Fix Python – How to capture stdout output from a Python function call?


Asked By – Nico Schlömer

I’m using a Python library that does something to an object


and changes it. While doing so, it prints some statistics to stdout, and I’d like to get a grip on this information. The proper solution would be to change do_something() to return the relevant information,

out = do_something(my_object)

but it will be a while before the devs of do_something() get to this issue. As a workaround, I thought about parsing whatever do_something() writes to stdout.

How can I capture stdout output between two points in the code, e.g.,

out = end_capturing()


Now we will see solution for issue: How to capture stdout output from a Python function call?


Try this context manager:

from io import StringIO 
import sys

class Capturing(list):
    def __enter__(self):
        self._stdout = sys.stdout
        sys.stdout = self._stringio = StringIO()
        return self
    def __exit__(self, *args):
        del self._stringio    # free up some memory
        sys.stdout = self._stdout


with Capturing() as output:

output is now a list containing the lines printed by the function call.

Advanced usage:

What may not be obvious is that this can be done more than once and the results concatenated:

with Capturing() as output:
    print('hello world')

print('displays on screen')

with Capturing(output) as output:  # note the constructor argument
    print('hello world2')

print('output:', output)


displays on screen                     
output: ['hello world', 'hello world2']

Update: They added redirect_stdout() to contextlib in Python 3.4 (along with redirect_stderr()). So you could use io.StringIO with that to achieve a similar result (though Capturing being a list as well as a context manager is arguably more convenient).

This question is answered By – kindall

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0