my bash environment when executing a bash command from IPython using the ! operator, thus allowing me access to my defined bash functions:
In[2]: !<my_fancy_bash_function> <function_argument>
currently IPython is sourcing sh rather than bash:
In[3]: !declare -F
sh: 1: declare: not found
How to source bash and set my environment settings from IPython?
Fernando Perez, creator of IPython, suggests this:
In [1]: %%bash
. ~/.bashrc
<my_fancy_bash_function> <function_argument>
This works on the current stable version (0.13.2). He admits that's a bit clunky, and welcomes pull requests. . .
If the ! implementation uses IPython.utils._process_posix.system under the hood, then it is going to use whatever which sh returns as the processing shell. This could be a true implementation of Bourne shell - it is likely Bash in some compatibility mode on many Linuxes. On my MacBook Pro it looks like it is a raw Bash shell:
In [12]: !declare -F
In [13]: !echo $BASH
/bin/sh
In [14]: !echo $BASH_VERSION
3.2.48(1)-release
In [15]: import os
In [16]: os.environ['SHELL']
Out[16]: '/bin/zsh'
I was hoping that it would use the $SHELL environment variable but it does not seem to today. You can probably branch the github repo, modify the ProcessHandler.sh property implementation to peek into os.environ['SHELL'] and use this if it is set instead of calling pexpect.which('sh'). Then issue a pull request.
! implementation (linked to in your answer)? Doing a quick 'ghetto' test of symlinking /bin/bash to /bin/sh and issuing ! source ~/.bashrc && declare -F yields nothing, whereas ! env && declare -F correctly prints out my declared variables, but the bash functions are still not accessible...any ideas what is going on there?
bash variable expansion tricks to expand a variable declared in python! For example extension=".txt" followed by !ls *"$extension" prints all .txt files!