Is There A Python Equivalent To The 'which' Command
Solution 1:
Python 3.3 added shutil.which()
to provide a cross-platform means of discovering executables:
http://docs.python.org/3.3/library/shutil.html#shutil.which
Return the path to an executable which would be run if the given cmd was called. If no cmd would be called, return None.
Sample calls:
>>> shutil.which("python")
'/usr/local/bin/python'>>> shutil.which("python")
'C:\\Python33\\python.EXE'
Unfortunately, this has not been backported to 2.7.x.
Solution 2:
An option for Python 2 and 3:
from distutils.spawn import find_executable
find_executable('python') # '/usr/bin/python'
find_executable('does_not_exist') # None
find_executable(executable, path=None)
simply tries to find 'executable' in the directories listed in 'path'. Defaults to os.environ['PATH']
if 'path' is None
. Returns the complete path to 'executable' or None
if not found.
Keep in mind that unlike which
, find_executable
does not actually check that the result is marked as executable. You may want to call os.access(path, os.X_OK)
to check that on your own if you want to be certain that subprocess.Popen
will be able to execute the file.
Also of note, shutil.which
of Python 3.3+ has been backported and made available for Python 2.6, 2.7, and 3.x via the 3rd-party module whichcraft.
It is available for installation via the aforementioned GitHub page (i.e. pip install git+https://github.com/pydanny/whichcraft.git
) or the Python package index (i.e. pip install whichcraft
). It can be used like such:
from whichcraft import whichwhich('wget') # '/usr/bin/wget'
Solution 3:
I believe there is none in the python libraries
>>> def which(pgm):
path=os.getenv('PATH')
for p inpath.split(os.path.pathsep):
p=os.path.join(p,pgm)
ifos.path.exists(p) andos.access(p,os.X_OK):
return p
>>> os.which=which
>>> os.which('ls.exe')
'C:\\GNUwin32\\bin\\ls.exe'
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