I would like to know if there is a way I could check from my python code if matlab exists on a system. So far the only thing I can come up with is: exists = os.system("matlab") and then parse the exists for a command not found. But I'm almost sure this will:
Start matlab in case it exists on the system. I don't want this.
The response may vary depending on the system I'm running ?
So is there any way I could check if a matlab installation is available on the system from python ?
Regards,
Bogdan
Assuming your system call works, you can check the path for matlab.exe like this:
import os
def matlab_installed():
for path in os.environ["PATH"].split(";"):
if os.path.isfile(os.path.join(path, "matlab.exe")):
return True
return False
For Unix, you have to change split(";") to split(":") and "matlab.exe" to whatever the matlab executable is called under Unix.
Another way would be to to use shutil
import shutil
mt = shutil.which("matlab")
If 'matlab' is found it returns the path where it was found, else it returns 'NoneType'. You may check for 'matlab' or 'matlab.exe' depending on the OS.
Related
Is there a way to obtain the original argv value from python after execution code like this:
sys.argv[0]=""
In fact I search for the unwritable version of sys.argv.
I would prefer a OS independent solution, but I'm working on windows, so a windows only solution would be sufficient
On Linux you could use:
open("/proc/{}/cmdline".format(os.getpid())).read()[:-1].split("\x00")[1:]
How can i get the current directory to which I am in? like the use of
os.getcwd()
First, I presume you're not asking about a particular subprocess that exists simply to tell you the current working directory and do nothing else (Apducer's answer). If that were the case you could simply as os.getcwd() and forget the subprocess. You clearly already know that. So you must be dealing with some other (arbitrary?) subprocess.
Second, I presume you understand, via dr1fter's answer, that you have control over the working directory in which the subprocess starts. I suspect that's not enough for you.
Rather, I suspect you're thinking that the subprocess might, according to its own internal logic, have changed its working directory sometime since its launch, that you can't predict where it has ended up, and you want to be able to send some sort of signal to the subprocess at an arbitrary time, to interrogate it about where it's currently working. In general, this is only possible if the process has been specifically programmed with the logic that receives such a signal (through whatever route) and issues such a response. I think that's what SuperStew meant by the comment, "isn't that going to depend on the subprocess?"
I say "in general" because there are platform-specific approaches. For example, see:
windows batch command to determine working directory of a process
How do I print the current working directory of another user in linux?
by default, subprocesses you spawn inherit your PWD. you can however, specify the cwd argument to the subprocess.Popen c'tor to set a different initial PWD.
Unix (Linux, MacOS):
import subprocess
arguments = ['pwd']
directory = subprocess.check_output(arguments)
Windows:
import subprocess
arguments = ['cd']
directory = subprocess.check_output(arguments)
If you want to run in both types of OS, you'll have to check the machine OS:
import os
import subprocess
if os.name == 'nt': # Windows
arguments = ['cd']
else: # other (unix)
arguments = ['pwd']
directory = subprocess.check_output(arguments)
I am writing a script in Python 2.7.
It needs to be able to go whoever the current users profile in Windows.
This is the variable and function I currently have:
import os
desired_paths = os.path.expanduser('HOME'\"My Documents")
I do have doubts that this expanduser will work though. I tried looking for Windows Env Variables to in Python to hopefully find a list and know what to convert it to. Either such tool doesn't exist or I am just not using the right search terms since I am still pretty new and learning.
You can access environment variables via the os.environ mapping:
import os
print(os.environ['USERPROFILE'])
This will work in Windows. For another OS, you'd need the appropriate environment variable.
Also, the way to concatenate strings in Python is with + signs, so this:
os.path.expanduser('HOME'\"My Documents")
^^^^^^^^^^^^^^^^^^^^^
should probably be something else. But to concatenate paths you should be more careful, and probably want to use something like:
os.sep.join(<your path parts>)
# or
os.path.join(<your path parts>)
(There is a slight distinction between the two)
If you want the My Documents directory of the current user, you might try something like:
docs = os.path.join(os.environ['USERPROFILE'], "My Documents")
Alternatively, using expanduser:
docs = os.path.expanduser(os.sep.join(["~","My Documents"]))
Lastly, to see what environment variables are set, you can do something like:
print(os.environ.keys())
(In reference to finding a list of what environment vars are set)
Going by os.path.expanduser , using a ~ would seem more reliable than using 'HOME'.
Let's focus on one dll: C:\Windows\System32\wbem\wmiutils.dll. Why? Because it's the file in which I personally discovered Windows delivers a different dll depending on process architecture.
TLDR; Is there a way to programmatically determine the actual path of the dll that was returned by the file system redirector?
I understand that if launched as a x86 process, I get C:\Windows\SysWOW64\wbem\wmiutils.dll. And, if launched as a x64 process, I get C:\Windows\System32\wbem\wmiutils.dll.
I need to determine which wmiutils.dll I'm actually looking at. The redirector makes system32\wbem\wmiutils.dll look and feel identical but it's not. If I use parent path, I get C:\Windows\System32\wbem even though I may/may not be looking at C:\Windows\SysWOW64\wbem.
Any sweet python magic to make this happen? I can't seem to see anything from other languages I can port. Based on my use case, I've come up with a couple hacks but they're just that. Hoping somebody has found a solution as easy as parent path that actually works in this case.
import ctypes, hashlib
k32 = ctypes.windll.kernel32
oldValue = ctypes.c_long(0)
k32.Wow64DisableWow64FsRedirection(ctypes.byref(oldValue)) # Should open 32-bit
with open(r"C:\Windows\System32\wbem\wmiutil.dll", "rb") as f:
checksum32 = hashlib.md5(f.read()).hexdigest()
k32.Wow64RevertWow64FsRedirection(oldValue) # Should use what Windows thinks you need
with open(r"C:\Windows\System32\wbem\wmiutil.dll", "rb") as f:
checksum64 = hashlib.md5(f.read()).hexdigest()
if (checksum32 != checksum64):
print("You're running 64bit wmiutil dll")
I don't have Windows Python to test this, but it should work according to https://msdn.microsoft.com/en-us/library/windows/desktop/aa365745%28v=vs.85%29.aspx.
I think an easier way would be to just do some test like creating a struct and seeing if it's 8 bytes or 4 bytes. Then you can assume that Windows is using the 64-bit version of DLLs if it's 8 bytes.
I see that if we change the HOME (linux) or USERPROFILE (windows) environmental variable and run a python script, it returns the new value as the user home when I try
os.environ['HOME']
os.exp
Is there any way to find the real user home directory without relying on the environmental variable?
edit:
Here is a way to find userhome in windows by reading in the registry,
http://mail.python.org/pipermail/python-win32/2008-January/006677.html
edit:
One way to find windows home using pywin32,
from win32com.shell import shell,shellcon
home = shell.SHGetFolderPath(0, shellcon.CSIDL_PROFILE, None, 0)
I think os.path.expanduser(path) could be helpful.
On Unix and Windows, return the argument with an initial component of ~ or ~user replaced by that user‘s home directory.
On Unix, an initial ~ is replaced by the environment variable HOME if it is set; otherwise the current user’s home directory is looked up in the password directory through the built-in module pwd. An initial ~user is looked up directly in the password directory.
On Windows, HOME and USERPROFILE will be used if set, otherwise a combination of HOMEPATH and HOMEDRIVE will be used. An initial ~user is handled by stripping the last directory component from the created user path derived above.
If the expansion fails or if the path does not begin with a tilde, the path is returned unchanged.
So you could just do:
os.path.expanduser('~user')
from pathlib import Path
str(Path.home())
works in Python 3.5 and above. Path.home() returns a Path object providing an API I find very useful.
I think os.path.expanduser(path) is the best answer to your question, but there's an alternative that may be worth mentioning in the Unix world: the pwd package. e.g.
import os, pwd
pwd.getpwuid(os.getuid()).pw_dir
For windows;
import os
homepath = os.path.expanduser(os.getenv('USERPROFILE'))
will give you a handle to current user's home directory and
filepath = os.path.expanduser(os.getenv('USERPROFILE'))+'\\Documents\\myfile.txt'
will give you a handle to below file;
C:\Users\urUserName\Documents\myfile.txt
home_folder = os.getenv('HOME')
This should work on Windows and Mac OS too, works well on Linux.
Really, a change in environment variable indicates that the home must be changed. So every program/script should have the new home in context; also the consequences are up to the person who changed it.
I would still stick with
home = os.getenv('USERPROFILE') or os.getenv('HOME')
what exactly is required?
I realize that this is an old question that has been answered but I thought I would add my two cents. The accepted answer was not working for me. I needed to find the user directory and I wanted it to work with and without sudo. In Linux, my user directory is "/home/someuser" but my root directory is "/root/". However, on my Mac, the user directory is "/Users/someuser". Here is what I ended up doing:
_USERNAME = os.getenv("SUDO_USER") or os.getenv("USER")
_HOME = os.path.expanduser('~'+_USERNAME)
This worked both with and without sudo on Mac and Linux.
get (translated) user folder names on Linux:
from gi.repository import GLib
docs = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DOCUMENTS)
desktop = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DESKTOP)
pics = GLib.get_user_special_dir(GLib.USER_DIRECTORY_PICTURES)
videos = GLib.get_user_special_dir(GLib.USER_DIRECTORY_VIDEOS)
music = GLib.get_user_special_dir(GLib.USER_DIRECTORY_MUSIC)
downloads = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DOWNLOAD)
public = GLib.get_user_special_dir(GLib.USER_DIRECTORY_PUBLIC_SHARE)
templates = GLib.get_user_special_dir(GLib.USER_DIRECTORY_TEMPLATES)
print(docs)
print(desktop)
print(pics)
print(videos)
print(music)
print(downloads)
print(public)
print(templates)
On Linux and other UNIXoids you can always take a peek in /etc/passwd. The home directory is the sixth colon-separated field in there. No idea on how to do better than the environment variable on Windows though. There'll be a system call for it, but if it's available from Python, ...