I have a directory full of scripts (let"s say project/bin). I also have a library located in project/lib and want the scripts to automatically load it. This is what I normally use at the top of each script:
#!/usr/bin/python
from os.path import dirname, realpath, sep, pardir
import sys
sys.path.append(dirname(realpath(__file__)) + sep + pardir + sep + "lib")
# ... now the real code
import mylib
This is kind of cumbersome, ugly, and has to be pasted at the beginning of every file. Is there a better way to do this?
Really what I"m hoping for is something as smooth as this:
#!/usr/bin/python
import sys.path
from os.path import pardir, sep
sys.path.append_relative(pardir + sep + "lib")
import mylib
Or even better, something that wouldn"t break when my editor (or someone else who has commit access) decides to reorder the imports as part of its clean-up process:
#!/usr/bin/python --relpath_append ../lib
import mylib
That wouldn"t port directly to non-posix platforms, but it would keep things clean.
Python: Best way to add to sys.path relative to the current running script dirname: Questions
os.path.dirname(__file__) returns empty
2 answers
I want to get the path of the current directory under which a .py file is executed.
For example a simple file D: est.py with code:
import os
print os.getcwd()
print os.path.basename(__file__)
print os.path.abspath(__file__)
print os.path.dirname(__file__)
It is weird that the output is:
D:
test.py
D: est.py
EMPTY
I am expecting the same results from the getcwd() and path.dirname().
Given os.path.abspath = os.path.dirname + os.path.basename, why
os.path.dirname(__file__)
returns empty?
Answer #1
Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have
os.path.dirname(filename) + os.path.basename(filename) == filename
Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.
To get the dirname of the absolute path, use
os.path.dirname(os.path.abspath(__file__))
What is the difference between os.path.basename() and os.path.dirname()?
2 answers
What is the difference between os.path.basename() and os.path.dirname()?
I already searched for answers and read some links, but didn"t understand. Can anyone give a simple explanation?
Answer #1
Both functions use the os.path.split(path) function to split the pathname path into a pair; (head, tail).
The os.path.dirname(path) function returns the head of the path.
E.g.: The dirname of "/foo/bar/item" is "/foo/bar".
The os.path.basename(path) function returns the tail of the path.
E.g.: The basename of "/foo/bar/item" returns "item"
From: http://docs.python.org/3/library/os.path.html#os.path.basename
