drop into python interpreter while executing function

Question

i have a python module with a function:

def do_stuff(param1 = \'a\'):
    if type(param1) == int:
        # enter python interpreter here
        do_somethin         


        

Answer 1

If you want a standard interactive prompt (instead of the debugger, as shown by prestomation), you can do this:

import code
code.interact(local=locals())

See: the code module.

If you have IPython installed, and want an IPython shell instead, you can do this for IPython >= 0.11:

import IPython; IPython.embed()

or for older versions:

from IPython.Shell import IPShellEmbed
ipshell = IPShellEmbed()
ipshell(local_ns=locals())

Answer 2

Inserting

import pdb; pdb.set_trace()

will enter the python debugger at that point

See here: http://docs.python.org/library/pdb.html

Answer 3

If you want a default Python interpreter, you can do

import code
code.interact(local=dict(globals(), **locals()))

This will allow access to both locals and globals.

If you want to drop into an IPython interpreter, the IPShellEmbed solution is outdated. Currently what works is:

from IPython import embed
embed()