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...
Related
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')
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!
Is there an option I can pass open() that will cause an IOerror when trying to write a nonexistent file? I am using python to read and write block devices via symlinks, and if the link is missing I want to raise an error rather than create a regular file. I know I could add a check to see if the file exists and manually raise the error, but would prefer to use something built-in if it exists.
Current code looks like this:
device = open(device_path, 'wb', 0)
device.write(data)
device.close()
Yes.
open(path, 'r+b')
Specifying the "r" option means the file must exist and you can read.
Specifying "+" means you can write and that you will be positioned at the end.
https://docs.python.org/3/library/functions.html?#open
Use os.path.islink() or os.path.isfile() to check if the file exists.
Doing the check each time is a nuisance, but you can always wrap open():
import os
def open_if_exists(*args, **kwargs):
if not os.path.exists(args[0]):
raise IOError('{:s} does not exist.'.format(args[0]))
f = open(*args, **kwargs)
return f
f = open_if_exists(r'file_does_not_exist.txt', 'w+')
This is just quick and dirty, so it doesn't allow for usage as: with open_if_exists(...).
Update
The lack of a context manager was bothering me, so here goes:
import os
from contextlib import contextmanager
#contextmanager
def open_if_exists(*args, **kwargs):
if not os.path.exists(args[0]):
raise IOError('{:s} does not exist.'.format(args[0]))
f = open(*args, **kwargs)
try:
yield f
finally:
f.close()
with open_if_exists(r'file_does_not_exist.txt', 'w+') as f:
print('foo', file=f)
I am afraid you can't perform the check of file existence and raise error using the open() function.
Below is the signature of open() in python where name is the file_name, mode is the access mode and buffering to indicate if buffering is to be performed while accessing a file.
open(name[, mode[, buffering]])
Instead, you can check if the file exists or not.
>>> import os
>>> os.path.isfile(file_name)
This will return True or False depending on if the file exists. To test a file specifically, you can use this.
To test the existence of both files and directories, you can use:
>>> os.path.exists(file_path)
I have a Matlab application that writes in to a .csv file and a Python script that reads from it. These operations happen concurrently and at their own respective periods (not necessarily the same). All of this runs on Windows 7.
I wish to know :
Would the OS inherently provide some sort of locking mechanism so that only one of the two applications - Matlab or Python - have access to the shared file?
In the Python application, how do I check if the file is already "open"ed by Matlab application? What's the loop structure for this so that the Python application is blocked until it gets access to read the file?
I am not sure about window's API for locking files
Heres a possible solution:
While matlab has the file open, you create an empty file called "data.lock" or something to that effect.
When python tries to read the file, it will check for the lock file, and if it is there, then it will sleep for a given interval.
When matlab is done with the file, it can delete the "data.lock" file.
Its a programmatic solution, but it is simpler than digging through the windows api and finding the right calls in matlab and python.
If Python is only reading the file, I believe you have to lock it in MATLAB because a read-only open call from Python may not fail. I am not sure how to accomplish that, you may want to read this question atomically creating a file lock in MATLAB (file mutex)
However, if you are simply consuming the data with python, did you consider using a socket instead of a file?
In Windows on the Python side, CreateFile can be called (directly or indirectly via the CRT) with a specific sharing mode. For example, if the desired sharing mode is FILE_SHARE_READ, then the open will fail if the file is already open for writing. If the latter call instead succeeds, then a future attempt to open the file for writing will fail (e.g. in Matlab).
The Windows CRT function _wsopen_s allows setting the sharing mode. You can call it with ctypes in a Python 3 opener:
import sys
import os
import ctypes as ctypes
import ctypes.util
__all__ = ['shdeny', 'shdeny_write', 'shdeny_read']
_SH_DENYRW = 0x10 # deny read/write mode
_SH_DENYWR = 0x20 # deny write mode
_SH_DENYRD = 0x30 # deny read
_S_IWRITE = 0x0080 # for O_CREAT, a new file is not readonly
if sys.version_info[:2] < (3,5):
_wsopen_s = ctypes.CDLL(ctypes.util.find_library('c'))._wsopen_s
else:
# find_library('c') may be deprecated on Windows in 3.5, if the
# universal CRT removes named exports. The following probably
# isn't future proof; I don't know how the '-l1-1-0' suffix
# should be handled.
_wsopen_s = ctypes.CDLL('api-ms-win-crt-stdio-l1-1-0')._wsopen_s
_wsopen_s.argtypes = (ctypes.POINTER(ctypes.c_int), # pfh
ctypes.c_wchar_p, # filename
ctypes.c_int, # oflag
ctypes.c_int, # shflag
ctypes.c_int) # pmode
def shdeny(file, flags):
fh = ctypes.c_int()
err = _wsopen_s(ctypes.byref(fh),
file, flags, _SH_DENYRW, _S_IWRITE)
if err:
raise IOError(err, os.strerror(err), file)
return fh.value
def shdeny_write(file, flags):
fh = ctypes.c_int()
err = _wsopen_s(ctypes.byref(fh),
file, flags, _SH_DENYWR, _S_IWRITE)
if err:
raise IOError(err, os.strerror(err), file)
return fh.value
def shdeny_read(file, flags):
fh = ctypes.c_int()
err = _wsopen_s(ctypes.byref(fh),
file, flags, _SH_DENYRD, _S_IWRITE)
if err:
raise IOError(err, os.strerror(err), file)
return fh.value
For example:
if __name__ == '__main__':
import tempfile
filename = tempfile.mktemp()
fw = open(filename, 'w')
fw.write('spam')
fw.flush()
fr = open(filename)
assert fr.read() == 'spam'
try:
f = open(filename, opener=shdeny_write)
except PermissionError:
fw.close()
with open(filename, opener=shdeny_write) as f:
assert f.read() == 'spam'
try:
f = open(filename, opener=shdeny_read)
except PermissionError:
fr.close()
with open(filename, opener=shdeny_read) as f:
assert f.read() == 'spam'
with open(filename, opener=shdeny) as f:
assert f.read() == 'spam'
os.remove(filename)
In Python 2 you'll have to combine the above openers with os.fdopen, e.g.:
f = os.fdopen(shdeny_write(filename, os.O_RDONLY|os.O_TEXT), 'r')
Or define an sopen wrapper that lets you pass the share mode explicitly and calls os.fdopen to return a Python 2 file. This will require a bit more work to get the file mode from the passed in flags, or vice versa.
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)