Develop Reference

Open explorer on a network path

Let's say I have a network path like this below:
\\srv\teams\dir 1
How can I open it using subprocess? I am trying:
subprocess.Popen("explorer '\\srv\teams\dir 1'")
but it always leads me to my 'My Documents'. It works fine from cmd. I am using win7.
I also tried:
os.system("explorer '\\srv\teams\dir 1'")

Please see Mike Scotty for solution regarding os.system. If you use subprocess, please use a list of string for your command instead of a single string:
subprocess.call(['explorer', '\\\\srv\\teams\\dir 1'])
Note that I use subprocess.call instead of subprocess.Popen since this is a simple call, no need to overkill

There are two issues with your code:
1) Use a raw string or escape your \ characters
2) Use " instead of ' to enclose the path
os.system(r'explorer "\\srv\teams\dir 1"')

python: How does subprocess.check_output create it's calls?

I'm trying to read the duration of video files using mediainfo. This shell command works
mediainfo --Inform="Video;%Duration/String3%" file
and produces an output like
00:00:33.600
But when I try to run it in python with this line
subprocess.check_output(['mediainfo', '--Inform="Video;%Duration/String3%"', file])
the whole --Inform thing is ignored and I get the full mediainfo output instead.
Is there a way to see the command constructed by subprocess to see what's wrong?
Or can anybody just tell what's wrong?

Try:
subprocess.check_output(['mediainfo', '--Inform=Video;%Duration/String3%', file])
The " in your python string are likely passed on to mediainfo, which can't parse them and will ignore the option.
These kind of problems are often caused by shell commands requiring/swallowing various special characters. Quotes such as " are often removed by bash due to shell magic. In contrast, python does not require them for magic, and will thus replicate them the way you used them. Why would you use them if you wouldn't need them? (Well, d'uh, because bash makes you believe you need them).
For example, in bash I can do
$ dd of="foobar"
and it will write to a file named foobar, swallowing the quotes.
In python, if I do
subprocess.check_output(["dd", 'of="barfoo"', 'if=foobar'])
it will write to a file named "barfoo", keeping the quotes.

Run R script from python WINDOWS

I made this code and it is working but only in Linux.
import subprocess as sub
sub.Popen([r"Rscript","diccionari.R"])
Where "diccionari.R" is the name of my script in R.
Error text message: System can't found the specific file.
Can somebody help me and do that it works on windows please?
Thank you.

You should probably try the slashes the other way around as how I said it earlier.
Using full path to the .r script (e.g. "C:/myfolder/diccionari.R") instead of just the script file, and using OS independent slashes.

You should specify where Rscriptis located i.e
import subprocess as sub
cmd_line = [r"C:\\Program Files\\R\\R-3.6.0\\bin\\Rscript", "diccionari.R"]
sub.Popen(cmd_line)
watch for the \\ characters

Trouble extracting zip in python over ftp

I'm trying to unzip a file from an FTP site. I've tried it using 7z in a subprocess as well as using 7z in the older os.system format. I get closest however when I'm using the zipfile module in python so I've decided to stick with that. No matter how I edit this I seem to get one of two errors so here are both of them so y'all can see where I'm banging my head against the wall:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall()
NotImplementedError: compression type 6 (implode)
(I think this one is totally wrong, but figured I'd include.)
I seem to get the closest with the following:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
IOError: [Errno 2] No such file or directory: '\\\\svr-dc...'
The catch with this is that it is actually giving me the first file name in the zip. I can see the file AJ07242013.PRN at the end of the error so I feel closer because it's at least getting to the point of reading the contents of the zip file.
Pretty much any iteration of this that I try gets me one of those two errors, or a syntax error but that's easily addressed and not my primary concern.
Sorry for being so long winded. I'd love to get this working, so let me know what you think I need to do.
EDIT:
So 7z has finally been added to the path and is running through without any errors with both the subprocess as well as os.system. However, I still can't seem to get anything to unpack. It looks to me, from all I've read in the python documentation that I should be using the subprocess.communicate() module to extract this file but it just won't unpack. When I use os.system it keeps telling me that it cannot find the archive.
import subprocess
cmd = ['7z', 'e']
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
sp.communicate('r"\C:\Users\boster\Desktop\Data1.zip"')
I don't think that sp.communicate is right but if I add anything else to it I have too many arguments.

python's zipfile doesn't support compression type 6 (imploded) so its simply not going to work. In the first case, that's obvious from the error. In the second case, things are worse. The parameter for extractfile is an alternate unzip directory. Since you gave it the name of your zip file, a directory of the same name can't be found and zipfile gives up before getting to the not-supported problem.
Make sure you can do this with 7z on the command line, try implementing subprocess again and ask for help on that technique if you need it.
Here's a script that will look for 7z in the usual places:
import os
import sys
import subprocess
from glob import glob
print 'python version:', sys.version
subprocess.call('ver', shell=True)
print
if os.path.exists(r'C:\Program Files\7-Zip'):
print 'have standard 7z install'
if '7-zip' in os.environ['PATH'].lower():
print '...and its in the path'
else:
print '...but its not in the path'
print
print 'find in path...'
found = 0
for p in os.environ['PATH'].split(os.path.pathsep):
candidate = os.path.join(p, '7z.*')
for fn in glob(candidate):
print ' found', fn
found += 1
print
if found:
print '7z located, attempt run'
subprocess.call(['7z'])
else:
print '7z not found'

Accoring to the ZipFile documentation, you might be better off copying the zip first to your working directory. (http://docs.python.org/2/library/zipfile#zipfile.ZipFile.extract)
If you have problems copying, you might want to store the zip in a path with no spaces or protect your code against spaces by using os.path.
I made a small test in which I used os.path.abspath to make sure I had the proper path to my zip and it worked properly.
Also make sure that for extractall the path that you specify is the path where the zip content will be extracted. (If a folder that is specified is not created, it will be created automatically) Your files will be extracted in your current working directory (CWD) if no parameter is passed to extractall.
Cheers!

Managed to get this to work without using the PIPE functionality as subprocess.communicate wouldn't unpack the files. Here was the solution using subprocess.call. Hope this can help someone in the future.
def extract_data_one():
for item in sites:
os.chdir(r"\\svr-dc\ftp site\%s\Daily" % item)
subprocess.call(['7z', 'e', 'data1.zip', '*.*'])

How to clear output in Vte.Terminal?

How to clear all output in Vte.Terminal ?

use the os library
import os
os.system('clear')

It would only clear the text visible in the terminal, but, you can run the cls command or tinker with the feed() method and empty spaces.

I have been having some problems of my own with Vte so I don't know if I'm the right person to answer.
Have you tried replacing the old terminal with another in the container? It's not clearing, but there will be an empty terminal.

vte.feed('\033[2J')
For other escape sequences:
Bash Prompt HOWTO
ANSI escape code (Wikipedia)

I'm using
vte.fork_command("clear")