I have a large project consisting of sufficiently large number of modules, each printing something to the standard output. Now as the project has grown in size, there are large no. of print statements printing a lot on the std out which has made the program considerably slower.

So, I now want to decide at runtime whether or not to print anything to the stdout. I cannot make changes in the modules as there are plenty of them. (I know I can redirect the stdout to a file but even this is considerably slow.)

So my question is how do I redirect the stdout to nothing ie how do I make the print statement do nothing?

# I want to do something like this.
sys.stdout = None         # this obviously will give an error as Nonetype object does not have any write method.

Currently the only idea I have is to make a class which has a write method (which does nothing) and redirect the stdout to an instance of this class.

class DontPrint(object):
    def write(*args): pass

dp = DontPrint()
sys.stdout = dp

Is there an inbuilt mechanism in python for this? Or is there something better than this?

Cross-platform:

import os
import sys
f = open(os.devnull, 'w')
sys.stdout = f

On Windows:

f = open('nul', 'w')
sys.stdout = f

On Linux:

f = open('/dev/null', 'w')
sys.stdout = f

A nice way to do this is to create a small context processor that you wrap your prints in. You then just use is in a with-statement to silence all output.

Python 2:

import os
import sys
from contextlib import contextmanager

@contextmanager
def silence_stdout():
    old_target = sys.stdout
    try:
        with open(os.devnull, "w") as new_target:
            sys.stdout = new_target
            yield new_target
    finally:
        sys.stdout = old_target

with silence_stdout():
    print("will not print")

print("this will print")

Python 3.4+:

Python 3.4 has a context processor like this built-in, so you can simply use contextlib like this:

import contextlib

with contextlib.redirect_stdout(None):
    print("will not print")

print("this will print")

Running this code only prints the second line of output, not the first:

$ python test.py
this will print

This works cross-platform (Windows + Linux + Mac OSX), and is cleaner than the ones other answers imho.

If you’re in python 3.4 or higher, there’s a simple and safe solution using the standard library:

import contextlib

with contextlib.redirect_stdout(None):
  print("This won't print!")

(at least on my system) it appears that writing to os.devnull is about 5x faster than writing to a DontPrint class, i.e.

#!/usr/bin/python
import os
import sys
import datetime

ITER = 10000000
def printlots(out, it, st="abcdefghijklmnopqrstuvwxyz1234567890"):
   temp = sys.stdout
   sys.stdout = out
   i = 0
   start_t = datetime.datetime.now()
   while i < it:
      print st
      i = i+1
   end_t = datetime.datetime.now()
   sys.stdout = temp
   print out, "\n   took", end_t - start_t, "for", it, "iterations"

class devnull():
   def write(*args):
      pass


printlots(open(os.devnull, 'wb'), ITER)
printlots(devnull(), ITER)

gave the following output:

<open file '/dev/null', mode 'wb' at 0x7f2b747044b0> 
   took 0:00:02.074853 for 10000000 iterations
<__main__.devnull instance at 0x7f2b746bae18> 
   took 0:00:09.933056 for 10000000 iterations

If you’re in a Unix environment (Linux included), you can redirect output to /dev/null:

python myprogram.py > /dev/null

And for Windows:

python myprogram.py > nul

How about this:

from contextlib import ExitStack, redirect_stdout
import os

with ExitStack() as stack:
    if should_hide_output():
        null_stream = open(os.devnull, "w")
        stack.enter_context(null_stream)
        stack.enter_context(redirect_stdout(null_stream))
    noisy_function()

This uses the features in the contextlib module to hide the output of whatever command you are trying to run, depending on the result of should_hide_output(), and then restores the output behavior after that function is done running.

If you want to hide standard error output, then import redirect_stderr from contextlib and add a line saying stack.enter_context(redirect_stderr(null_stream)).

The main downside it that this only works in Python 3.4 and later versions.

Your class will work just fine (with the exception of the write() method name — it needs to be called write(), lowercase). Just make sure you save a copy of sys.stdout in another variable.

If you’re on a *NIX, you can do sys.stdout = open('/dev/null'), but this is less portable than rolling your own class.

You can just mock it.

import mock

sys.stdout = mock.MagicMock()

Why don’t you try this?

sys.stdout.close()
sys.stderr.close()

sys.stdout = None

It is OK for print() case. But it can cause an error if you call any method of sys.stdout, e.g. sys.stdout.write().

There is a note in docs:

Under some conditions stdin, stdout and stderr as well as the original values stdin, stdout and stderr can be None. It is usually the case for Windows GUI apps that aren’t connected to a console and Python apps started with pythonw.

Supplement to iFreilicht’s answer – it works for both python 2 & 3.

import sys

class NonWritable:
    def write(self, *args, **kwargs):
        pass

class StdoutIgnore:
    def __enter__(self):
        self.stdout_saved = sys.stdout
        sys.stdout = NonWritable()
        return self

    def __exit__(self, *args):
        sys.stdout = self.stdout_saved

with StdoutIgnore():
    print("This won't print!")

In my program, user inputs number n, and then inputs n number of strings, which get stored in a list.

I need to code such that if a certain list index exists, then run a function.

This is made more complicated by the fact that I have nested if statements about len(my_list).

Here’s a simplified version of what I have now, which isn’t working:

n = input ("Define number of actors: ")

count = 0

nams = []

while count < n:
    count = count + 1
    print "Define name for actor ", count, ":"
    name = raw_input ()
    nams.append(name)

if nams[2]: #I am trying to say 'if nams[2] exists, do something depending on len(nams)
    if len(nams) > 3:
        do_something
    if len(nams) > 4
        do_something_else

if nams[3]: #etc.

Could it be more useful for you to use the length of the list len(n) to inform your decision rather than checking n[i] for each possible length?

I need to code such that if a certain list index exists, then run a function.

This is the perfect use for a try block:

ar=[1,2,3]

try:
    t=ar[5]
except IndexError:
    print('sorry, no 5')   

# Note: this only is a valid test in this context 
# with absolute (ie, positive) index
# a relative index is only showing you that a value can be returned
# from that relative index from the end of the list...

However, by definition, all items in a Python list between 0 and len(the_list)-1 exist (i.e., there is no need for a try, except if you know 0 <= index < len(the_list)).

You can use enumerate if you want the indexes between 0 and the last element:

names=['barney','fred','dino']

for i, name in enumerate(names):
    print(i + ' ' + name)
    if i in (3,4):
        # do your thing with the index 'i' or value 'name' for each item...

If you are looking for some defined ‘index’ thought, I think you are asking the wrong question. Perhaps you should consider using a mapping container (such as a dict) versus a sequence container (such as a list). You could rewrite your code like this:

def do_something(name):
    print('some thing 1 done with ' + name)

def do_something_else(name):
    print('something 2 done with ' + name)        

def default(name):
    print('nothing done with ' + name)     

something_to_do={  
    3: do_something,        
    4: do_something_else
    }        

n = input ("Define number of actors: ")
count = 0
names = []

for count in range(n):
    print("Define name for actor {}:".format(count+1))
    name = raw_input ()
    names.append(name)

for name in names:
    try:
        something_to_do[len(name)](name)
    except KeyError:
        default(name)

Runs like this:

Define number of actors: 3
Define name for actor 1: bob
Define name for actor 2: tony
Define name for actor 3: alice
some thing 1 done with bob
something 2 done with tony
nothing done with alice

You can also use .get method rather than try/except for a shorter version:

>>> something_to_do.get(3, default)('bob')
some thing 1 done with bob
>>> something_to_do.get(22, default)('alice')
nothing done with alice

len(nams) should be equal to n in your code. All indexes 0 <= i < n “exist”.

It can be done simply using the following code:

if index < len(my_list):
    print(index, 'exists in the list')
else:
    print(index, "doesn't exist in the list")

I need to code such that if a certain list index exists, then run a function.

You already know how to test for this and in fact are already performing such tests in your code.

The valid indices for a list of length n are 0 through n-1 inclusive.

Thus, a list has an index i if and only if the length of the list is at least i + 1.

Using the length of the list would be the fastest solution to check if an index exists:

def index_exists(ls, i):
    return (0 <= i < len(ls)) or (-len(ls) <= i < 0)

This also tests for negative indices, and most sequence types (Like ranges and strs) that have a length.

If you need to access the item at that index afterwards anyways, it is easier to ask forgiveness than permission, and it is also faster and more Pythonic. Use try: except:.

try:
    item = ls[i]
    # Do something with item
except IndexError:
    # Do something without the item

This would be as opposed to:

if index_exists(ls, i):
    item = ls[i]
    # Do something with item
else:
    # Do something without the item

If you want to iterate the inserted actors data:

for i in range(n):
    if len(nams[i]) > 3:
        do_something
    if len(nams[i]) > 4:
        do_something_else

ok, so I think it’s actually possible (for the sake of argument):

>>> your_list = [5,6,7]
>>> 2 in zip(*enumerate(your_list))[0]
True
>>> 3 in zip(*enumerate(your_list))[0]
False

You can try something like this

list = ["a", "b", "C", "d", "e", "f", "r"]

for i in range(0, len(list), 2):
    print list[i]
    if len(list) % 2 == 1 and  i == len(list)-1:
        break
    print list[i+1];

Oneliner:

do_X() if len(your_list) > your_index else do_something_else()  

Full example:

In [10]: def do_X(): 
    ...:     print(1) 
    ...:                                                                                                                                                                                                                                      

In [11]: def do_something_else(): 
    ...:     print(2) 
    ...:                                                                                                                                                                                                                                      

In [12]: your_index = 2                                                                                                                                                                                                                       

In [13]: your_list = [1,2,3]                                                                                                                                                                                                                  

In [14]: do_X() if len(your_list) > your_index else do_something_else()                                                                                                                                                                      
1

Just for info. Imho, try ... except IndexError is better solution.

Do not let any space in front of your brackets.

Example:

n = input ()
         ^

Tip: You should add comments over and/or under your code. Not behind your code.

Have a nice day.

A lot of answers, not the simple one.

To check if a index ‘id’ exists at dictionary dict:

dic = {}
dic['name'] = "joao"
dic['age']  = "39"

if 'age' in dic

returns true if ‘age’ exists.