Python snippets

December 10, 2018

How to reliably get folder path of what script we are in right now:

here = pathlib . Path ( __file__ ). parent . absolute () 
here = os. path . abspath (os. path . dirname ( __file__ ))

Pathlib returns a path object , os returns a string:

WindowsPath('T:/ENV.558/scratch/merge-rgb-nir') 

T:\\ENV.558\\scratch\\merge-rgb-nir

from inspect import getsourcefile 
here = os.path.dirname(os.path.abspath(getsourcefile(lambda:0)))

How do I get the path of the current executed file in Python?
https://stackoverflow.com/a/18489147/14420

this solution is robust even in executables

import inspect, os.path 
filename = inspect.getframeinfo(inspect.currentframe()).filename 
path = os.path.dirname(os.path.abspath(filename))

@ José Crespo Barrios

https://stackoverflow.com/a/44592299/14420

#Py3 
HERE = pathlib.Path(__file__).parent.absolute() 
os.chdir(HERE) 

import inspect, os.path, pathlib 
filename = inspect.getframeinfo(inspect.currentframe()).filename 
filepath = os.path.dirname(os.path.abspath(filename)) 
filedir = pathlib.Path(__file__).parent.absolute()

…

import socket 
socket.gethostbyname_ex(socket.gethostname())

('ENV-Y225411.YNet.gov.yk.ca', [], ['199.247.217.109'])

From < https://github.com/jupyter/notebook/issues/1927 >

Leo snippets

# this doesn't work. not sure why. 
if not ctypes.windll.shell32.IsUserAnAdmin(): 
g.es_print("No admin privileges. Start Leo as administrator first and then run this script") 
subprocess.run([sys.executable, 
os.path.join(here, 'elevate.py'), 
os.path.abspath(getsourcefile(lambda:0)) 
])