I know that if you want to redirect stdout to a file, you can simply do it like this.
sys.stdout = open(fpath, 'w')
But how can I switch back stdout to write on the terminal?
You can assign it to variable and later assing it back
temp = sys.stdout
print('console')
sys.stdout = open('output.txt', 'w')
print('file')
sys.stdout = temp
print('console')
You can also find examples how to use it with context manager so you can change it using with
import sys
from contextlib import contextmanager
#contextmanager
def custom_redirection(fileobj):
old = sys.stdout
sys.stdout = fileobj
try:
yield fileobj
finally:
sys.stdout = old
# ---
print('console')
with open('output.txt', 'w') as out:
with custom_redirection(out):
print('file')
print('console')
Code from: Python 101: Redirecting stdout
Currently you can even find redirect_stdout in contextlib
import sys
from contextlib import redirect_stdout
print('console')
with open('output.txt', 'w') as out:
with redirect_stdout(out):
print('file')
print('console')
BTW: if you want to redirect all text to file then you can use system/shell for this
$ python script.py > output.txt
A better bet is to simply write to the file when you want.
with open('samplefile.txt', 'w') as sample:
print('write to sample file', file=sample)
print('write to console')
reassigning the stdout would mean you need to track the previous file descriptor and assign it back whenever you want to send text to the console.
If you really must reassign you could do it like this.
holder = sys.stdout
sys.stdout = open(fpath, 'w')
print('write something to file')
sys.stdout = holder
print('write something to console')
Related
I tried this code posted 2 years ago:
import subprocess
with open("output.txt", "wb") as f:
subprocess.check_call(["python", "file.py"], stdout=f)
import sys
import os.path
orig = sys.stdout
with open(os.path.join("dir", "output.txt"), "wb") as f:
sys.stdout = f
try:
execfile("file.py", {})
finally:
sys.stdout = orig
It hangs up the terminal until I ctl-z and then it crashes the terminal but prints the output.
I'm new to coding and am not sure how to resolve. I'm obviously doing something wrong. Thanks for your help.
You can simply open and write to the file with write.
with open('output.txt', 'w') as f:
f.write('output text') # You can use a variable from other data you collect instead if you would like
Since you are new to coding, i'll just let you know that opening a file using with will actually close it automatically after the indented code is ran. Good luck with your project!
I'm having trouble writing the terminal output (all print statements) to a textfile then reading that textfile in the same script. I keep getting an I/O error if I close the program to finish writing to the file and then re-open the file to read it, or no output for the final print(file_contents) statement.
Here's my code:
import sys
filename = open("/Users/xxx/documents/python/dump.txt", 'r+')
filename.truncate()
sys.stdout = filename
print('Hello')
print('Testing')
filename.close()
with open("/Users/xxx/documents/python/dump.txt") as file:
data = file.read()
print(file)
Any suggestions would be great! I'm planning to use this to print output's from some longer scripts to a slack channel.
Thanks!
The error you get is:
IOError: [Errno 2] No such file or directory: '/Users/xxx/documents/python/dump.txt' because:
file open mode r+ does not create a file. Use mode w like this:
You have to reattach stdout to console again to print in console.
import sys
filename = open('/Users/xxx/documents/python/dump.txt', 'w')
# filename.truncate() # mode 'w' truncates file
sys.stdout = filename
print('Hello')
print('Testing')
filename.close()
# reattach stdout to console
sys.stdout = sys.__stdout__
with open('/Users/xxx/documents/python/dump.txt') as file:
data = file.read()
print(data)
will print:
Hello
Testing
The problem is you redirect sys.stdout to filename, and then you close the file. Afterwards you can't print anything anymore, since the file is closed.
sys.stdout = filename
..
..
filename.close()
with open("/Users/xxx/documents/python/dump.txt") as file:
data = file.read()
print(file)
The last print statement tries to print output to sys.stdout, which is a closed file.
If you want to get the old behavior back, you need to keep a reference to sys.stdout. This will solve it:
sys_out = sys.stdout
sys.stdout = filename
..
..
filename.close()
sys.stdout = sys_out
with open("/Users/xxx/documents/python/dump.txt") as file:
data = file.read()
print(file)
import sys
filename = open("/Users/xxx/documents/python/dump.txt", 'w')
sys_out = sys.stdout
sys.stdout = filename
print('Hello')
print('Testing')
print('Test')
filename.close()
sys.stdout = sys_out
with open("/Users/xxx/documents/python/dump.txt", 'r') as file:
data = file.read()
print(data)
I have a code use print() to write in file:
with open('test.xml', "w+") as outfile:
sys.stdout = outfile
Now I want write to console after this code, how can I do this?
You can restore sys.stdout from sys.__stdout__:
with open('test.xml', "w+") as outfile:
sys.stdout = outfile
sys.stdout = sys.__stdout__
or you could store the original up-front:
orig_stdout = sys.stdout
with open('test.xml', "w+") as outfile:
sys.stdout = outfile
sys.stdout = orig_stdout
You could use a context manager here:
from contextlib import contextmanager
#contextmanager
def redirect_stdout(filename):
orig_stdout = sys.stdout
try:
with open(filename, "w+") as outfile:
sys.stdout = outfile
yield
finally:
sys.stdout = orig_stdout
then use that in your code:
with redirect_stdout('test.xml'):
# stdout is redirected in this block
# stdout is restored afterwards
store stdout in a variable
stdout = sys.stdout
with open('test.xml', 'w+') as outfile:
sys.stdout = outfile
print("<text>Hello World</text>") # print to outfile instead of stdout
sys.stdout = stdout # now its back to normal
While this works really you should just be writing to the file directly
with open('test.xml', 'w+') as outfile
outfile.write("<text>Hello World</text">)
I always open and write into files using with statement:
with open('file_path', 'w') as handle:
print >>handle, my_stuff
However, there is one instance where I need to be able to be more flexible, and write to sys.stdout (or other types of streams), if that is provided instead of file path:
So, my question is this: Is there a way for using with statement both with real files and with sys.stdout?
Note that I can use the following code, but I think this defeats the purpose of using with:
if file_path != None:
outputHandle = open(file_path, 'w')
else:
outputHandle = sys.stdout
with outputHandle as handle:
print >>handle, my_stuff
You can create a context manager and use it like this
import contextlib, sys
#contextlib.contextmanager
def file_writer(file_name = None):
# Create writer object based on file_name
writer = open(file_name, "w") if file_name is not None else sys.stdout
# yield the writer object for the actual use
yield writer
# If it is file, then close the writer object
if file_name != None: writer.close()
with file_writer("Output.txt") as output:
print >>output, "Welcome"
with file_writer() as output:
print >>output, "Welcome"
If you don't pass any input to file_writer it will use sys.stdout.
Thing is, you don't need to use a context processor with stdout, because you're not opening or closing it. A less fancy way of abstracting this is:
def do_stuff(file):
# Your real code goes here. It works both with files or stdout
return file.readline()
def do_to_stdout():
return do_stuff(sys.stdout)
def do_to_file(filename):
with open(filename) as f:
return do_stuff(f)
print do_to_file(filename) if filename else do_to_stdout()
The simplest way is to simply use "old school" streamed filenames, that way your code doesn't have to change. In Unix this is "/dev/tty" or in Windows this is "con" (although there are other choices for both platforms).
if default_filename is None:
default_filename = "/dev/tty"
with open(default_filename, 'w') as handle:
handle.write("%s\n" % my_stuff)
This code tested in Python 2.7.3 and 3.3.5
With python3 optional closefd argument is recognized.
If set to False, resulting IO object won't close underlying fd:
if file_path != None:
outputHandle = open(file_path, 'w')
else:
outputHandle = open(sys.stdout.fileno(), 'w', closefd=False)
with outputHandle as handle:
print(my_stuff, file=handle)
I have a script that accepts as an argument a filename than opens it and writes some stuff.
I use the with statement:
with open(file_name, 'w') as out_file:
...
out_file.write(...)
Now what if I want to write to sys.stdout if no file_name is provided?
Do I necessarily need to wrap all actions in a function and put a condition before?
if file_name is None:
do_everything(sys.stdout)
else:
with open(file_name, 'w') as out_file:
do_everything(out_file)
from contextlib import contextmanager
#contextmanager
def file_or_stdout(file_name):
if file_name is None:
yield sys.stdout
else:
with open(file_name, 'w') as out_file:
yield out_file
Then you can do
with file_or_stdout(file_name) as wfile:
do_stuff_writing_to(wfile)
How do you handle command line arguments? If you use argparse you could use the type and default parameters of add_argument to handle this. For example, try something like the following:
import sys
import argparse
def main(argv=None):
if argv is None:
argv=sys.argv[1:]
parser = argparse.ArgumentParser()
parser.add_argument('infile', nargs='?',
type=argparse.FileType('w'),
default=sys.stdin)
args = parser.parse_args(argv)
print args.infile
return 0
if __name__=="__main__":
sys.exit(main(sys.argv[1:]))
If a file name is present as an argument the the script argparse will automatically open and close this file and args.infile will be a handle to this file. Otherwise args.infile will simply be sys.stdin.
You could write your own context manager. I'll post sample code later if noone else does
if file_name is None:
fd = sys.stdout
else:
fd = open(file_name, 'w')
# write to fd
if fd != sys.stdout:
fd.close();
Using the with ... as construct is useful to close the file automatically. This means that using it with sys.stdout, as I guess you know, would crash your program, because it would attempt at closing the system stdout!
This means something like with open(name, 'w') if name else sys.stdout as: would not work.
This make me say there isn't any simple-nice way to write your snippet better... but there are probably better ways to think on how to construct such a code!
The main point to clarify is when you need to open (and, more importantly, close) the filehandler for file_name, when file_name exists.
Personally I would simply drop the with .. as and take care of opening the file - and, more importantly, close it! - somewhere else. Mileage for that might vary depending on how your software is working.
This means you can simply do:
out_file = open(file_name, 'w') if file_name else sys.stdout
and work with out_file throughout your program.
When you close, remember to check if it's a file or not :)
And have you thought about simply using the logging module? That easily allows you to add different handlers, print to file, print to stdout...