Very simple question. I am using the IDLE Python shell to run my Python scripts. I use the following type of structure
import sys
sys.argv = ['','fileinp']
execfile('mypythonscript.py')
Is there a simple way of outputting the results to a file? Something like
execfile('mypythonscript.py') > 'output.dat'
Thanks
$ python
Python 2.6.5 (r265:79063, Apr 16 2010, 13:57:41)
[GCC 4.4.3] on linux2
>>> import sys
>>> sys.displayhook(stdout)
<open file '<stdout>', mode 'w' at 0x7f8d3197a150>
>>> x=open('myFile','w')
>>> sys.displayhook(x)
<open file 'myFile', mode 'w' at 0x7fb729060c00>
>>> sys.stdout=x
>>> print 'changed stdout!'
>>> x.close()
$ cat myFile
changed stdout!
NOTE: Changing these objects doesn’t affect the standard I/O streams of processes executed by os.popen(), os.system() or the exec*() family of functions in the os module.
So
>>> import os
>>> os.system("./x")
1 #<-- output from ./x
0 #<-- ./x's return code
>>> quit()
$
sys.displayhook(stdout) be sys.displayhook(sys.stdout)? When I try to run that line I get "NameError: name 'stdout' is not defined"So sayeth the documentation:
Standard output is defined as the file object named stdout in the built-in module sys.
So you can change stdout like so:
import sys
sys.stdout = open("output.dat", "w")
