I am getting an Access is denied error while I am trying to run the .pyo file by double click or from the command prompt.
Lets say I have abc.py (keeping main method entry point) which imports files xyz.py and imports wx etc.
I generate the .pyo file. But once I try to run abc.pyo I get the access is denied error.
I am not getting why this happening? Any help will really appreciated.
(I am using windows xp as os).
I am making .pyo from .py as following.
I am having a .bat file CompileAllToPyo.bat which have
python -O Compileall.py
The Compileall.py keep the follwoing things
import os
import compileall
os.popen3(cmdLine, 'b')
compileall.compile_dir('.', force=1)
This is all the info
Thanks
You can tell the system that your hw.pyo file is "executable", for example (in Linux, MacOSX, or any other Unix-y system) by executing the command chmod +w hw.pyo at the terminal shell prompt. Consider, for example, the following short and simple shell session:
$ cat >hw.py
print('hello world')
$ python2.5 -O -c'import hw'
hello world
$ ./hw.pyo
bash: ./hw.pyo: Permission denied
$ chmod +x hw.pyo
$ ./hw.pyo
hello world
$
By default, .pyo (and .pyc) files are not marked as executable because they're mostly meant to be imported, not directly executed (indeed, note that we're explicitly using a Python import statement to create the .pyo file!); however, as this example shows, it's quite easy to make one of them "executable as the main script". BTW, observe also:
$ cat >hw.py
print('hello world from ' + __name__)
$ python2.5 -O -c'import hw'
hello world from hw
$ chmod +x hw.pyo
$ ./hw.pyo
hello world from __main__
$
The __name__ is what tells the module whether it's being imported (so the first "hello world" says "from hw") or run as the main script (so the second one says "from __main__"). That's the reason modules that are designed to be used both ways normally end with if __name__ == '__main__': main() or the like, where main is a function that, this way, gets called iff the module's running as the main script (it's always best to have all substantial code execute in a function, not at a module's top level).
You don't "run" a .pyo file, as it's not an executable. You can give it to the python interpreter in lieu of the .py file, but in general, you should use a .py file as your entry point, so that the .pyc or .pyo file can be recreated when necessary.
$ python imported.pyo
Success!
$ ./imported.pyo
bash: ./imported.pyo: Permission denied
Related
In my perl program which runs the python script
I have provided the PYTHONPATH env param with the path for the lib and i have run the python script. I am getting
ImportError: No module named "....."
my $script = "/path/pythonscript.py";
$ENV{'PYTHONPATH'} = "/path/lib";
system("python $script");
Whereas when i run the same python script on command line in the same directory where the script executes in my perl program, it is working.
Can anyone give me some pointers on why this is happening.
Try printing the contents of sys.path and compare the difference e.g. change your python script to
import sys
print(sys.path)
Most likely there is a difference here and this is causing the module to not be found.
I once had a similar problem. I solved it by creating an executable script (chmod) and making that script run instead of the python script. The script simply contained a cd to the directory and a python3 program. py
my current bash shell is loaded with python3 but occasionally I tend to run old python2 scripts and I get this error
python/2.7.10/bin/python: error while loading shared libraries: libpython2.7.so.1.0: cannot open shared object file: No such file or directory"
so every-time I had to load python2 back and forth. Is there a way to run python2 script in python3 environment by just changing any environmental variables?
Are you open to virtual envs? Otherwise you can run python2 or python3 from the command line.
As an example, create the following file, hello_world.py:
#!/usr/bin/env python
print "Hello, World!"
Then you can run from the bash shell:
python2 hello_world.py
That should work no problem. However, if you run:
python3 hello_world.py
You will run into the error:
SyntaxError: Missing parentheses in call to 'print'. Did you mean
print("Hello, world!")?
I have a bash script which changes the path on my command line,
This one,
#!/usr/bin/env python
cd /mnt/vvc/username/deployment/
I have a python script which i wish to run after the path changes to the desired path,
The script,
#!/usr/bin/env python
import subprocess
import os
subprocess.call(['/home/username/new_file.sh'])
for folder in os.listdir(''):
print ('deploy_predict'+' '+folder)
I get this
File "/home/username/new_file.sh", line 2
cd /mnt/vvc/username/deployment/
^
SyntaxError: invalid syntax
Any suggestions on how can i fix this?thanks in advance
You need to explicitly tell subprocess which shell to run the sh file with. Probably one of the following:
subprocess.call(['sh', '/home/username/new_file.sh'])
subprocess.call(['bash', '/home/username/new_file.sh'])
However, this will not change the python program's working directory as the command is run in a separate context.
You want to do this to change the python program's working directory as it runs:
os.chdir('/mnt/vvc/username/deployment/')
But that's not really great practice. Probably better to just pass the path into os.listdir, and not change working directories:
os.listdir('/mnt/vvc/username/deployment/')
I'm facing of a strange issue, and after a couple of hour of research I'm looking for help / explanation about the issue.
It's quite simple, I wrote a cgi server in python and I'm working with some libs including pynetlinux for instance.
When I'm starting the script from terminal with any user, it works fine, no bug, no dependency issue. But when I'm trying to start it using a script in rc.local, the following code produce an error.
import sys, cgi, pynetlinux, logging
it produce the following error :
Traceback (most recent call last):
File "/var/simkiosk/cgi-bin/load_config.py", line 3, in
import cgi, sys, json, pynetlinux, loggin
ImportError: No module named pynetlinux
Other dependencies produce similar issue.I suspect some few things like user who executing the script in rc.local (root normaly) and trying some stuff found on the web without success.
Somebody can help me ?
Thanx in advance.
Regards.
Ollie314
First of all, you need to make sure if the module you want to import is installed properly. You can check if the name of the module exists in pip list
Then, in a python shell, check what the paths are where Python is looking for modules:
import sys
sys.path
In my case, the output is:
['', '/usr/lib/python3.4', '/usr/lib/python3.4/plat-x86_64-linux-gnu', '/usr/lib/python3.4/lib-dynload', '/usr/local/lib/python3.4/dist-packages', '/usr/lib/python3/dist-packages']
Finally, append those paths to $PATH variable in /etc/rc.local. Here is an example of my rc.local:
#!/bin/sh -e
#
# rc.local
#
# This script is executed at the end of each multiuser runlevel.
# Make sure that the script will "exit 0" on success or any other
# value on error.
#
# In order to enable or disable this script just change the execution
# bits.
#
# By default this script does nothing
export PATH="$PATH:/usr/lib/python3.4:/usr/lib/python3.4/plat-x86_64-linux-gnu:/usr/lib/python3.4/lib-dynload:/usr/local/lib/python3.4/dist-packages:/usr/lib/python3/dist-packages"
# Do stuff
exit 0
The path where your modules are install is probably normally sourced by .bashrc or something similar. .bashrc doesn't get sourced when it's not an interactive shell. /etc/profile is one place that you can put system wide path changes. Depending on what Linux version/distro it may use /etc/profile.d/ in which case /etc/profile runs all the scripts in /etc/profile.d, add a new shell script there with execute permissions and a .sh extention.
I'm on Mac OS 10.6 Snow Leopard and I'm trying to add a directory to my PATH variable so I can run a tiny script I wrote by just typing: python alarm.py at the terminal prompt.
I put the path in my .profile file and it seems to show up when I echo $PATH, but python still can't find script the that I've put in that directory.
Here's the contents of my .profile file in my home directory:
~ toby$ vim .profile
export PATH=/Users/tobylieven/Documents/my_scripts:$PATH
Here's the output of echo $PATH, where all seems well:
~ toby$ echo $PATH
/Users/tobylieven/Documents/my_scripts:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/X11/bin
Here's the script I'm trying to run:
~ toby$ ls /Users/tobylieven/Documents/my_scripts
-rwxrwxrwx# 1 tobylieven staff 276 17 Jan 21:17 alarm.py
Here's the command I'm trying to use to run the script and the fail message I'm getting instead:
~ toby$ python alarm.py
python: can't open file 'alarm.py': [Errno 2] No such file or directory
If anyone has an idea what I might be doing wrong, that'd be great.
Thanks a lot.
PATH is only for executables, not for python scripts. Add the following to the beginning of your Python script:
#!/usr/bin/env python
and run
sudo chmod a+x /Users/tobylieven/Documents/my_scripts/alarm.py
Then, you can type just alarm.py to execute your program.
Which python are you targeting?
Did you install it with brew? It uses a different path.
which python3 or which python
Choose the one you want
Copy that output
Paste it at the top of your python file
add a #! in front of that path so it looks something like
#!/usr/local/bin/python3
Make sure to change the file permissions
chmod +x filename
Put that file in a folder that is in your path
Not sure if your folder is in your path?
echo $path
How to add that folder to your path?
Find your path first
echo $HOME
If you are using bash or zsh you might have something like this
In ~/.bash_profile or ~/.bashrc or ~/.zshrc at the bottom of your file
export PYTHON_UTILS="$HOME/code/python/utils"
export PATH="$PYTHON_UTILS:$PATH"
Consider removing the .py from your file bc it is not needed in this case
Close and open your terminal, which is sourcing your file by its path
And now you should be able to treat your python file similar to a bash command
You don't need to use python3 filename.py to run the file, you can just use filename
From anywhere on your filesystem!
change alarm.py to include:
#!/bin/python
as the very first line in the file.
(or /usr/bin/python, depending on where you python interpreter is located. You can figure this out by typing: which python in the terminal.)
You can then just run alarm.py instead of python alarm.py.
e.g.:
~ toby$ alarm.py
And phihag who beat me by a few seconds is right, you need to add execute permissions (via chmod) to alarm.py.
You need to modify the Python specific path variable: PYTHONPATH.
So:
export PYTHONPATH=/Users/tobylieven/Documents/my_scripts
should get you working.
See: Python Module Search path
Something of interest that I really struggled with on OS X coming from Window, is that you its very hard to get the directory of your current script.
I found this.
#! /bin/zsh
cd "${0:h}"
Now you could execute a python file relative to the executed script instead of having to know the exact path where your python file is.
This might or might not help but I use this a lot to make my scripts and .command files work better.