Question

Is there a way in Python to silence stdout without wrapping a function call like following?

Original Broken Code:

from sys import stdout
from copy import copy
save_stdout = copy(stdout)
stdout = open('trash','w')
foo()
stdout = save_stdout

Edit: Corrected code from Alex Martelli

import sys
save_stdout = sys.stdout
sys.stdout = open('trash', 'w')
foo()
sys.stdout = save_stdout

That way works but appears to be terribly inefficient. There has to be a better way. Any ideas?

Answer 1

Assigning the stdout variable as you’re doing has no effect whatsoever, assuming foo contains print statements — yet another example of why you should never import stuff from inside a module (as you’re doing here), but always a module as a whole (then use qualified names). The copy is irrelevant, by the way. The correct equivalent of your snippet is:

import sys
save_stdout = sys.stdout
sys.stdout = open('trash', 'w')
foo()
sys.stdout = save_stdout

Now, when the code is correct, is the time to make it more elegant or fast. For example, you could use an in-memory file-like object instead of file ‘trash’:

import sys
import io
save_stdout = sys.stdout
sys.stdout = io.BytesIO()
foo()
sys.stdout = save_stdout

for elegance, a context is best, e.g:

import contextlib
import io
import sys

@contextlib.contextmanager
def nostdout():
    save_stdout = sys.stdout
    sys.stdout = io.BytesIO()
    yield
    sys.stdout = save_stdout

once you have defined this context, for any block in which you don’t want a stdout,

with nostdout():
    foo()

More optimization: you just need to replace sys.stdout with an object that has a no-op write method. For example:

import contextlib
import sys

class DummyFile(object):
    def write(self, x): pass

@contextlib.contextmanager
def nostdout():
    save_stdout = sys.stdout
    sys.stdout = DummyFile()
    yield
    sys.stdout = save_stdout

to be used the same way as the previous implementation of nostdout. I don’t think it gets any cleaner or faster than this;-).