I want to write a function that accepts either a path as a string or a file object. So far I have:
def awesome_parse(path_or_file):
if isinstance(path_or
The odd thing about your code is that if it is passed an open file, it will close it. This isn't good. Whatever code opened the file should be responsible for closing it. This makes the function a bit more complex though:
def awesome_parse(path_or_file):
if isinstance(path_or_file, basestring):
f = file_to_close = open(path_or_file, 'rb')
else:
f = path_or_file
file_to_close = None
try:
return do_stuff(f)
finally:
if file_to_close:
file_to_close.close()
You can abstract this away by writing your own context manager:
@contextlib.contextmanager
def awesome_open(path_or_file):
if isinstance(path_or_file, basestring):
f = file_to_close = open(path_or_file, 'rb')
else:
f = path_or_file
file_to_close = None
try:
yield f
finally:
if file_to_close:
file_to_close.close()
def awesome_parse(path_or_file):
with awesome_open(path_or_file) as f:
return do_stuff(f)
You could do:
def awesome_parse(do_stuff):
"""Decorator to open a filename passed to a function
that requires an open file object"""
def parse_path_or_file(path_or_file):
"""Use a ternary expression to either open the file from the filename
or just pass the extant file object on through"""
with (open(path_or_file, 'rb')
if isinstance(path_or_file, basestring)
else path_or_file) as f:
return do_stuff(f)
return parse_path_or_file
And then when you declare any function that does stuff on an open file object:
@awesome_parse
def do_things(open_file_object):
"""This will always get an open file object even if passed a string"""
pass
@awesome_parse
def do_stuff(open_file_object):
"""So will this"""
pass
Edit 2: More detailed info on the decorator.