How to handle both with open(...) and sys.stdout nicely?
Often I need to output data either to file or, if file is not specified, to stdout. I use the following snippet:
if target: with open(target, 'w') as h: h.write(content) else: sys.stdout.write(content)
I would like to rewrite it and handle both targets uniformly.
In ideal case it would be:
with open(target, 'w') as h: h.write(content)
but this will not work well because sys.stdout is be closed when leaving
with block and I don't want that. I neither want to
stdout = open(target, 'w') ...
because I would need to remember to restore original stdout.
- Redirect stdout to a file in Python?
- Handling Exceptions - interesting article about handling exceptions in Python, as compared to C++
I know that I can wrap
target, define separate function or use context manager. I look for a simple, elegant, idiomatic solution fitting that wouldn't require more than 5 lines
Just thinking outside of the box here, how about a custom
import sys import contextlib @contextlib.contextmanager def smart_open(filename=None): if filename and filename != '-': fh = open(filename, 'w') else: fh = sys.stdout try: yield fh finally: if fh is not sys.stdout: fh.close()
Use it like this:
# writes to some_file with smart_open('some_file') as fh: print >>fh, 'some output' # writes to stdout with smart_open() as fh: print >>fh, 'some output' # writes to stdout with smart_open('-') as fh: print >>fh, 'some output'